Free math worksheets

- Read each row as an equation. - The left side is already in identity-matrix form. - Match each pivot row with its variable.

The first row gives \(x=3\), and the second row gives \(y=-2\).

\((x, y)=(3, -2)\)

- Keep the variable order the same in both equations. - The coefficient columns come before the augmentation bar. - Put each equation on its own row.

Place the coefficients of \(x\) and \(y\) in the first two columns and the constants in the augmented column: \(\begin{pmatrix}1&2&\mid&5\\3&-1&\mid&4\end{pmatrix}\).

\(\begin{pmatrix}1&2&\mid&5\\3&-1&\mid&4\end{pmatrix}\)

- Only the second row changes. - Apply the row operation to every entry in the row, including the augmented entry. - Keep the first row exactly as it is.

1. Keep row 1 unchanged. 2. Replace row 2 by \((2,-1\mid4)-2(1,2\mid5)=(0,-5\mid-6)\). 3. The resulting matrix is \(\left[\begin{array}{cc|c}1&2&5\\0&-5&-6\end{array}\right]\).

\(\left[\begin{array}{cc|c}1&2&5\\0&-5&-6\end{array}\right]\)

- Identify the pivot columns. - A non-pivot variable is free. - A free variable in a consistent system produces infinitely many solutions.

1. The first two columns contain pivots, so \(x_1\) and \(x_2\) are leading variables. 2. The third variable \(x_3\) has no pivot, so \(x_3\) is free. 3. There is no contradictory row, and a free variable remains. Therefore, the system has infinitely many solutions.

\(x_3\) is free, and the system has infinitely many solutions.

- Set up the equation \(\mathbf{c}=r\mathbf{a}+s\mathbf{b}\). - Write the component equations as an augmented matrix. - Row-reduce to solve for the coefficients. - A nontrivial linear combination relating the vectors shows linear dependence.

1. Test whether \(\mathbf{c}=r\mathbf{a}+s\mathbf{b}\). This gives the system \(2r+s=5\), \(-r+4s=2\), and \(3r-2s=4\), represented by the augmented matrix \(\left[\begin{array}{cc|c}2&1&5\\-1&4&2\\3&-2&4\end{array}\right]\). 2. Multiply the second row by \(-1\) and use it to eliminate \(r\) from the other rows. This gives an equivalent matrix \(\left[\begin{array}{cc|c}1&-4&-2\\0&9&9\\0&10&10\end{array}\right]\). Dividing the second row by \(9\) gives \(s=1\), and then the first row gives \(r=2\). The third row is consistent with these values. 3. Therefore, \(\mathbf{c}=2\mathbf{a}+\mathbf{b}\), so the vectors are linearly dependent.

The vectors are linearly dependent, and \(\mathbf{c}=2\mathbf{a}+\mathbf{b}\).

- Identify which coordinates are zero on the y-axis. - Use those coordinate conditions to form a system in \(a\) and \(\lambda\). - Place the coefficients in an augmented matrix. - Use the solved parameter to find the remaining coordinate.

1. A point on the y-axis has \(x=0\) and \(z=0\). Therefore \(a+2\lambda=0\) and \(4-2\lambda=0\). 2. For variables \(a\) and \(\lambda\), the augmented matrix is \(\left[\begin{array}{cc|c}1&2&0\\0&-2&-4\end{array}\right]\). 3. The second row gives \(\lambda=2\). The first row then gives \(a+4=0\), so \(a=-4\). 4. The y-coordinate is \(1+\lambda=3\). Thus the intersection point is \(S=(0, 3, 0)\).

a) \(a+2\lambda=0\) and \(4-2\lambda=0\) b) \(a=-4\), and the intersection point is \(S=(0, 3, 0)\).

- Equating the three coordinates produces three linear equations in the two line parameters. - Keep the augmented column visible while row-reducing; consistency is the key question. - If the system is consistent, substitute the solved parameters into either original line to recover the point.

1. Equating coordinates gives \(2r+s=3\), \(r-2s=-1\), and \(-3r+4s=1\). 2. Row-reduce \(\left[\begin{array}{cc|c}2&1&3\\1&-2&-1\\-3&4&1\end{array}\right]\). Its reduced form is \(\left[\begin{array}{cc|c}1&0&1\\0&1&1\\0&0&0\end{array}\right]\). 3. Thus \(r=1\) and \(s=1\). Substitution into either line gives \((3, 2, 1)\).

The lines intersect at \((3, 2, 1)\), with \(r=1\) and \(s=1\).

- Treat the line parameters as the two unknowns of a linear system. - A contradiction row after row reduction proves that no common parameter pair exists. - After establishing nonintersection, compare the direction vectors before deciding between parallel and skew.

1. Equating coordinates gives \(\lambda-3\mu=2\), \(2\lambda+\mu=-3\), and \(2\lambda-\mu=4\). 2. Row-reducing \(\left[\begin{array}{cc|c}1&-3&2\\2&1&-3\\2&-1&4\end{array}\right]\) produces a contradiction row of the form \(\left[\begin{array}{cc|c}0&0&1\end{array}\right]\), so the system is inconsistent. 3. The direction vectors \(\langle 1,2,2\rangle\) and \(\langle 3,-1,1\rangle\) are not scalar multiples, so the lines are not parallel. Nonparallel lines in space that do not intersect are skew.

The augmented system is inconsistent, so the lines do not intersect. Their direction vectors are not parallel, so the lines are skew.

- Put the vectors into the columns of a \(3\times3\) matrix. - Row-reduce and check whether every column has a pivot. - Three independent vectors in \(\mathbb{R}^3\) form a basis.

1. Put the vectors into the columns of the coefficient matrix \(A=\begin{pmatrix}1&0&2\\2&1&0\\0&-1&3\end{pmatrix}\). 2. Row-reduce: \(\begin{pmatrix}1&0&2\\2&1&0\\0&-1&3\end{pmatrix}\to\begin{pmatrix}1&0&2\\0&1&-4\\0&-1&3\end{pmatrix}\to\begin{pmatrix}1&0&2\\0&1&-4\\0&0&-1\end{pmatrix}\). 3. There is a pivot in every column, so the homogeneous system \(A\mathbf{x}=\mathbf{0}\) has only the trivial solution. Thus the three vectors are linearly independent. 4. Three linearly independent vectors in \(\mathbb{R}^3\) form a basis, so these vectors form a basis of \(\mathbb{R}^3\).

Yes. The row-reduced coefficient matrix has a pivot in every column, so the vectors are linearly independent and therefore form a basis of \(\mathbb{R}^3\).

- Use the two fully known coordinates to form a system for \(\lambda\) and \(\mu\). - Row-reduce the augmented matrix. - Substitute the parameter values into the remaining coordinate equation.

1. The known y- and z-coordinates give the system \(2\lambda+4\mu=10\) and \(2\lambda+\mu=8\). 2. Its augmented matrix is \(\left[\begin{array}{cc|c}2&4&10\\2&1&8\end{array}\right]\). 3. Row reduction gives \(\mu=\frac23\) and \(\lambda=\frac{11}{3}\). 4. Use the x-coordinate equation: \(k=2+\lambda-3\mu=2+\frac{11}{3}-2=\frac{11}{3}\).

\(\lambda=\frac{11}{3}\), \(\mu=\frac23\), and \(k=\frac{11}{3}\)

- Put the three vectors in the columns of a coefficient matrix and augment with the zero vector. - Row-reduce the homogeneous system. - What does a pivot in every variable column imply about the solution and linear independence?

1. Put the vectors in the coefficient columns of the homogeneous system: \(\left[\begin{array}{ccc|c}1&0&1&0\\2&1&0&0\\3&2&1&0\end{array}\right]\). 2. Apply \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-3R_1\): \(\left[\begin{array}{ccc|c}1&0&1&0\\0&1&-2&0\\0&2&-2&0\end{array}\right]\). 3. Apply \(R_3\leftarrow R_3-2R_2\): \(\left[\begin{array}{ccc|c}1&0&1&0\\0&1&-2&0\\0&0&2&0\end{array}\right]\). There is a pivot in every variable column, so the homogeneous system has only the trivial solution. 4. Therefore, the vectors are linearly independent and form a basis of \(\mathbb{R}^3\).

Yes. The vectors form a basis of \(\mathbb{R}^3\).

- Enter both plane equations in an augmented matrix. - Expect one free variable because there are two independent equations in three variables. - Choose a convenient multiple for the free variable. - Separate the constant and parameter terms into vector form.

1. The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&1&6\\2&-1&3&9\end{array}\right]\). 2. Row reduction gives the equivalent equations \(3x+4z=15\) and \(x+y+z=6\). 3. Choose \(z=3t\). Then \(3x+12t=15\), so \(x=5-4t\). 4. Substitution into the first plane gives \(y=1+t\). 5. Therefore \(g:\mathbf{r}(t)=\langle 5,1,0\rangle+t\langle -4,1,3\rangle\).

\(g:\mathbf{r}(t)=\langle 5,1,0\rangle+t\langle -4,1,3\rangle\)

- Use row operations to create zeros below the leading entries. - Work from the last row upward after reaching row-echelon form. - Check the solution in the original equations.

1. Form the augmented matrix: \(\left(\begin{array}{ccc|c}1&2&1&3\\2&-1&3&14\\3&1&-2&-1\end{array}\right)\). 2. Replace row 2 by \(R_2-2R_1\) and row 3 by \(R_3-3R_1\): \(\left(\begin{array}{ccc|c}1&2&1&3\\0&-5&1&8\\0&-5&-5&-10\end{array}\right)\). 3. Replace row 3 by \(R_3-R_2\): \(\left(\begin{array}{ccc|c}1&2&1&3\\0&-5&1&8\\0&0&-6&-18\end{array}\right)\). 4. Back-substitution gives \(x_3=3\), \(x_2=-1\), and \(x_1=2\).

\((x_1, x_2, x_3)=(2, -1, 3)\)

- Eliminate the same variable from all equations below the first. - Compare equations with identical left sides. - A contradictory row means the solution set is empty.

1. Eliminate \(x_1\) from the last three equations using the first equation. The resulting equations are \(-5x_2+3x_3=-3\), \(-5x_2+3x_3=-3\), and \(-5x_2+3x_3=3\). 2. The first and third reduced equations have identical left sides but different right sides. Subtracting them gives \(0=6\), a contradiction. 3. Therefore, the system is inconsistent and has no solution.

\(\varnothing\)

- Begin with the general quadratic form. - Substitute each point to create one equation. - Row-reduce the coefficient system for \(a\), \(b\), and \(c\).

1. Let \(f(x)=ax^2+bx+c\). Substituting the three points gives \(4a-2b+c=14\), \(a+b+c=2\), and \(9a+3b+c=4\). 2. Put the second equation first in the augmented matrix: \(\left(\begin{array}{ccc|c}1&1&1&2\\4&-2&1&14\\9&3&1&4\end{array}\right)\). Use \(R_2-4R_1\) and \(R_3-9R_1\), then \(R_3-R_2\), to obtain \(\left(\begin{array}{ccc|c}1&1&1&2\\0&-6&-3&6\\0&0&-5&-20\end{array}\right)\). 3. Back-substitution gives \(c=4\), \(b=-3\), and \(a=1\). Therefore, \(f(x)=x^2-3x+4\).

\(f(x)=x^2-3x+4\)

- Compare the number of independent equations with the number of variables. - Use one free variable as a parameter. - Express the remaining variables in terms of that parameter.

1. Form the augmented matrix: \(\left(\begin{array}{ccc|c}1&2&-1&4\\2&-1&3&3\end{array}\right)\). 2. Replace row 2 by \(R_2-2R_1\): \(\left(\begin{array}{ccc|c}1&2&-1&4\\0&-5&5&-5\end{array}\right)\). Divide the second row by \(-5\): \(x_2-x_3=1\). 3. Let \(x_3=t\). Then \(x_2=1+t\). Substituting into the first equation gives \(x_1=2-t\). Therefore, \(\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}2\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\1\\1\end{pmatrix}\), where \(t\in\mathbb R\).

