Midterm 2 Study Guide

Let \[ A = \begin{bmatrix*}[r] 1 & 2 & -1 & 0 & 3\\ 2 & 4 & -1 & 1 & 4\\ 1 & 2 & 0 & 1 & 1 \end{bmatrix*}. \]

1. Find a basis for $\Nul(A)$.

   Solution

   We row reduce $A$.

   \[ \begin{aligned} &\left[\begin{array}{ccccc} 1 & 2 & \llap{$-$}1 & 0 & 3 \\ 2 & 4 & \llap{$-$}1 & 1 & 4 \\ 1 & 2 & 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{ccccc} 1 & 2 & \llap{$-$}1 & 0 & 3 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 \\ 1 & 2 & 0 & 1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{1}} \left[\begin{array}{ccccc} 1 & 2 & \llap{$-$}1 & 0 & 3 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccccc} 1 & 2 & \llap{$-$}1 & 0 & 3 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} + R_{2}} \left[\begin{array}{ccccc} 1 & 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{aligned} \] The pivots are in columns 1 and 3, so $x_2 = r$, $x_4 = s$, and $x_5 = t$ are free. The nonzero rows say $x_1 = -2r - s - t$ and $x_3 = -s + 2t$, so the solutions of $Ax = 0$ are \[ r \begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix*} + s \begin{bmatrix*}[r] -1 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix*} + t \begin{bmatrix*}[r] -1 \\ 0 \\ 2 \\ 0 \\ 1 \end{bmatrix*}. \] These three vectors form a basis for $\Nul(A)$.

2. Find a basis for $\col(A)$, and find $\rank(A)$ and $\nullity(A)$.

   Solution

   The pivot columns of $A$ are columns 1 and 3, so a basis for $\col(A)$ is \[ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix},\qquad \begin{bmatrix*}[r] -1 \\ -1 \\ 0 \end{bmatrix*}. \] So $\rank(A) = 2$ and $\nullity(A) = 3$ (and $2 + 3 = 5$, the number of columns).

3. Is $T_A$ injective? Is it surjective? Explain.

   Solution

   $T_A$ is not injective, since only 2 of the 5 columns have pivots (so $\Nul(A) \neq \{0\}$). $T_A$ is not surjective, since only 2 of the 3 rows have pivots (so $\col(A)$ is a plane, not all of $\RR^3$).

4. Is $b = \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$ in $\col(A)$?

   Solution

   Yes. We look for $x_1, x_2$ with $x_1 (1, 2, 1) + x_2 (-1, -1, 0) = (1, 3, 2)$. The third coordinate gives $x_1 = 2$, the first then gives $x_2 = 1$, and the second coordinate checks: $4 - 1 = 3$. So $b$ is the first column of $A$ times 2 plus the third column of $A$.

5. Find a vector in $\RR^3$ that is not in $\col(A)$.

   Solution

   We find exactly which $b$ are in $\col(A)$ by row reducing $[\,a_1\ a_3 \mid b\,]$, where $a_1, a_3$ are the pivot columns: \[ \left[\begin{array}{rr|c} 1 & -1 & b_1 \\ 2 & -1 & b_2 \\ 1 & 0 & b_3 \end{array}\right] \xrightarrow[R_3 \to R_3 - R_1]{R_2 \to R_2 - 2R_1} \left[\begin{array}{rr|c} 1 & -1 & b_1 \\ 0 & 1 & b_2 - 2b_1 \\ 0 & 1 & b_3 - b_1 \end{array}\right] \xrightarrow{R_3 \to R_3 - R_2} \left[\begin{array}{rr|c} 1 & -1 & b_1 \\ 0 & 1 & b_2 - 2b_1 \\ 0 & 0 & b_1 - b_2 + b_3 \end{array}\right]. \] So $b$ is in $\col(A)$ exactly when $b_1 - b_2 + b_3 = 0$. For example, $(1, 0, 0)$ is not in $\col(A)$.

Let $A$ be the matrix from Problem 1, and let $b = \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$.

1. Find the general solution of $Ax = b$ in parametric vector form (that is, in the form $v_0 + t_1 v_1 + \cdots + t_k v_k$).

   Solution

   We row reduce $[\,A \mid b\,]$.