\(\left\{\begin{pmatrix}2\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\1\\1\end{pmatrix}: t\in\mathbb R\right\}\)

- Write the coefficients and constants in an augmented matrix. - Use the first row to eliminate the first entry of row 2. - Back-substitute after the matrix is in row-echelon form.

1. Form the augmented matrix \(\left[\begin{array}{cc|c}1&1&5\\2&-1&1\end{array}\right]\). 2. Apply \(R_2\leftarrow R_2-2R_1\): \(\left[\begin{array}{cc|c}1&1&5\\0&-3&-9\end{array}\right]\). 3. The second row gives \(y=3\). 4. Substitute into the first row: \(x+3=5\), so \(x=2\).

\((x,y)=(2,3)\)

- Use row 1 to eliminate \(x\) from the two rows below it. - Combine the two reduced rows to eliminate \(y\). - Once one variable is known, back-substitute upward.

1. Form \(\left[\begin{array}{ccc|c}1&1&1&6\\2&-1&1&3\\1&2&-1&2\end{array}\right]\). 2. Apply \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-R_1\): \(\left[\begin{array}{ccc|c}1&1&1&6\\0&-3&-1&-9\\0&1&-2&-4\end{array}\right]\). 3. Replace row 2 by \(R_2+3R_3\): \(\left[\begin{array}{ccc|c}1&1&1&6\\0&0&-7&-21\\0&1&-2&-4\end{array}\right]\). 4. Thus \(z=3\). Row 3 gives \(y-6=-4\), so \(y=2\). 5. Row 1 gives \(x+2+3=6\), so \(x=1\).

\((x,y,z)=(1,2,3)\)

- Equating the three coordinates produces three linear equations in \(r\), \(s\), and \(k\). - Keep the variable order \(r,s,k\) fixed when you build the augmented matrix. - Use row reduction to solve the complete system before substituting into either line.

1. Equating coordinates gives \(r-2s=2\), \(r-s=2\), and \(2r-k=0\). 2. The augmented matrix is \(\left[\begin{array}{ccc|c}1&-2&0&2\\1&-1&0&2\\2&0&-1&0\end{array}\right]\), with columns for \(r,s,k\). 3. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&0\\0&0&1&4\end{array}\right]\). Therefore, \(r=2\), \(s=0\), and \(k=4\). 4. Substituting \(r=2\) into \(g\) gives \(S=(4, 3, 4)\).

\(k=4\), and the lines intersect at \(S=(4, 3, 4)\).

- Interpret the entries of a solution to \(A\mathbf{x}=\mathbf{0}\) as coefficients in a linear combination of the columns. - If a nonzero homogeneous solution exists, add it to a solution of \(A\mathbf{x}=\mathbf{b}\). - In part c, look for coefficients that make the third column cancel the first two. - For part d, consider a matrix with a zero column.

1. Statement a is true. The equation \(A\mathbf{x}=\mathbf{0}\) records the coefficient relation among the columns of \(A\). Having only the trivial solution is exactly the definition of linear independence of those columns. 2. Statement b is true. Suppose \(A\mathbf{x}=\mathbf{b}\) has the unique solution \(\mathbf{x}_0\). If \(A\mathbf{z}=\mathbf{0}\) had a nonzero solution \(\mathbf{z}\), then \(A(\mathbf{x}_0+\mathbf{z})=\mathbf{b}\), giving a second solution. Therefore the homogeneous system has only the trivial solution, so the columns are independent. 3. Statement c is true. The coefficient vector \(\langle 1,-2,-1\rangle\) is nonzero and satisfies \(\mathbf{a}-2\mathbf{b}-(\mathbf{a}-2\mathbf{b})=\mathbf{0}\). Thus \(A\mathbf{x}=\mathbf{0}\) has a nontrivial solution, so the columns are dependent. 4. Statement d is false. For example, let the columns be \(\mathbf{a},\mathbf{b},\mathbf{0}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are linearly independent. The columns are dependent because the homogeneous system has the nontrivial solution \((0,0,1)\), but \(\mathbf{a}\) cannot be written as a linear combination of \(\mathbf{b}\) and \(\mathbf{0}\).

a) True b) True c) True d) False

- Put the first three vectors into the columns of a matrix \(A\). - Make \(A\mathbf{x}=\mathbf{0}\) have a nonzero solution so the set is dependent. - Then choose \(\mathbf{v}_4\) so the augmented system \(A\mathbf{x}=\mathbf{v}_4\) is inconsistent.

1. Choose \(\mathbf{v}_1=\langle 1,0,0\rangle\), \(\mathbf{v}_2=\langle 2,0,0\rangle\), \(\mathbf{v}_3=\langle 0,1,0\rangle\), and \(\mathbf{v}_4=\langle 0,0,1\rangle\). 2. Put the first three vectors into the columns of \(A=\begin{pmatrix}1&2&0\\0&0&1\\0&0&0\end{pmatrix}\). The homogeneous system \(A\mathbf{x}=\mathbf{0}\) has the nontrivial solution \(\mathbf{x}=\langle -2,1,0\rangle\). Therefore the first three vectors, and hence the full set of four vectors, are linearly dependent. 3. To test whether \(\mathbf{v}_4\) is in the span of the first three vectors, solve \(A\mathbf{x}=\mathbf{v}_4\). The augmented matrix is \(\left[\begin{array}{ccc|c}1&2&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]\). Its last row represents \(0=1\), so the system is inconsistent. 4. Therefore \(\mathbf{v}_4\) is not a linear combination of \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\), while the four-vector set is linearly dependent.

One example is \(\mathbf{v}_1=\langle 1,0,0\rangle\), \(\mathbf{v}_2=\langle 2,0,0\rangle\), \(\mathbf{v}_3=\langle 0,1,0\rangle\), and \(\mathbf{v}_4=\langle 0,0,1\rangle\). The homogeneous system for the first three vectors has a nontrivial solution, while the augmented system for expressing \(\mathbf{v}_4\) as their linear combination is inconsistent.

- Put \(\mathbf{a},\mathbf{b},\mathbf{c}\) into the columns of a matrix and look for a nonzero solution of the homogeneous system. - For part 2, append \(\mathbf{d}\) as the right-hand side and row-reduce. - What does a dependent subset imply about the larger set?

1. Put \(\mathbf{a},\mathbf{b},\mathbf{c}\) into the columns of \(A=\begin{pmatrix}2&4&0\\1&2&0\\0&0&1\end{pmatrix}\). The homogeneous system \(A\mathbf{x}=\mathbf{0}\) has the nontrivial solution \(\mathbf{x}=\langle -2,1,0\rangle\), because \(-2\mathbf{a}+\mathbf{b}=\mathbf{0}\). Therefore the three vectors are linearly dependent. 2. To test whether \(\mathbf{d}\) is in their span, row-reduce the augmented matrix \(\left[\begin{array}{ccc|c}2&4&0&0\\1&2&0&1\\0&0&1&0\end{array}\right]\). Replacing the first row by the first row minus \(2\) times the second row gives a row \(\left[\begin{array}{ccc|c}0&0&0&-2\end{array}\right]\), which represents a contradiction. Thus the system is inconsistent, so \(\mathbf{d}\) is not a linear combination of \(\mathbf{a},\mathbf{b},\mathbf{c}\). 3. The four-vector set contains the dependent subset \(\{\mathbf{a},\mathbf{b}\}\), so the full set is linearly dependent. This also agrees with the fact that four vectors in \(\mathbb{R}^3\) cannot be linearly independent.

1. The vectors are dependent because \(A\mathbf{x}=\mathbf{0}\) has the nontrivial solution \(\mathbf{x}=\langle -2,1,0\rangle\). 2. No. The augmented system is inconsistent. 3. The full set is dependent because it contains the dependent pair \(\mathbf{a},\mathbf{b}\).

- For part 1, append \(\mathbf{c}\) as the right-hand side of the coefficient matrix with columns \(\mathbf{a}\) and \(\mathbf{b}\). - For part 2, look for a nonzero solution of the homogeneous system with columns \(\mathbf{a},\mathbf{b},\mathbf{c}\). - A three-vector set can be dependent because of a relation involving only two of its vectors.

1. To test whether \(r\mathbf{a}+s\mathbf{b}=\mathbf{c}\), form the augmented matrix \(\left[\begin{array}{cc|c}6&-2&1\\-3&1&2\\9&-3&1\end{array}\right]\). Replacing the first row by the first row plus \(2\) times the second row gives \(\left[\begin{array}{cc|c}0&0&5\end{array}\right]\), which represents a contradiction. Therefore the system is inconsistent, so \(\mathbf{c}\) is not a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\). 2. Put all three vectors into the columns of \(A\). Since \(\mathbf{a}+3\mathbf{b}=\mathbf{0}\), the homogeneous system \(A\mathbf{x}=\mathbf{0}\) has the nontrivial solution \(\mathbf{x}=\langle 1,3,0\rangle\). Thus the three-vector set is linearly dependent. 3. The claim is false. Here \(\mathbf{c}\) lies outside the span of \(\mathbf{a}\) and \(\mathbf{b}\), but \(\mathbf{a}\) and \(\mathbf{b}\) are already dependent because \(\mathbf{a}=-3\mathbf{b}\).

1. No. The augmented system is inconsistent. 2. The set is linearly dependent because \(A\mathbf{x}=\mathbf{0}\) has the nontrivial solution \(\mathbf{x}=\langle 1,3,0\rangle\). 3. The claim is false.

- Interpret a nonzero solution of a homogeneous system as a dependence relation among the coefficient columns. - A solution of \(x\mathbf{u}+y\mathbf{v}=\mathbf{w}\) can be rearranged into a homogeneous relation involving all three vectors. - In part c, consider separately whether the coefficient of \(\mathbf{w}\) is zero.

1. Statement a is true. A nontrivial solution \((p,q)\) to \(p\mathbf{u}+q\mathbf{v}=\mathbf{0}\) gives the nontrivial solution \((p,q,0)\) to \(p\mathbf{u}+q\mathbf{v}+0\mathbf{w}=\mathbf{0}\). Thus the three-column homogeneous system has a nontrivial solution for every \(\mathbf{w}\). 2. Statement b is true. If \(x\mathbf{u}+y\mathbf{v}=\mathbf{w}\) is consistent, then for some \(x,y\), \(x\mathbf{u}+y\mathbf{v}-\mathbf{w}=\mathbf{0}\). The coefficient triple \((x,y,-1)\) is nonzero, so the three vectors are linearly dependent. 3. Statement c is true. Suppose \(p\mathbf{u}+q\mathbf{v}+r\mathbf{w}=\mathbf{0}\). If \(r\ne0\), then \(\mathbf{w}=-\frac{p}{r}\mathbf{u}-\frac{q}{r}\mathbf{v}\), which would make \(x\mathbf{u}+y\mathbf{v}=\mathbf{w}\) consistent, a contradiction. Hence \(r=0\). The homogeneous system for \(\mathbf{u},\mathbf{v}\) then forces \(p=q=0\). Therefore the three-column homogeneous system has only the trivial solution, so the vectors are linearly independent.

a) True b) True c) True

- Assume there are two solutions to the same matrix equation. - Subtract the two equations to obtain a homogeneous system. - What does independence of the columns imply about solutions of that homogeneous system?

1. Suppose \(A\mathbf{r}=\mathbf{x}\) has two solutions, \(\mathbf{r}\) and \(\mathbf{s}\). Then \(A\mathbf{r}=\mathbf{x}\) and \(A\mathbf{s}=\mathbf{x}\). 2. Subtract the matrix equations: \(A(\mathbf{r}-\mathbf{s})=\mathbf{0}\). 3. Because the columns of \(A\) are linearly independent, the homogeneous system \(A\mathbf{y}=\mathbf{0}\) has only the trivial solution. Therefore \(\mathbf{r}-\mathbf{s}=\mathbf{0}\). 4. Hence \(\mathbf{r}=\mathbf{s}\), so every consistent system \(A\mathbf{r}=\mathbf{x}\) has exactly one solution.

If \(\mathbf{r}\) and \(\mathbf{s}\) both satisfy \(A\mathbf{r}=\mathbf{x}\), then \(A(\mathbf{r}-\mathbf{s})=\mathbf{0}\). Column independence makes the homogeneous solution trivial, so \(\mathbf{r}=\mathbf{s}\). Thus the solution is unique.

- Add a multiple of a homogeneous solution to the given solution of \(A\mathbf{x}=\mathbf{v}\). - Choose the multiple so the first coordinate becomes \(0\). - Interpret the final coordinate vector as coefficients of \(\mathbf{a},\mathbf{b},\mathbf{c}\).

1. If \(\mathbf{x}_0=\langle 4,2,-3\rangle\) satisfies \(A\mathbf{x}_0=\mathbf{v}\) and \(\mathbf{z}=\langle 1,-2,1\rangle\) satisfies \(A\mathbf{z}=\mathbf{0}\), then every vector of the form \(\mathbf{x}_0+t\mathbf{z}\) is also a solution of \(A\mathbf{x}=\mathbf{v}\). 2. Its first entry is \(4+t\). Set \(4+t=0\), giving \(t=-4\). 3. Then \(\mathbf{x}=\langle 4,2,-3\rangle-4\langle 1,-2,1\rangle=\langle 0,10,-7\rangle\). 4. Therefore \(\mathbf{v}=0\mathbf{a}+10\mathbf{b}-7\mathbf{c}\).

\(\mathbf{x}=\langle 0,10,-7\rangle\), so \(\mathbf{v}=0\mathbf{a}+10\mathbf{b}-7\mathbf{c}\).

- Put the three vectors into the columns of a \(2\times3\) coefficient matrix. - Row-reduce the homogeneous system and choose a free parameter. - For part b, compare the maximum number of pivots with the number of variables.

1. The homogeneous system has coefficient matrix \(\begin{pmatrix}1&-2&5\\4&3&1\end{pmatrix}\). Row-reducing gives \(\begin{pmatrix}1&-2&5\\0&11&-19\end{pmatrix}\). 2. The second row gives \(11s-19t=0\), so \(s=\frac{19}{11}t\). The first row then gives \(r=2s-5t=-\frac{17}{11}t\). 3. Let \(t=11k\). Then every solution has the form \((r,s,t)=k(-17,19,11)\), where \(k\in\mathbb{R}\). 4. In general, the coefficient matrix for three vectors in \(\mathbb{R}^2\) has \(2\) rows and \(3\) columns. Its row-echelon form can have at most \(2\) pivots, so at least one variable in the homogeneous system is free. Therefore a nontrivial solution always exists, and the three vectors are linearly dependent.

a) \((r,s,t)=k(-17,19,11)\), where \(k\in\mathbb{R}\) b) A \(2\times3\) coefficient matrix has at most \(2\) pivots, so its homogeneous system has a free variable and a nontrivial solution. Thus any three vectors in \(\mathbb{R}^2\) are linearly dependent.

- Interpret spanning in terms of consistency of \(A\mathbf{x}=\mathbf{d}\). - Turn a solution of that system into a homogeneous equation involving all four vectors. - The coefficient of \(\mathbf{d}\) can be chosen as \(-1\).

1. Because \(\{\mathbf{a},\mathbf{b},\mathbf{c}\}\) is a basis of \(\mathbb{R}^3\), its vectors span \(\mathbb{R}^3\). Therefore the system \(A\mathbf{x}=\mathbf{d}\) is consistent for every \(\mathbf{d}\in\mathbb{R}^3\). 2. Let one solution be \(\mathbf{x}=\langle \lambda_1,\lambda_2,\lambda_3\rangle\). Then \(\mathbf{d}=\lambda_1\mathbf{a}+\lambda_2\mathbf{b}+\lambda_3\mathbf{c}\). 3. Rearranging gives \(\lambda_1\mathbf{a}+\lambda_2\mathbf{b}+\lambda_3\mathbf{c}-\mathbf{d}=\mathbf{0}\). Thus the homogeneous system whose coefficient columns are \(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\) has the nontrivial solution \((\lambda_1,\lambda_2,\lambda_3,-1)\). 4. Therefore the four vectors are linearly dependent.

a) The system \(A\mathbf{x}=\mathbf{d}\) is consistent because the columns of \(A\) form a basis and therefore span \(\mathbb{R}^3\). b) If \(A\mathbf{x}=\mathbf{d}\) has solution \((\lambda_1,\lambda_2,\lambda_3)\), then \((\lambda_1,\lambda_2,\lambda_3,-1)\) is a nontrivial homogeneous dependence relation for \(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\).

- Points in the yz-plane have x-coordinate \(0\). - Points on the y-axis have both x- and z-coordinates equal to \(0\). - Use those two coordinate conditions to form a system in \(k\) and \(\lambda\). - Solve the system with an augmented matrix.

1. A point in the yz-plane has \(x=0\). Thus \(6+3\lambda=0\), so \(\lambda=-2\). 2. The remaining coordinates are \(y=k-2(-2)=k+4\) and \(z=k-2+(-2)=k-4\). Therefore \(S_k=(0, k+4, k-4)\). 3. A point on the y-axis also has \(z=0\). The equations are \(3\lambda=-6\) and \(k+\lambda=2\). 4. For variables \(k\) and \(\lambda\), an augmented matrix is \(\left[\begin{array}{cc|c}0&3&-6\\1&1&2\end{array}\right]\). 5. Row reduction gives \(\lambda=-2\) and \(k=4\). 6. The y-coordinate is \(k-2\lambda=4+4=8\), so the intersection point is \(P=(0, 8, 0)\).

a) \(S_k=(0, k+4, k-4)\) b) \(k=4\), and the intersection point is \(P=(0, 8, 0)\).

- Translate line intersection into \(A\langle r,s\rangle=\mathbf{w}\). - Compare coefficient and augmented ranks for each condition. - Equal rank \(2\) gives a unique parameter pair; rank mismatch gives no intersection; equal rank \(1\) gives the identical-line case here.

1. If \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly independent, then \(\operatorname{rank}(A)=2\) but \(\operatorname{rank}([\,A\mid\mathbf{w}\,])=3\). The system is inconsistent, and the nonparallel lines are skew. 2. If \(\mathbf{u}\) and \(\mathbf{v}\) are independent while the three vectors are dependent, then \(\operatorname{rank}(A)=\operatorname{rank}([\,A\mid\mathbf{w}\,])=2\). The system has a unique solution, so the lines intersect. 3. If both pairs are dependent, then \(\mathbf{v}\) and \(\mathbf{w}\) are scalar multiples of the nonzero direction vector \(\mathbf{u}\). Hence \(\operatorname{rank}(A)=\operatorname{rank}([\,A\mid\mathbf{w}\,])=1\), so the lines describe the same set of points and are identical.

1. skew 2. intersecting 3. identical

- Translate point membership into one linear equation for each coordinate. - Put the coefficients in a consistent column order for \(a\), \(\lambda\), and \(\mu\). - Row-reduce the augmented matrix rather than solving the equations by successive substitution.

1. Equating coordinates gives \(a-\lambda=0\), \(a+\lambda-2\mu=2\), and \(2\lambda+\mu=8\). 2. With variables ordered as \(a,\lambda,\mu\), the augmented matrix is \(\left[\begin{array}{ccc|c}1&-1&0&0\\1&1&-2&2\\0&2&1&8\end{array}\right]\). 3. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&3\\0&0&1&2\end{array}\right]\). Thus \(a=3\), \(\lambda=3\), and \(\mu=2\). 4. These values satisfy all three coordinate equations, so the point lies in the plane when \(a=3\).

\(a=3\)

- A point lies in a plane when its coordinates satisfy the plane equation. - Equivalent plane equations may differ by a nonzero scalar multiple, so choose one coefficient conveniently. - After fixing that coefficient, write the remaining equations as an augmented matrix and row-reduce. - Use the solved coefficients in \(ax+by+cz+d=0\).

1. Substitution gives the equations \(3a+d=0\), \(2c+d=0\), and \(a-b+c+d=0\). 2. The system is underdetermined because multiplying all four coefficients by the same nonzero constant gives an equivalent plane equation. Choose \(d=-6\). 3. For variables \(a,b,c\), the remaining system has augmented matrix \(\left[\begin{array}{ccc|c}3&0&0&6\\0&0&2&6\\1&-1&1&6\end{array}\right]\). Row reduction gives \(a=2\), \(b=-1\), and \(c=3\). 4. Therefore, one equation is \(E:2x-y+3z-6=0\).

a) Substitution gives \(3a+d=0\), \(2c+d=0\), and \(a-b+c+d=0\). b) One possible equation is \(E:2x-y+3z-6=0\).

- Enter the plane coefficients and constants in an augmented matrix. - Row-reduce until the variables are determined. - For line membership, compare all three coordinates using one parameter. - The same parameter value must satisfy every coordinate equation.

1. The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&1&6\\2&-1&1&3\\1&2&-1&2\end{array}\right]\). 2. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&2\\0&0&1&3\end{array}\right]\). 3. Therefore the common point is \(S=(1, 2, 3)\). 4. For \(S\) to lie on \(h\), the coordinates must satisfy \(1=t\), \(2=1+t\), and \(3=4-t\). 5. All three equations give \(t=1\), so \(S\) lies on \(h\).

a) \(S=(1, 2, 3)\) b) Yes. The point lies on \(h\) when \(t=1\).

- Use an augmented matrix for the three plane equations. - A pivot in every variable column gives a unique solution. - A plane parallel to the xy-plane has a fixed z-coordinate. - Use the z-coordinate of \(S\).

1. a) The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&2&3\\2&-1&1&0\\-1&2&2&5\end{array}\right]\). 2. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&-1\\0&1&0&0\\0&0&1&2\end{array}\right]\). 3. There is a pivot in every variable column, so the system has exactly one solution: \(S(-1, 0, 2)\). 4. b) Every plane parallel to the xy-plane has an equation of the form \(z=d\). 5. Since \(S\) lies in \(H\), \(d=2\). Thus \(H:z=2\).

a) \(S(-1, 0, 2)\) b) \(H:z=2\)

- Collect the terms containing \(a\). - For an identity in \(a\), both the coefficient of \(a\) and the remaining expression must be zero. - Put the two coordinate equations in an augmented matrix. - Assign a parameter to the free variable.

1. Rewrite the equation as \(a(x+y)+(x-2z-4)=0\). 2. For a point to lie in every plane, it must satisfy \(x+y=0\) and \(x-2z=4\). 3. The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&0&0\\1&0&-2&4\end{array}\right]\). 4. Row reduction gives \(x-2z=4\) and \(y+2z=-4\). Let \(z=t\). 5. Then \(x=4+2t\) and \(y=-4-2t\), so \(s:\mathbf{r}(t)=\langle 4,-4,0\rangle+t\langle 2,-2,1\rangle\).

\(s:\mathbf{r}(t)=\langle 4,-4,0\rangle+t\langle 2,-2,1\rangle\)

- Write the coefficients and constants in an augmented matrix. - Use row operations to identify any dependent row. - Assign a parameter to the free variable. - Translate the parametric solution into a geometric intersection.