   \[ \begin{aligned} &\left[\begin{array}{ccccc|c} 1 & 2 & \llap{$-$}1 & 0 & 3 & 1 \\ 2 & 4 & \llap{$-$}1 & 1 & 4 & 3 \\ 1 & 2 & 0 & 1 & 1 & 2 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{ccccc|c} 1 & 2 & \llap{$-$}1 & 0 & 3 & 1 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 & 1 \\ 1 & 2 & 0 & 1 & 1 & 2 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{1}} \left[\begin{array}{ccccc|c} 1 & 2 & \llap{$-$}1 & 0 & 3 & 1 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 & 1 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccccc|c} 1 & 2 & \llap{$-$}1 & 0 & 3 & 1 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} + R_{2}} \left[\begin{array}{ccccc|c} 1 & 2 & 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 & \llap{$-$}2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{aligned} \] With $x_2 = r$, $x_4 = s$, $x_5 = t$ free, the nonzero rows say $x_1 = 2 - 2r - s - t$ and $x_3 = 1 - s + 2t$, so \[ x = \begin{bmatrix} 2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + r \begin{bmatrix*}[r] -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix*} + s \begin{bmatrix*}[r] -1 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix*} + t \begin{bmatrix*}[r] -1 \\ 0 \\ 2 \\ 0 \\ 1 \end{bmatrix*}. \] In other words, $x = v_0 + \Nul(A)$ with $v_0 = (2, 0, 1, 0, 0)$, using the basis of $\Nul(A)$ from Problem 1.

2. Is the solution set of $Ax = b$ a subspace of $\RR^5$? Explain.

   Solution

   No. Every subspace contains $0$, but $x = 0$ is not a solution, since $A0 = 0 \neq b$.

For each collection of vectors, decide whether it is linearly independent. If it is not, find a linear dependence relation among the vectors ($c_1v_1 + c_2v_2 + c_3v_3 = 0$ with $c_1, c_2, c_3$ not all zero).

1. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix},\quad v_2 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix},\quad v_3 = \begin{bmatrix} 0 \\ 3 \\ 6 \end{bmatrix} } \)

   Solution

   We row reduce the matrix with columns $v_1, v_2, v_3$.

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 3 \\ 3 & 0 & 6 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}3 & 3 \\ 3 & 0 & 6 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - 3\,R_{1}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}3 & 3 \\ 0 & \llap{$-$}6 & 6 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - 2\,R_{2}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}3 & 3 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{2} \to -\frac{1}{3}\,R_{2}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} - 2\,R_{2}} \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] Column 3 has no pivot, so the vectors are linearly dependent. Setting the free variable $x_3 = 1$ gives $x_1 = -2$ and $x_2 = 1$, so \[ -2v_1 + v_2 + v_3 = 0. \]

2. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix},\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix},\quad v_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix} } \)

   Solution

   We row reduce the matrix with columns $v_1, v_2, v_3$ to echelon form.

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \xrightarrow{R_{4} \to R_{4} - R_{1}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 2 \\ 0 & 1 & 0 \end{array}\right] \xrightarrow{R_{4} \to R_{4} - R_{2}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \leftrightarrow R_{4}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{array}\right] \xrightarrow{R_{4} \to R_{4} - 2\,R_{3}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] Every column has a pivot, so the vectors are linearly independent.

1. Is $\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ a basis of $\RR^3$?

   Solution

   Yes. Three vectors in $\RR^3$ form a basis exactly when the matrix with these columns has nonzero determinant. Expanding along the first row, \[ \det \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} = 1 \cdot (1 - 0) - 2 \cdot (0 - 1) + 0 = 3 \neq 0. \]

2. Is $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ a basis of $\RR^3$?

   Solution

   No. We row reduce the matrix with these columns.

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & 3 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}1 & 1 \\ 2 & 3 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}1 & 1 \\ 0 & \llap{$-$}1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & \llap{$-$}1 & 1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] There are only 2 pivots, so the vectors are linearly dependent (in fact $-2v_1 + v_2 + v_3 = 0$) and do not form a basis.

3. The vectors $v_1 = (1, 1, 0, 0)$ and $v_2 = (0, 1, 1, 0)$ are linearly independent. Find a basis of $\RR^4$ that contains $v_1$ and $v_2$.

   Solution

   There are many answers. One is $v_1, v_2, e_3, e_4$. To check it, we row reduce the matrix with these columns: \[ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 \to R_2 - R_1} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_3 \to R_3 - R_2} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. \] There are 4 pivots, so these four vectors are linearly independent and form a basis of $\RR^4$.

Let $V = \sp\{v_1, v_2\}$, where \[ v_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix},\qquad v_2 = \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*}. \] The vectors $v_1, v_2$ are not multiples of each other, so $\cB = \{v_1, v_2\}$ is a basis of $V$.

1. The vector $w = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$ is in $V$. Find $[w]_\cB$.

   Solution

   We solve $x_1 v_1 + x_2 v_2 = w$.

   \[ \begin{aligned} &\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 2 & \llap{$-$}1 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & \llap{$-$}1 & \llap{$-$}3 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] So $x_1 = 2$ and $x_2 = 3$, and \[ [w]_\cB = \begin{bmatrix} 2 \\ 3 \end{bmatrix}. \]

2. Is $u = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ in $V$?

   Solution

   No. We try to solve $x_1 v_1 + x_2 v_2 = u$.

   \[ \begin{aligned} &\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 2 & \llap{$-$}1 & 3 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & \llap{$-$}1 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{array}\right] \end{aligned} \] The last row says $0 = 3$, so the system is inconsistent and $u$ is not in $V$.