1. Write the augmented matrix: \(\left[\begin{array}{ccc|c}1&1&1&3\\1&-1&2&1\\2&0&3&4\end{array}\right]\). 2. Replace row 2 by row 2 minus row 1 and row 3 by row 3 minus twice row 1: \(\left[\begin{array}{ccc|c}1&1&1&3\\0&-2&1&-2\\0&-2&1&-2\end{array}\right]\). 3. Replace row 3 by row 3 minus row 2: \(\left[\begin{array}{ccc|c}1&1&1&3\\0&-2&1&-2\\0&0&0&0\end{array}\right]\). 4. There is one free variable. Let \(z=t\). The second row gives \(-2y+t=-2\), so \(y=1+\frac12t\). 5. The first row gives \(x+1+\frac12t+t=3\), so \(x=2-\frac32t\). 6. The solution set is \(\mathbf{r}(t)=\langle 2,1,0\rangle+t\left\langle-\frac32,\frac12,1\right\rangle\). 7. Geometrically, the three planes share this line. The third equation is dependent on the first two, since its row becomes zero during row reduction.

The system has infinitely many solutions: \(\mathbf{r}(t)=\langle 2,1,0\rangle+t\left\langle-\frac32,\frac12,1\right\rangle\). The three planes intersect in this common line.

- Compare the direction vectors first. - Equate corresponding coordinates to form a system in \(\lambda\), \(\mu\), and \(a\). - Solve the system with an augmented matrix. - If nonparallel lines in space do not intersect, they are skew.

1. The direction vectors \(\langle 1,-1,2\rangle\) and \(\langle 0,1,1\rangle\) are not scalar multiples, so the lines cannot be parallel. 2. Equating corresponding coordinates gives \(\lambda=-1\), \(-\lambda-\mu-a=-1\), and \(2\lambda-\mu=-1\). 3. For variables \(\lambda\), \(\mu\), and \(a\), an augmented matrix is \(\left[\begin{array}{ccc|c}1&0&0&-1\\-1&-1&-1&-1\\2&-1&0&-1\end{array}\right]\). 4. Row reduction gives \(\lambda=-1\), \(\mu=-1\), and \(a=3\). 5. Substituting \(\lambda=-1\) into \(g\) gives the intersection point \(S(1, 2, 1)\). 6. For \(a\neq3\), the coordinate system is inconsistent. Since the lines are not parallel, they are skew.

For \(a=3\), the lines intersect at \(S(1, 2, 1)\). For every \(a\neq3\), the lines are skew.

- Use one common initial point so all three displacement vectors can be compared in one linear-combination equation. - Put the coefficients of \(r\) and \(s\) into the first two matrix columns and \(\mathbf{w}\) into the augmented column. - Consistency of the matrix system determines whether the fourth displacement lies in the span of the first two.

1. \(\mathbf{u}=\langle 3,1,-2\rangle\), \(\mathbf{v}=\langle -1,3,-1\rangle\), and \(\mathbf{w}=\langle 5,5,-5\rangle\). 2. The system for \(r\mathbf{u}+s\mathbf{v}=\mathbf{w}\) is \(3r-s=5\), \(r+3s=5\), and \(2r+s=5\). 3. Row-reducing \(\left[\begin{array}{cc|c}3&-1&5\\1&3&5\\2&1&5\end{array}\right]\) gives \(\left[\begin{array}{cc|c}1&0&2\\0&1&1\\0&0&0\end{array}\right]\). 4. Thus \(r=2\) and \(s=1\), so \(\mathbf{w}=2\mathbf{u}+\mathbf{v}\). Therefore, \(D\) lies in the plane through \(A\), \(B\), and \(C\), and the four points are coplanar.

\(r=2\) and \(s=1\), so \(\mathbf{w}=2\mathbf{u}+\mathbf{v}\). The four points are coplanar.

- Set the two parametric expressions equal coordinate by coordinate. - Put the parameter equations in an augmented matrix. - Choose one free parameter. - Substitute the parameter relationships back into either plane equation.

1. Equating corresponding coordinates gives \(r-t=1\), \(r+s=-1\), and \(s-u=0\). 2. The augmented matrix for variables \(r,s,t,u\) is \(\left[\begin{array}{cccc|c}1&0&-1&0&1\\1&1&0&0&-1\\0&1&0&-1&0\end{array}\right]\). 3. Let \(r=q\). Then \(t=q-1\), \(s=-1-q\), and \(u=-1-q\). 4. Substitute into \(E_1\): \(\mathbf{r}=\langle 1,1,1\rangle+q\langle 1,1,0\rangle+(-1-q)\langle 0,1,1\rangle\). 5. Simplifying gives \(g:\mathbf{r}(q)=\langle 1,0,0\rangle+q\langle 1,0,-1\rangle\).

\(g:\mathbf{r}(q)=\langle 1,0,0\rangle+q\langle 1,0,-1\rangle\)

- Put \(\mathbf{a}\) and \(\mathbf{b}\) into the coefficient columns and \(\mathbf{c}_k\) into the augmented column. - Row-reduce until the consistency condition involving \(k\) appears. - The vectors are dependent when the system expressing \(\mathbf{c}_k\) as a combination of the first two is consistent.

1. Solve \(r\mathbf{a}+s\mathbf{b}=\mathbf{c}_k\) with the augmented matrix \(\left[\begin{array}{cc|c}2&1&k\\1&0&2\\4&-1&10\end{array}\right]\). 2. Swap the first two rows, then eliminate the first-column entries below the pivot: \(\left[\begin{array}{cc|c}1&0&2\\0&1&k-4\\0&-1&2\end{array}\right]\). 3. Add the second row to the third row. The last row becomes \(\left[\begin{array}{cc|c}0&0&k-2\end{array}\right]\). 4. The system is consistent exactly when \(k-2=0\), so \(k=2\). For this value, \(\mathbf{c}_k\) is in the span of \(\mathbf{a}\) and \(\mathbf{b}\), making the three vectors linearly dependent.

\(k=2\)

- Substitute each point into the general plane equation. - Scalar multiples describe the same plane, so choose one coefficient conveniently. - Write the remaining equations as an augmented matrix and row-reduce. - Use the solved coefficients in \(ax+by+cz=d\).

1. Substitution gives \(a+2b=d\), \(4b+c=d\), and \(2a+b+3c=d\). 2. Scalar multiples of all four coefficients describe the same plane, so choose \(d=15\). The system for \(a,b,c\) has augmented matrix \(\left[\begin{array}{ccc|c}1&2&0&15\\0&4&1&15\\2&1&3&15\end{array}\right]\). 3. Row reduction gives \(a=7\), \(b=4\), and \(c=-1\). 4. Therefore, one Cartesian equation is \(E:7x+4y-z=15\).

\(E:7x+4y-z=15\)

- Express the coordinates of a general point in the parametric plane. - Substitute them into the standard equation. - Interpret the resulting parameter equation as a one-row augmented matrix. - Use the free parameter to write the intersection line.

1. A point of \(E_2\) has coordinates \(x=1+2r+s\), \(y=1+s\), and \(z=2+r\). 2. Substitution into \(E_1\) gives \(3(1+2r+s)-(1+s)+2(2+r)=10\). 3. Simplifying gives \(8r+2s=4\). The augmented matrix \(\left[\begin{array}{cc|c}8&2&4\end{array}\right]\) has one free variable, so the planes intersect in a line. 4. Let \(r=t\). Then \(s=2-4t\). 5. Substitute into \(E_2\): \(\mathbf{r}=\langle 1,1,2\rangle+t\langle 2,0,1\rangle+(2-4t)\langle 1,1,0\rangle\). 6. Therefore the intersection line is \(g:\mathbf{r}(t)=\langle 3,3,2\rangle+t\langle -2,-4,1\rangle\).

The planes intersect in \(g:\mathbf{r}(t)=\langle 3,3,2\rangle+t\langle -2,-4,1\rangle\).

- Form three connecting vectors from one common point and place them in the columns of a homogeneous system. - Row-reduce and look for a nontrivial solution. - What does a nontrivial dependence relation among the three connecting vectors imply geometrically?

1. Using \(A\) as the common starting point gives \(\overrightarrow{AB}=\langle 2,1,1\rangle\), \(\overrightarrow{AC}=\langle -1,2,2\rangle\), and \(\overrightarrow{AD}=\langle 1,3,3\rangle\). 2. Test linear dependence by solving \(r\overrightarrow{AB}+s\overrightarrow{AC}+t\overrightarrow{AD}=\mathbf{0}\). The augmented matrix is \(\left[\begin{array}{ccc|c}2&-1&1&0\\1&2&3&0\\1&2&3&0\end{array}\right]\). 3. The last two rows are equal, and row reduction gives the equations \(2r-s+t=0\) and \(r+2s+3t=0\). Solving gives \(r=-t\) and \(s=-t\). Taking \(t=1\) gives the nontrivial relation \(-\overrightarrow{AB}-\overrightarrow{AC}+\overrightarrow{AD}=\mathbf{0}\). 4. Thus \(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AC}\), so the three connecting vectors are linearly dependent and the four points are coplanar.

Yes. The points are coplanar because \(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AC}\).

- Form three connecting vectors from the same point. - Put the two spanning vectors in the coefficient columns and the third vector in the augmented column. - Row-reduce until the consistency condition involving \(k\) appears.

1. Using \(S\) as the common starting point gives \(\overrightarrow{SQ}=\langle 0,2,-2\rangle\), \(\overrightarrow{SR}=\langle -2,0,1\rangle\), and \(\overrightarrow{SP}=\langle -1,k-1,-1\rangle\). 2. Coplanarity requires the system \(r\overrightarrow{SQ}+s\overrightarrow{SR}=\overrightarrow{SP}\) to be consistent. Its augmented matrix is \(\left[\begin{array}{cc|c}0&-2&-1\\2&0&k-1\\-2&1&-1\end{array}\right]\). 3. Swap the first two rows, then apply \(R_3\leftarrow R_3+R_1\). This gives \(\left[\begin{array}{cc|c}2&0&k-1\\0&-2&-1\\0&1&k-2\end{array}\right]\). 4. Apply \(R_2\leftarrow R_2+2R_3\). The second row becomes \([\,0\ 0\mid 2k-5\,]\). The system is consistent exactly when \(2k-5=0\), so \(k=\frac{5}{2}\).

\(k=\frac{5}{2}\)

- Convert the matrix equation to an augmented matrix. - Determine the pivot and free variables after row reduction. - Write the solution as a particular vector plus vectors multiplied by free parameters.

1. Form the augmented matrix and use row operations: \(\left(\begin{array}{cccc|c}1&-1&2&0&4\\2&0&2&4&6\\1&1&0&4&2\end{array}\right)\). Replace row 2 by \(R_2-2R_1\) and row 3 by \(R_3-R_1\): \(\left(\begin{array}{cccc|c}1&-1&2&0&4\\0&2&-2&4&-2\\0&2&-2&4&-2\end{array}\right)\). 2. The last two rows are identical, so one equation is redundant. Dividing the second row by \(2\) gives \(x_2-x_3+2x_4=-1\). 3. Let \(x_3=s\) and \(x_4=t\). Then \(x_2=-1+s-2t\). From the first equation, \(x_1=3-s-2t\). 4. Therefore, \(\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}3\\-1\\0\\0\end{pmatrix}+s\begin{pmatrix}-1\\1\\1\\0\end{pmatrix}+t\begin{pmatrix}-2\\-2\\0\\1\end{pmatrix}\), where \(s, t\in\mathbb R\).

\(\left\{\begin{pmatrix}3\\-1\\0\\0\end{pmatrix}+s\begin{pmatrix}-1\\1\\1\\0\end{pmatrix}+t\begin{pmatrix}-2\\-2\\0\\1\end{pmatrix}: s, t\in\mathbb R\right\}\)

- Treat \(t\) as a constant while performing row operations. - Check whether the final row creates a restriction on \(t\). - Use back-substitution after reaching row-echelon form.

1. The augmented matrix is \(\left(\begin{array}{ccc|c}1&0&-2&2t+4\\1&1&0&3t\\1&1&1&4t\\1&2&5&7t-4\end{array}\right)\). 2. Replace rows 2, 3, and 4 by \(R_2-R_1\), \(R_3-R_1\), and \(R_4-R_1\): \(\left(\begin{array}{ccc|c}1&0&-2&2t+4\\0&1&2&t-4\\0&1&3&2t-4\\0&2&7&5t-8\end{array}\right)\). 3. Replace row 3 by \(R_3-R_2\) and row 4 by \(R_4-2R_2\): \(\left(\begin{array}{ccc|c}1&0&-2&2t+4\\0&1&2&t-4\\0&0&1&t\\0&0&3&3t\end{array}\right)\). Then \(R_4-3R_3\) gives a zero row for every \(t\). 4. Back-substitution gives \(x_3=t\), \(x_2+2t=t-4\), so \(x_2=-t-4\), and \(x_1-2t=2t+4\), so \(x_1=4t+4\).

\(\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}4t+4\\-t-4\\t\end{pmatrix}\)

- Use the first row to eliminate \(x_1\) from the other rows. - Substitute the solution into the additional equation. - Verify each original equation separately.

1. Form the augmented matrix: \(\left(\begin{array}{ccc|c}1&1&1&4\\2&-1&3&9\\-1&2&-1&-1\end{array}\right)\). 2. Replace row 2 by \(R_2-2R_1\) and row 3 by \(R_3+R_1\): \(\left(\begin{array}{ccc|c}1&1&1&4\\0&-3&1&1\\0&3&0&3\end{array}\right)\). 3. The third row gives \(x_2=1\). The second row then gives \(x_3=4\), and the first row gives \(x_1=-1\). 4. The additional equation is satisfied because \(3\cdot(-1)+1+2\cdot4=6\). Therefore, it does not remove the solution. 5. Verification: \(-1+1+4=4\), \(2\cdot(-1)-1+3\cdot4=9\), \(-(-1)+2\cdot1-4=-1\), and \(3\cdot(-1)+1+2\cdot4=6\).

The solution set is \(\{(-1, 1, 4)\}\). The additional equation is also satisfied, so it imposes no further restriction.

- Eliminate entries one column at a time. - Scale a row when it simplifies later calculations. - Use back-substitution after reaching row-echelon form.

1. Form the augmented matrix: \(\left(\begin{array}{cccc|c}1&1&1&1&2\\2&-1&2&-1&-2\\1&3&-1&2&8\\3&2&1&-3&6\end{array}\right)\). 2. Use \(R_2-2R_1\), \(R_3-R_1\), and \(R_4-3R_1\): \(\left(\begin{array}{cccc|c}1&1&1&1&2\\0&-3&0&-3&-6\\0&2&-2&1&6\\0&-1&-2&-6&0\end{array}\right)\). 3. Divide row 2 by \(-3\), then eliminate the second-column entries below it: \(\left(\begin{array}{cccc|c}1&1&1&1&2\\0&1&0&1&2\\0&0&-2&-1&2\\0&0&-2&-5&2\end{array}\right)\). 4. Replace row 4 by \(R_4-R_3\): \(\left(\begin{array}{cccc|c}1&1&1&1&2\\0&1&0&1&2\\0&0&-2&-1&2\\0&0&0&-4&0\end{array}\right)\). 5. Back-substitution gives \(x_4=0\), \(x_3=-1\), \(x_2=2\), and \(x_1=1\).

\((x_1, x_2, x_3, x_4)=(1, 2, -1, 0)\)

- Reduce the system as usual even though it has more equations than variables. - Check whether the reduced equations are consistent with one another. - Verify the final point in all four original equations.

1. Eliminate \(x_1\) using the first equation. The other three equations become \(-x_2-5x_3=3\), \(-4x_2+x_3=-9\), and \(-x_2-2x_3=0\). 2. Eliminate \(x_2\). Combining the first two reduced equations gives \(21x_3=-21\), so \(x_3=-1\). Combining the first and third reduced equations also gives \(3x_3=-3\), confirming consistency. 3. Substitute \(x_3=-1\) into \(-x_2-5x_3=3\) to get \(x_2=2\). 4. The first original equation then gives \(x_1=3\).

\(\{(3, 2, -1)\}\)

- Use the first equation to eliminate \(x_1\) from the fourth. - Continue eliminating one variable at a time. - Back-substitute from the last equation.

1. The augmented matrix is \(\left(\begin{array}{cccc|c}2&1&0&0&4\\0&1&2&0&8\\0&0&1&2&10\\2&0&0&1&7\end{array}\right)\). 2. Replace row 4 by \(R_4-R_1\): \(R_4=(0, -1, 0, 1\mid3)\). Then replace row 4 by \(R_4+R_2\): \(R_4=(0, 0, 2, 1\mid11)\). 3. Replace row 4 by \(R_4-2R_3\). This gives \(-3x_4=-9\), so \(x_4=3\). 4. Back-substitution gives \(x_3=4\), \(x_2=0\), and \(x_1=2\).

\(\{(2, 0, 4, 3)\}\)

- Choose an equation with a coefficient of \(1\) as the first pivot equation. - Check whether the extra equation creates a contradiction or confirms the same solution. - Verify the point in all four equations.

1. Use the fourth equation to eliminate \(x_1\) from the first three. The resulting equations are \(-x_2-3x_3=-5\), \(2x_2+x_3=0\), and \(-4x_2-x_3=2\). 2. Eliminate \(x_2\). The reduced equations consistently give \(x_3=2\). 3. Substitute into \(2x_2+x_3=0\): \(2x_2+2=0\), so \(x_2=-1\). 4. Substitute into the fourth original equation: \(x_1-1+2=3\), so \(x_1=2\).

\(\{(2, -1, 2)\}\)

- Look for dependent equations during row reduction. - When there are fewer pivots than variables, choose a free parameter. - Select a variable that is easy to use in the original equations.

1. Eliminate \(x_1\) from the second and third equations using \(R_2-2R_1\) and \(R_3-3R_1\). Both reduced equations are \(-3x_2+5x_3=-3\). 2. The duplicate equation produces a zero row, so the system has one free variable and infinitely many solutions. 3. Let \(x_1=t\). From the third original equation, \(3t-x_3=3\), so \(x_3=3t-3\). 4. Substitute into the first equation: \(t+x_2-2(3t-3)=2\), which gives \(x_2=5t-4\).

\(\{(t, 5t-4, 3t-3): t\in\mathbb R\}\)

- Two independent equations are enough to determine one point in two variables. - For part b), think of three lines that intersect pairwise but are not concurrent. - A contradiction after row reduction shows that the full system has no solution.

1. For part a), choose three lines that all pass through \((3, 2)\), with at least two distinct lines. One example is \(x_1+x_2=5\), \(x_1-x_2=1\), and \(x_1=3\). The first two equations have the unique solution \((3, 2)\), and the third equation is also satisfied there. 2. For part b), choose three nonparallel lines with three different pairwise intersection points. One example is \(x_1=0\), \(x_2=0\), and \(x_1+x_2=1\). Any two lines intersect in exactly one point, but no point lies on all three. Geometrically, the three lines form the sides of a triangle.

a) One possible system is \(x_1+x_2=5\), \(x_1-x_2=1\), and \(x_1=3\). b) One possible system is \(x_1=0\), \(x_2=0\), and \(x_1+x_2=1\). The three lines intersect pairwise at different points and form a triangle.

- Row-reduce each system and watch for pivot rows, zero rows, and contradictions. - A row equivalent to \(0=c\), where \(c\ne0\), means no solution. - Introduce a parameter when a variable is free.

1. For system a), eliminate \(x_1\). The reduced equations are \(-3x_2-x_3=-9\) and \(x_2-2x_3=-4\). Solving gives \(x_3=3\), \(x_2=2\), and \(x_1=1\). The system has exactly one solution. 2. For system b), adding the first two equations gives \(3x_1+3x_2=9\), but the third equation states \(3x_1+3x_2=10\). This contradiction means the system has no solution. 3. For system c), eliminate \(x_1\). Both remaining equations reduce to \(3x_2-5x_3=-3\). Let \(x_3=t\). Then \(x_2=\frac53t-1\) and \(x_1=2-\frac13t\). The system has infinitely many solutions.

a) Exactly one solution: \(\{(1, 2, 3)\}\). b) No solution: \(\varnothing\). c) Infinitely many solutions: \(\left\{\left(2-\frac13t, \frac53t-1, t\right): t\in\mathbb R\right\}\).

- Compare the number of equations with the number of variables, but still test for dependence and consistency. - In part b), identify two free variables. - Substitute the parameterized solution into both equations.

1. In system a), the second equation gives \(x_1=x_2\). The first and third equations then both give \(2x_1+x_3=4\). The fourth equation becomes \(3x_1+x_3=6\). Subtracting gives \(x_1=2\), so \(x_2=2\) and \(x_3=0\). The overdetermined system is consistent and has one solution. 2. For system b), add the two equations: \(3x_1+3x_4=6\), so \(x_1+x_4=2\). Let \(x_4=r\) and \(x_3=s\). Then \(x_1=2-r\). Substituting into the first equation gives \(x_2=s\). Thus, there are two free parameters.

a) \(\{(2, 2, 0)\}\). b) \(\{(2-r, s, s, r): r, s\in\mathbb R\}\).

- Begin with the general cubic form. - Use the point with \(x=0\) to determine the constant term immediately. - Use the other three points to form a system for the remaining coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). 2. The point \((0, -2)\) gives \(d=-2\). The remaining points give \(a+b+c=2\), \(8a+4b+2c=14\), and \(-a+b-c=-4\). 3. Form the augmented matrix \(\left(\begin{array}{ccc|c}1&1&1&2\\8&4&2&14\\-1&1&-1&-4\end{array}\right)\). Replace row 2 by \(R_2-8R_1\) and row 3 by \(R_3+R_1\): \(\left(\begin{array}{ccc|c}1&1&1&2\\0&-4&-6&-2\\0&2&0&-2\end{array}\right)\). Replace row 2 by \(R_2+2R_3\): \(\left(\begin{array}{ccc|c}1&1&1&2\\0&0&-6&-6\\0&2&0&-2\end{array}\right)\). 4. Back-substitution gives \(b=-1\), \(c=1\), and \(a=2\). Therefore, \(f(x)=2x^3-x^2+x-2\). 5. Substitution verifies all four points.

\(f(x)=2x^3-x^2+x-2\)

- Row-reduce and identify duplicate or zero rows. - Count the free variables after determining the pivot columns. - Write the solution as a particular vector plus parameter vectors.

1. The augmented matrix is \(\left(\begin{array}{cccc|c}1&1&1&1&10\\1&-1&2&0&4\\2&0&3&1&14\end{array}\right)\). 2. Replace row 2 by \(R_2-R_1\) and row 3 by \(R_3-2R_1\): \(\left(\begin{array}{cccc|c}1&1&1&1&10\\0&-2&1&-1&-6\\0&-2&1&-1&-6\end{array}\right)\). The last two rows are identical, so there are two pivot variables and two free variables. 3. Let \(x_2=r\) and \(x_4=s\). The second row gives \(x_3=2r+s-6\). The first row then gives \(x_1=16-3r-2s\). 4. Thus, \(\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\begin{pmatrix}16\\0\\-6\\0\end{pmatrix}+r\begin{pmatrix}-3\\1\\2\\0\end{pmatrix}+s\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}\), where \(r, s\in\mathbb R\).

\(\left\{\begin{pmatrix}16\\0\\-6\\0\end{pmatrix}+r\begin{pmatrix}-3\\1\\2\\0\end{pmatrix}+s\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}: r, s\in\mathbb R\right\}\)

- Start with the general cubic polynomial. - Each point provides one linear equation in the coefficients. - One coefficient remains free because there are fewer point conditions than coefficients. - Exclude a zero leading coefficient if the degree must be exactly \(3\).

1. Let \(f(x)=ax^3+bx^2+cx+d\). Substitution gives \(d=1\), \(a+b+c=3\), and \(-a+b-c=1\). 2. Adding the last two equations gives \(2b=4\), so \(b=2\). 3. Then \(a+c=1\), so \(c=1-a\). 4. Therefore, \(f_a(x)=ax^3+2x^2+(1-a)x+1\). To ensure the polynomial has degree exactly \(3\), require \(a\ne0\).

\(f_a(x)=ax^3+2x^2+(1-a)x+1\), where \(a\in\mathbb R\setminus\{0\}\).

- Represent the number as \(100x+10y+z\). - Translate each condition into a linear equation. - When one condition is removed, keep the restrictions that \(x\), \(y\), and \(z\) are digits.

1. Let \(x\), \(y\), and \(z\) be the hundreds, tens, and ones digits. The conditions give \(x+y+z=12\), \(y+z=3x\), and \(100y+10x+z=100x+10y+z-180\). The third equation simplifies to \(x-y=2\). 2. Combining the first two equations gives \(4x=12\), so \(x=3\). Then \(x-y=2\) gives \(y=1\), and the digit sum gives \(z=8\). The number is \(318\). 3. Without the third condition, \(x=3\) and \(y+z=9\). Since \(y\) and \(z\) are digits, the possibilities are \(309, 318, 327, 336, 345, 354, 363, 372, 381, 390\).

The number is \(318\). Without the third condition, the possible numbers are \(309, 318, 327, 336, 345, 354, 363, 372, 381, 390\).

- Assign one variable to each digit. - Add and subtract the first two equations to find two digit-pair sums. - Express the remaining digits in terms of one variable. - Verify that each solution value is a valid digit.

1. Let \(a\), \(b\), \(c\), and \(d\) be the thousands, hundreds, tens, and ones digits. Then \(a+b+c+d=16\), \(a+d=b+c\), \(a=3b\), and \(c-d=4\). 2. The first two equations together give \(a+d=b+c\) and \((a+d)+(b+c)=16\). Therefore, \(a+d=8\) and \(b+c=8\). 3. Since \(a=3b\), the equation \(a+d=8\) gives \(d=8-3b\). Also, \(c=d+4\). 4. Substitute into \(b+c=8\): \(b+(8-3b+4)=8\). Thus, \(-2b=-4\), so \(b=2\). 5. Therefore, \(a=6\), \(d=2\), and \(c=6\). All are valid digits, so the number is \(6262\).

\(6262\)

- Parametrize the line containing the segment. - Equate corresponding coordinates to form a system. - Solve the system with an augmented matrix. - Check whether the segment parameter lies between \(0\) and \(1\).

1. Parametrize the line through \(A\) and \(B\): \(\mathbf{r}(r)=\langle 2,0,1\rangle+r\langle 4,4,8\rangle\). The segment corresponds to \(0\le r\le1\). 2. Equating this expression with \(h\) gives \(4r-t=-1\), \(4r+t=3\), and \(8r-t=0\). 3. The augmented matrix is \(\left[\begin{array}{cc|c}4&-1&-1\\4&1&3\\8&-1&0\end{array}\right]\), with variables \(r\) and \(t\). 4. Adding the first two equations gives \(8r=2\), so \(r=\frac14\). The second equation then gives \(t=2\). These values also satisfy the third equation. 5. Because \(\frac14\in[0,1]\), the common point lies on the segment. 6. Substitution gives \(S=\langle 2,0,1\rangle+\frac14\langle 4,4,8\rangle=\langle 3,1,3\rangle\), so \(S=(3, 1, 3)\).

Yes. Line \(h\) intersects segment \(\overline{AB}\) at \(S=(3, 1, 3)\).

- Write line intersection as the matrix system \(A\langle r,s\rangle=\mathbf{w}\). - Compare the rank of the coefficient matrix with the rank of the augmented matrix. - What rank conditions give no solution, one solution, or infinitely many parameter solutions?

1. In part a, linear dependence of \(\mathbf{u}\) and \(\mathbf{v}\) gives \(\operatorname{rank}(A)=1\). Because \(\mathbf{u}\) and \(\mathbf{w}\) are linearly independent, \(\mathbf{w}\) is not in the column space of \(A\), so \(\operatorname{rank}([\,A\mid\mathbf{w}\,])=2\). 2. Since the coefficient and augmented ranks differ, the intersection system is inconsistent. The direction vectors are parallel, but the lines have no common point, so the lines are distinct and parallel. 3. For exactly one intersection point, the system must be consistent and the two direction columns must be independent. Thus \(\operatorname{rank}(A)=\operatorname{rank}([\,A\mid\mathbf{w}\,])=2\). The three columns \(\mathbf{u}\), \(-\mathbf{v}\), and \(\mathbf{w}\) then have rank \(2\), so \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly dependent. 4. Dependence of the three vectors alone is not sufficient. If \(\operatorname{rank}(A)=1\) and the augmented rank is \(2\), the lines are distinct and parallel; if both ranks are \(1\), the lines are identical. Neither case has exactly one intersection point.

a) The lines are distinct and parallel. b) Linear dependence of \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) is necessary because a unique intersection requires \(\operatorname{rank}(A)=\operatorname{rank}([\,A\mid\mathbf{w}\,])=2\). It is not sufficient because rank-\(1\) direction matrices can describe distinct parallel or identical lines.

- Build one system in the two line parameters, but keep \(a\) in the augmented column. - Row reduction should produce one row whose consistency depends only on \(a\). - Once the system is consistent, use the solved line parameters to recover the common point.

1. Equating coordinates gives \(\lambda-3\mu=a-1\), \(\lambda+\mu=1\), and \(\lambda=-1\). 2. The augmented matrix is \(\left[\begin{array}{cc|c}1&-3&a-1\\1&1&1\\1&0&-1\end{array}\right]\). Subtract the third row from the first two to obtain rows \([0,-3\mid a]\) and \([0,1\mid2]\). 3. Adding three times the second of these rows to the first gives \([0,0\mid a+6]\). Consistency therefore requires \(a+6=0\), so \(a=-6\). 4. Then \(\lambda=-1\) and \(\mu=2\). Substitution gives the intersection point \((0, 6, 1)\).

\(a=-6\), and the lines intersect at \((0, 6, 1)\).

- Substitute \(k=2\) and \(k=3\) first. - Row-reduce the two-equation system and assign a parameter to the free variable. - Substitute the resulting line into the general family equation. - A plane perpendicular to a vector has a normal vector parallel to that vector.

1. The two planes are \(E_2:y+z=4\) and \(E_3:x+2y+3z=12\). 2. Their augmented matrix is \(\left[\begin{array}{ccc|c}0&1&1&4\\1&2&3&12\end{array}\right]\). Row reduction gives \(x+z=4\) and \(y+z=4\). 3. Let \(z=r\). Then \(x=4-r\) and \(y=4-r\), so \(s:\mathbf{r}(r)=\langle 4,4,0\rangle+r\langle -1,-1,1\rangle\). 4. A point on \(s\) has coordinates \((4-r,4-r,r)\). Substitution into \(E_k\) gives \((k-2)(4-r)+2(4-r)+kr=4k\), which simplifies to \(4k=4k\). Thus every point of \(s\) lies in every \(E_k\). 5. A normal vector to \(E_k\) is \(\mathbf{n}_k=\langle k-2,2,k\rangle\). For the plane to be perpendicular to \(\mathbf{v}\), its normal must be parallel to \(\mathbf{v}\). 6. Write \(\langle k-2,2,k\rangle=c\langle 1,2,3\rangle\). The second component gives \(c=1\), and the first and third components then both give \(k=3\).

a) \(s:\mathbf{r}(r)=\langle 4,4,0\rangle+r\langle -1,-1,1\rangle\) b) Substitution produces an identity for every \(k\) and \(r\). c) \(k=3\)

- Use two adjacent sides to parametrize the rectangular region. - Equate the beam coordinates with the rectangle coordinates. - Solve the three-parameter system with an augmented matrix. - Check whether both rectangle parameters lie in \([0,1]\).

1. Parametrize the sign using \(\overrightarrow{AB}=\langle 4,0,0\rangle\) and \(\overrightarrow{AD}=\langle 0,3,1\rangle\): \(\mathbf{r}(r,s)=\langle 2,1,3\rangle+r\langle 4,0,0\rangle+s\langle 0,3,1\rangle\), where \(0\le r\le1\) and \(0\le s\le1\). 2. Equating this point with the beam gives \(4r+2t=6\), \(3s+2t=6\), and \(s-t=-1\). 3. The augmented matrix is \(\left[\begin{array}{ccc|c}4&0&2&6\\0&3&2&6\\0&1&-1&-1\end{array}\right]\), with variables \(r,s,t\). 4. Row reduction gives \(r=\frac35\), \(s=\frac45\), and \(t=\frac95\). 5. The intersection point is \(S=\langle 8,7,2\rangle+\frac95\langle -2,-2,1\rangle=\left\langle\frac{22}{5},\frac{17}{5},\frac{19}{5}\right\rangle\), so \(S=\left(\frac{22}{5},\frac{17}{5},\frac{19}{5}\right)\). 6. Since \(0\le\frac35\le1\) and \(0\le\frac45\le1\), the point lies inside the rectangular sign.

Yes. The beam hits the sign at \(S=\left(\frac{22}{5},\frac{17}{5},\frac{19}{5}\right)\).

- Parametrize the triangle from one vertex and two side vectors. - Equate the projectile and plane coordinates. - Solve the parameter system with an augmented matrix. - For a triangle, check \(r\ge0\), \(s\ge0\), and \(r+s\le1\).

1. Parametrize the plane of the triangle: \(\mathbf{r}(r,s)=\langle 1,0,2\rangle+r\langle 4,0,0\rangle+s\langle 0,4,4\rangle\). 2. A point lies in the triangular region when \(r\ge0\), \(s\ge0\), and \(r+s\le1\). 3. Equating the plane point with the projectile gives \(4r+k=4\), \(4s+k=3\), and \(4s-2k=-1\). 4. Row reduction of the augmented system gives \(r=\frac23\), \(s=\frac{5}{12}\), and \(k=\frac43\). 5. The line meets the plane at \(S=\langle 5,3,1\rangle+\frac43\langle -1,-1,2\rangle=\left(\frac{11}{3}, \frac53, \frac{11}{3}\right)\). 6. However, \(r+s=\frac23+\frac{5}{12}=\frac{13}{12}>1\). Therefore the point lies outside the triangular sail.

No. The projectile meets the plane at \(S\left(\frac{11}{3}, \frac53, \frac{11}{3}\right)\), but this point lies outside the triangular region.

- Think about pivots, free variables, and contradictory rows in an augmented matrix. - A free variable can take more than two values. - For part c), choose two coordinate planes and a third plane that conflicts with their common line. - Check that no pair of normal vectors is parallel.

1. a) Row reduction can produce a pivot in every variable column, which gives one solution; a free variable with no contradictory row, which gives infinitely many solutions; or a row of the form \([0\ 0\ 0\mid 1]\), which gives no solution. 2. Row reduction cannot produce exactly two isolated solutions. If a free variable exists, it can take infinitely many real values, so the system has infinitely many solutions rather than two. 3. b) Geometrically, the common solution set can be empty, one point, one line, or one plane. 4. c) Consider \(E_1:x=0\), \(E_2:y=0\), and \(E_3:x+y=1\). Their normal vectors \(\langle 1,0,0\rangle\), \(\langle 0,1,0\rangle\), and \(\langle 1,1,0\rangle\) are pairwise nonparallel. 5. The augmented matrix is \(\left[\begin{array}{ccc|c}1&0&0&0\\0&1&0&0\\1&1&0&1\end{array}\right]\). Replacing row 3 by row 3 minus rows 1 and 2 gives \(\left[\begin{array}{ccc|c}1&0&0&0\\0&1&0&0\\0&0&0&1\end{array}\right]\). 6. The final row represents \(0=1\), so the three planes have no common point.

a) A three-variable linear system can have no solution, one solution, or infinitely many solutions, but not exactly two. b) The solution set can be empty, a point, a line, or a plane. c) One example is \(x=0\), \(y=0\), and \(x+y=1\). Row reduction produces a contradictory row, so there is no common point.

- Parallel planes have the same two-dimensional direction space. - Express each direction vector of one plane as a linear combination of the directions of the other. - Use consistency of each augmented matrix to determine the parameters. - Test one point to distinguish identical planes from strictly parallel planes.

1. For the planes to be parallel, each direction of \(F\) must lie in the span of the two directions of \(E\). 2. Write \(\langle 2,3,5\rangle=\alpha\langle 1,a,2\rangle+\beta\langle 0,1,1\rangle\). The augmented matrix is \(\left[\begin{array}{cc|c}1&0&2\\a&1&3\\2&1&5\end{array}\right]\). 3. The first and third rows give \(\alpha=2\) and \(\beta=1\). Consistency of the second row requires \(2a+1=3\), so \(a=1\). 4. Now write \(\langle 1,0,b\rangle=\gamma\langle 1,1,2\rangle+\delta\langle 0,1,1\rangle\). The augmented matrix is \(\left[\begin{array}{cc|c}1&0&1\\1&1&0\\2&1&b\end{array}\right]\). 5. The first two rows give \(\gamma=1\) and \(\delta=-1\). Consistency of the third row requires \(b=2-1=1\). 6. Thus the planes are parallel when \(a=1\) and \(b=1\). 7. To test whether they are identical, substitute the initial point \((2, 0, 3)\) of \(F\) into \(E\). The first two coordinate equations give \(r=1\) and \(s=-2\), but the resulting z-coordinate is \(1+2(1)-2=1\), not \(3\). 8. Therefore, the planes are strictly parallel.

\(a=1\) and \(b=1\); the planes are strictly parallel.

- Write the three equations as an augmented matrix. - Compare the third row with the sum of the first two rows. - Interpret a zero row and an inconsistent row geometrically. - To compare the pairwise intersection directions, find a vector orthogonal to all three normals.

1. The augmented matrix is \(\left[\begin{array}{ccc|c}1&-1&2&3\\2&3&-1&1\\3&2&1&k\end{array}\right]\). 2. Replace row 3 with row 3 minus row 1 minus row 2. This gives \(\left[\begin{array}{ccc|c}1&-1&2&3\\2&3&-1&1\\0&0&0&k-4\end{array}\right]\). 3. When \(k=4\), the last row is zero. The system has rank \(2\) with three variables, so it has infinitely many solutions with one free variable. 4. Geometrically, all three planes contain the same intersection line. 5. When \(k\neq4\), the last row represents \(0=k-4\), so the system is inconsistent and has no solution. 6. No two plane normals are scalar multiples, so each pair of planes intersects in a line. The vector \(\langle 1,-1,-1\rangle\) is orthogonal to all three normals, so those three pairwise intersection lines are parallel.

For \(k=4\), the system has infinitely many solutions, and the three planes share one intersection line. For \(k\neq4\), the system has no solution, and the three pairwise intersection lines are parallel.

- Write each plane equation as one row of an augmented matrix. - Replace the third row by \(R_3-2R_1+R_2\). - What does a row of the form \([\,\mathbf{0}\mid 5\,]\) mean for consistency? - Relate the same left-hand side with different right-hand sides to parallel planes.

1. Write the augmented rows of \(E_1\) and \(E_2\) as \([\,\mathbf{a}_1\mid d_1\,]\) and \([\,\mathbf{a}_2\mid d_2\,]\). Then the row for \(E_3\) is \([\,2\mathbf{a}_1-\mathbf{a}_2\mid 2d_1-d_2+5\,]\). 2. Apply the row operation \(R_3\leftarrow R_3-2R_1+R_2\). The third row becomes \([\,\mathbf{0}\mid 5\,]\), which represents the contradiction \(0=5\). Therefore, the three-plane system has no solution. 3. Geometrically, every point on \(s\) satisfies \(E_1\) and \(E_2\), so it satisfies \((2\mathbf{a}_1-\mathbf{a}_2)\cdot\mathbf{x}=2d_1-d_2\). That plane contains \(s\). 4. Plane \(E_3\) has the same normal vector but its right-hand side is larger by \(5\), so it is a distinct parallel plane. Hence line \(s\) is parallel to \(E_3\) and does not intersect it.

Line \(s\) is parallel to \(E_3\) and does not lie in \(E_3\). Therefore, the three-equation system has no solution.

- Row-reduce the augmented matrix while keeping \(k\) and \(m\) symbolic. - A missing pivot in the coefficient matrix prevents a unique solution. - For infinitely many solutions, the remaining rows must be consistent. - Use one free variable to parameterize the common line.

1. The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&1&4\\2&1&k&6\\1&2&1&m\end{array}\right]\). 2. Apply \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-R_1\): \(\left[\begin{array}{ccc|c}1&1&1&4\\0&-1&k-2&-2\\0&1&0&m-4\end{array}\right]\). 3. The coefficient matrix loses a pivot when \(k-2=0\), so \(k=2\). 4. With \(k=2\), the last two rows are consistent only when \(m-4=2\), so \(m=6\). 5. Then \(y=2\), and the first equation gives \(x+z=2\). Let \(z=t\). The common line is \(g:\mathbf{r}(t)=\langle 2,2,0\rangle+t\langle -1,0,1\rangle\). 6. If \(k=2\) but \(m\neq6\), the last two rows contradict each other, so the system has no solution. 7. No two plane normals are scalar multiples. Their pairwise intersection lines all have direction \(\langle -1,0,1\rangle\), so the three lines are parallel.

a) \(k=2\) b) \(m=6\), and \(g:\mathbf{r}(t)=\langle 2,2,0\rangle+t\langle -1,0,1\rangle\) c) The system has no solution; the planes intersect pairwise in three parallel lines.

- Compare the first two plane equations as scalar multiples. - Normalize \(F_1\) before forming the augmented matrix. - Subtract the repeated-looking rows and examine the factor \(b-1\). - Plane angles depend on normal vectors.

1. a) The normals of \(E_b\) and \(F_1\) are \(\langle 1,b,2\rangle\) and \(\langle 2,2,4\rangle\). 2. They are parallel when \(\langle 2,2,4\rangle=2\langle 1,b,2\rangle\), which gives \(b=1\). 3. For \(b=1\), the entire equation of \(F_1\) is twice the equation of \(E_1\), so the planes are identical. 4. b) Divide the equation of \(F_1\) by \(2\). An augmented matrix for the three planes is \(\left[\begin{array}{ccc|c}1&b&2&3\\1&1&2&3\\1&-1&1&0\end{array}\right]\). 5. Apply \(R_1\leftarrow R_1-R_2\). The first row becomes \(\left[\begin{array}{ccc|c}0&b-1&0&0\end{array}\right]\). 6. If \(b\neq1\), this row supplies a pivot, and the three independent equations have one common point. 7. If \(b=1\), the first two plane equations are identical. The system has one free variable, so the common solution set is the intersection line of \(E_1=F_1\) with \(F_2\). 8. c) The angle between \(F_1\) and \(F_2\) depends only on their fixed normal vectors \(\langle 2,2,4\rangle\) and \(\langle 1,-1,1\rangle\). Neither vector contains \(b\), so the angle is independent of \(b\).

a) \(b=1\); \(E_1\) and \(F_1\) are identical. b) For \(b\neq1\), the three planes meet at one point. For \(b=1\), they share the intersection line of \(E_1=F_1\) and \(F_2\). c) The normals of \(F_1\) and \(F_2\) do not depend on \(b\), so their angle is constant.

- Row-reduce the augmented matrix first. - Interpret the three variable columns as line and plane direction vectors after rearrangement. - Use the known intersection point and the solved line parameter to find the line's initial point. - Use the right-side vector to relate the two initial points.

1. The augmented matrix is \(\left[\begin{array}{ccc|c}1&-1&1&4\\2&1&-1&2\\0&2&3&11\end{array}\right]\). 2. Row reduction gives \(r=2\), \(s=1\), and \(t=3\). 3. In an equation of the form \(\mathbf{p}+r\mathbf{u}=\mathbf{a}+s\mathbf{v}+t\mathbf{w}\), rearrangement gives \(r\mathbf{u}-s\mathbf{v}-t\mathbf{w}=\mathbf{a}-\mathbf{p}\). 4. Reading the system by coordinate columns, choose \(\mathbf{u}=\langle 1,2,0\rangle\), \(\mathbf{v}=\langle 1,-1,-2\rangle\), \(\mathbf{w}=\langle -1,1,-3\rangle\), and \(\mathbf{a}-\mathbf{p}=\langle 4,2,11\rangle\). 5. Since \(S\) occurs when \(r=2\), choose \(\mathbf{p}=\overrightarrow{OS}-2\mathbf{u}=\langle 3,-3,2\rangle\). 6. Then \(\mathbf{a}=\mathbf{p}+\langle 4,2,11\rangle=\langle 7,-1,13\rangle\). 7. One valid construction is \(g:\mathbf{r}(r)=\langle 3,-3,2\rangle+r\langle 1,2,0\rangle\) and \(E:\mathbf{r}=\langle 7,-1,13\rangle+s\langle 1,-1,-2\rangle+t\langle -1,1,-3\rangle\). 8. Substituting \((r, s, t)=(2, 1, 3)\) gives \(S(5, 1, 2)\) in both equations.

a) \((r, s, t)=(2, 1, 3)\) b) One possible pair is \(g:\mathbf{r}(r)=\langle 3,-3,2\rangle+r\langle 1,2,0\rangle\) and \(E:\mathbf{r}=\langle 7,-1,13\rangle+s\langle 1,-1,-2\rangle+t\langle -1,1,-3\rangle\).

- Use row operations to eliminate \(x_1\) from the last two equations. - Identify parameter values that make the final coefficient zero. - Treat contradictions and free variables as separate cases.

1. Subtract the first equation from the second and third equations. This gives \(x_2+(k-1)x_3=2\) and \((k^2-4)x_3=k-2\). The last equation factors as \((k-2)(k+2)x_3=k-2\). 2. If \(k=-2\), the last equation becomes \(0=-4\), so the system has no solution. 3. If \(k=2\), the last equation becomes \(0=0\). The second reduced equation gives \(x_2+x_3=2\). Let \(x_3=t\). Then \(x_2=2-t\), and the first equation gives \(x_1=1\). Thus, \((x_1, x_2, x_3)=(1, 2-t, t)\), where \(t\in\mathbb R\). 4. If \(k\notin\{-2, 2\}\), then \(x_3=\frac1{k+2}\). Back-substitution gives \(x_2=2-\frac{k-1}{k+2}=\frac{k+5}{k+2}\) and \(x_1=3-x_2-x_3=\frac{2k}{k+2}\).

For \(k=-2\): no solution. For \(k=2\): \(\{(1, 2-t, t): t\in\mathbb R\}\). For \(k\in\mathbb R\setminus\{-2, 2\}\): \(\left\{\left(\frac{2k}{k+2}, \frac{k+5}{k+2}, \frac1{k+2}\right)\right\}\).

- Eliminate the first entry from rows 2 and 3, then compare the resulting rows. - A homogeneous square system has a unique solution when every variable column contains a pivot. - For part c), apply the same row combination to the first three components only.

1. Replace row 2 by \(R_2-R_1\) and row 3 by \(R_3-R_1\): \(R_2=(0, 1, -2, b-a)\) and \(R_3=(0, 2, -4, c-a)\). 2. Replace row 3 by \(R_3-2R_2\). This gives \(R_3=(0, 0, 0, a-2b+c)\). Swap rows 3 and 4 to obtain the row-echelon form \(\begin{pmatrix}1&0&2&a\\0&1&-2&b-a\\0&0&1&d\\0&0&0&a-2b+c\end{pmatrix}\). 3. A homogeneous system with four variables has only the trivial solution exactly when there is a pivot in every column. Here, that occurs when \(a-2b+c\ne0\). This condition does not involve \(d\), so uniqueness is independent of \(d\). 4. The same row combination applied to the first three components gives \(\mathbf{v}_3-2\mathbf{v}_2+\mathbf{v}_1=\mathbf{0}\). Since this is a nontrivial linear relation, the three vectors are linearly dependent.

a) \(\begin{pmatrix}1&0&2&a\\0&1&-2&b-a\\0&0&1&d\\0&0&0&a-2b+c\end{pmatrix}\). b) The solution is unique exactly when \(a-2b+c\ne0\), which is independent of \(d\). c) \(\mathbf{v}_3-2\mathbf{v}_2+\mathbf{v}_1=\mathbf{0}\), so the vectors are linearly dependent.

- Subtract the first row from the other two rows. - Eliminate the second variable from the third row. - A nonzero final coefficient gives a unique solution. - A row of the form \(0=c\), with \(c\neq0\), makes the system inconsistent.

1. Start with the augmented matrix and apply \(R_2\leftarrow R_2-R_1\) and \(R_3\leftarrow R_3-R_1\): \(\left[\begin{array}{ccc|c}1&1&1&2\\0&1&k-1&1\\0&k-1&3&k-2\end{array}\right]\). 2. Apply \(R_3\leftarrow R_3-(k-1)R_2\). The last row becomes \(\left[\begin{array}{ccc|c}0&0&-k^2+2k+2&-1\end{array}\right]\). 3. a) The system has a pivot in every variable when \(-k^2+2k+2\neq0\). 4. Solving \(-k^2+2k+2=0\) gives \(k=1-\sqrt3\) or \(k=1+\sqrt3\). 5. Therefore, the system has exactly one solution for \(k\in\mathbb{R}\setminus\{1-\sqrt3,1+\sqrt3\}\). 6. b) When \(k=2\), row reduction gives \(z=-\frac12\), \(y=\frac32\), and \(x=1\). 7. Thus the three planes intersect at the single point \(\left(1, \frac32, -\frac12\right)\). 8. c) At either critical value, the last row becomes \(0=-1\). Therefore, the system has no solution for \(k=1-\sqrt3\) and \(k=1+\sqrt3\).

a) \(k\in\mathbb{R}\setminus\{1-\sqrt3,1+\sqrt3\}\) b) \(\left(1, \frac32, -\frac12\right)\); the three planes meet at one point. c) \(k=1-\sqrt3\) or \(k=1+\sqrt3\)

- Row-reduce the augmented matrix while keeping \(k\) symbolic. - Focus on the final pivot coefficient \(k-5\). - A zero row creates a free variable. - Interpret a single solution as a point and one free variable as a line.

1. Begin with the augmented matrix and apply \(R_2\leftarrow R_2-R_1\) and \(R_3\leftarrow R_3-3R_1\): \(\left[\begin{array}{ccc|c}1&1&1&2\\0&-2&1&-1\\0&-4&k-3&-2\end{array}\right]\). 2. Apply \(R_3\leftarrow R_3-2R_2\). The last row becomes \(\left[\begin{array}{ccc|c}0&0&k-5&0\end{array}\right]\). 3. If \(k\neq5\), then \(z=0\). Back-substitution gives \(y=\frac12\) and \(x=\frac32\). 4. Thus the solution set is \(\left\{\left(\frac32, \frac12, 0\right)\right\}\), and the three planes meet at one point. 5. If \(k=5\), the last row is zero, so one variable is free. Let \(z=t\). 6. The second row gives \(y=\frac12+\frac12t\), and the first row gives \(x=\frac32-\frac32t\). 7. The solution set is the line \(\mathbf{r}(t)=\left\langle \frac32,\frac12,0\right\rangle+t\left\langle -\frac32,\frac12,1\right\rangle\). 8. Geometrically, all three planes share this line.

a) For \(k\neq5\), the system has one solution. For \(k=5\), it has infinitely many solutions. b) For \(k\neq5\), the solution set is \(\left\{\left(\frac32, \frac12, 0\right)\right\}\). For \(k=5\), the solution set is \(\mathbf{r}(t)=\left\langle \frac32,\frac12,0\right\rangle+t\left\langle -\frac32,\frac12,1\right\rangle\). c) For \(k\neq5\), the three planes meet at one point. For \(k=5\), the three planes share the stated line.

- Use a dot product between plane normals in part a. - A plane parallel to the xy-plane has a vertical normal. - Row-reduce the parameterized augmented matrix. - Separate the case in which the final pivot coefficient is zero.

1. The normals are \(\mathbf{n}_a=\langle a+1,1-a,a\rangle\) and \(\mathbf{n}_H=\langle 2,-2,1\rangle\). 2. Perpendicular planes have orthogonal normals: \(\mathbf{n}_a\cdot\mathbf{n}_H=2(a+1)-2(1-a)+a=5a\). 3. This equals zero only when \(a=0\), so exactly one plane in the family is perpendicular to \(H\). 4. A plane parallel to the xy-plane has a normal parallel to \(\langle 0,0,1\rangle\). This would require \(a+1=0\) and \(1-a=0\) simultaneously. 5. These equations require \(a=-1\) and \(a=1\), so no plane in the family is parallel to the xy-plane. 6. The augmented matrix is \(\left[\begin{array}{ccc|c}1&1&0&2\\1&-1&1&0\\2&0&t&4\end{array}\right]\). 7. Apply \(R_2\leftarrow R_2-R_1\), \(R_3\leftarrow R_3-2R_1\), and then \(R_3\leftarrow R_3-R_2\). This gives a final row \(\left[\begin{array}{ccc|c}0&0&t-1&2\end{array}\right]\). 8. If \(t\neq1\), then \(z=\frac{2}{t-1}\). Back-substitution gives \(y=\frac{t}{t-1}\) and \(x=\frac{t-2}{t-1}\). 9. The three planes meet at the single point \(S\left(\frac{t-2}{t-1}, \frac{t}{t-1}, \frac{2}{t-1}\right)\). 10. If \(t=1\), the last row becomes \(0=2\), so the system has no solution. The planes have no common point; their pairwise intersection lines are parallel.

a) \(a=0\) b) No value of \(a\) c) For \(t\neq1\), the unique solution is \(S\left(\frac{t-2}{t-1}, \frac{t}{t-1}, \frac{2}{t-1}\right)\). For \(t=1\), the system has no solution, and the three pairwise intersection lines are parallel.

- Substitute \(a=0\), then test a general point of the line. - Separate the family equation into a part independent of \(a\) and a coefficient of \(a\). - Solve the two resulting plane equations with an augmented matrix. - View all planes through the common line as nonzero linear combinations of those two equations.

1. Setting \(a=0\) gives \(F_0:x-y+z=1\). 2. A general point of \(g\) is \((1+\lambda, \lambda, 0)\). Substitution gives \((1+\lambda)-\lambda+0=1\), so \(g\subset F_0\). 3. Rewrite the family equation as \(x-y+z-1+a(x+2y-z-3)=0\). 4. A point lies in every plane when it satisfies \(x-y+z=1\) and \(x+2y-z=3\). 5. An augmented matrix for this system is \(\left[\begin{array}{ccc|c}1&-1&1&1\\1&2&-1&3\end{array}\right]\). 6. Row reduction gives \(2x+y=4\). Let \(x=t\). Then \(y=4-2t\) and \(z=5-3t\). 7. Therefore, \(l:\mathbf{r}(t)=\langle 0,4,5\rangle+t\langle 1,-2,-3\rangle\). 8. Every plane through \(l\) has an equation \(\alpha P_1+\beta P_2=0\), where \(P_1=x-y+z-1\) and \(P_2=x+2y-z-3\), with \((\alpha,\beta)\neq(0,0)\). 9. The given family uses \(P_1+aP_2=0\), which represents all cases with \(\alpha\neq0\). The one missing case has \(\alpha=0\). 10. Thus the missing plane is \(P_2=0\), or \(x+2y-z=3\).

a) \(F_0:x-y+z=1\), and \(g\subset F_0\). b) \(l:\mathbf{r}(t)=\langle 0,4,5\rangle+t\langle 1,-2,-3\rangle\) c) \(x+2y-z=3\)

- Compare the rows of the augmented matrix. - Identical or proportional equations leave two free variables. - A contradictory row gives no solution. - Two independent plane equations normally leave one free variable.

1. Choose \(E_2:4x-2y+6z=12\). Its row is twice the row of \(E_1\), so row reduction leaves one nonzero row and two free variables. 2. Let \(y=s\) and \(z=t\). Then \(2x=6+s-3t\), so \(x=3+\frac12s-\frac32t\). 3. The solution set is \(\mathbf{r}=\langle 3,0,0\rangle+s\left\langle\frac12,1,0\right\rangle+t\left\langle-\frac32,0,1\right\rangle\). 4. Choose \(E_2:2x-y+3z=10\). The augmented matrix contains rows \([2\ -1\ 3\mid6]\) and \([2\ -1\ 3\mid10]\). Subtracting the first row from the second gives \([0\ 0\ 0\mid4]\), so the system is inconsistent. 5. Choose \(E_2:y=0\). The augmented matrix is \(\left[\begin{array}{ccc|c}2&-1&3&6\\0&1&0&0\end{array}\right]\). 6. Row reduction gives \(y=0\) and \(x+\frac32z=3\). Letting \(z=t\) gives \(\mathbf{r}=\langle 3,0,0\rangle+t\left\langle-\frac32,0,1\right\rangle\), a line through \(P\).

a) One choice is \(E_2:4x-2y+6z=12\), with solution set \(\mathbf{r}=\langle 3,0,0\rangle+s\left\langle\frac12,1,0\right\rangle+t\left\langle-\frac32,0,1\right\rangle\). b) One choice is \(E_2:2x-y+3z=10\). c) One choice is \(E_2:y=0\).

- Start with the general fourth-degree polynomial. - Use the point with \(x=0\) to find the constant term. - Reduce the remaining coefficient system and choose one free parameter. - Require a nonzero leading coefficient.

1. Let \(f(x)=ax^4+bx^3+cx^2+dx+e\). The point \((0, 0)\) gives \(e=0\). The other points give \(a+b+c+d=1\), \(a-b+c-d=-3\), and \(8a+4b+2c+d=9\). 2. Form the augmented matrix \(\left(\begin{array}{cccc|c}1&1&1&1&1\\1&-1&1&-1&-3\\8&4&2&1&9\end{array}\right)\). Replace row 2 by \(R_2-R_1\), divide it by \(-2\), and replace row 3 by \(R_3-8R_1\): \(\left(\begin{array}{cccc|c}1&1&1&1&1\\0&1&0&1&2\\0&-4&-6&-7&1\end{array}\right)\). Then replace row 3 by \(R_3+4R_2\): \(\left(\begin{array}{cccc|c}1&1&1&1&1\\0&1&0&1&2\\0&0&-6&-3&9\end{array}\right)\). 3. Let \(d\) be free. The last two rows give \(b=2-d\) and \(c=-\frac{3+d}{2}\). The first row gives \(a=\frac{1+d}{2}\). Equivalently, use \(a\) as the parameter, so \(d=2a-1\), \(b=3-2a\), and \(c=-a-1\). 4. Therefore, \(f_a(x)=ax^4+(3-2a)x^3-(a+1)x^2+(2a-1)x\). To ensure degree exactly \(4\), require \(a\ne0\).

\(f_a(x)=ax^4+(3-2a)x^3-(a+1)x^2+(2a-1)x\), where \(a\in\mathbb R\setminus\{0\}\).