3. Find the vector $x$ with $[x]_\cB = \begin{bmatrix*}[r] -1 \\ 4 \end{bmatrix*}$.

   Solution

   \[ x = -v_1 + 4v_2 = \begin{bmatrix*}[r] -1 \\ 0 \\ -2 \end{bmatrix*} + \begin{bmatrix*}[r] 0 \\ 4 \\ -4 \end{bmatrix*} = \begin{bmatrix*}[r] -1 \\ 4 \\ -6 \end{bmatrix*}. \]

1. $A$ is a $5 \times 7$ matrix with $\rank(A) = 4$. What is $\nullity(A)$?

   Solution

   By the Rank-Nullity Theorem, $\nullity(A) = 7 - 4 = 3$.

2. $A$ is a $4 \times 4$ matrix, and the solution set of $Ax = 0$ is a line through the origin. What is $\rank(A)$?

   Solution

   $\nullity(A) = 1$, so $\rank(A) = 4 - 1 = 3$. (In particular, $A$ is not invertible.)

3. $A$ is a $3 \times 5$ matrix, and $Ax = b$ is consistent for every $b$ in $\RR^3$. Describe the solution set of $Ax = 0$.

   Solution

   $\col(A) = \RR^3$, so $\rank(A) = 3$ and $\nullity(A) = 5 - 3 = 2$. The solution set of $Ax = 0$ is a plane through the origin in $\RR^5$.

4. $A$ is a $4 \times 7$ matrix. Explain why $\Nul(A)$ cannot be a plane through the origin.

   Solution

   $A$ has 4 rows, so it has at most 4 pivots and $\rank(A) \leq 4$. So $\nullity(A) = 7 - \rank(A) \geq 3$, but a plane is 2-dimensional.

In each part, $T_A : \RR^2 \to \RR^2$ is a linear transformation. Find the matrix $A$.

1. $T_A(e_1) = (2, -1)$ and $T_A(e_2) = (0, 3)$.

   Solution

   The columns of $A$ are $T_A(e_1)$ and $T_A(e_2)$: \[ A = \begin{bmatrix*}[r] 2 & 0 \\ -1 & 3 \end{bmatrix*}. \]

2. $T_A$ rotates the plane counterclockwise by $90^\circ$.

   Solution

   The rotation sends $e_1$ to $e_2 = (0, 1)$ and $e_2$ to $-e_1 = (-1, 0)$, so \[ A = \begin{bmatrix*}[r] 0 & -1 \\ 1 & 0 \end{bmatrix*}. \]

3. $T_A$ first rotates the plane counterclockwise by $90^\circ$ and then reflects it across the line $y = x$.

   Solution

   The reflection across $y = x$ swaps the coordinates, so its matrix is $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Doing the rotation first and the reflection second gives the product with the reflection on the left: \[ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix*}[r] 0 & -1 \\ 1 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}. \] So $T_A$ is the reflection across the $x$-axis.

4. $T_A(1, 2) = (3, 0)$ and $T_A(1, 1) = (1, 1)$.

   Solution

   Let $U = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}$ and $W = \begin{bmatrix} 3 & 1 \\ 0 & 1 \end{bmatrix}$. Then $AU = W$, so $A = WU^{-1}$. By the $2 \times 2$ inverse formula, \[ U^{-1} = \frac{1}{1 - 2} \begin{bmatrix*}[r] 1 & -1 \\ -2 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 1 \\ 2 & -1 \end{bmatrix*}, \qquad A = \begin{bmatrix} 3 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix*}[r] -1 & 1 \\ 2 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 2 \\ 2 & -1 \end{bmatrix*}. \] Check: $A(1, 2) = (3, 0)$ and $A(1, 1) = (1, 1)$.

The face on the left sits on the square with corners at the origin, $u_1 = (2, -1)$, $u_2 = (1, 2)$, and $u_1 + u_2 = (3, 1)$. The matrix $A$ is such that $T_A$ maps the face on the left to the face on the right.

1. Find $A$.

   Solution

   Reading off the picture, $T_A(u_1) = w_1 = (-2, 1)$ and $T_A(u_2) = w_2 = (4, 3)$. With $U = \begin{bmatrix} u_1 & u_2 \end{bmatrix}$ and $W = \begin{bmatrix} w_1 & w_2 \end{bmatrix}$ we have $AU = W$, so $A = WU^{-1}$: \[ U^{-1} = \frac{1}{4 + 1} \begin{bmatrix*}[r] 2 & -1 \\ 1 & 2 \end{bmatrix*}, \qquad A = \frac{1}{5} \begin{bmatrix*}[r] -2 & 4 \\ 1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -1 \\ 1 & 2 \end{bmatrix*} = \frac{1}{5} \begin{bmatrix} 0 & 10 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 1 & 1 \end{bmatrix}. \]

2. Find $\det(A)$. How can you see the sign of $\det(A)$ in the picture?

   Solution

   $\det(A) = (0)(1) - (2)(1) = -2$. The determinant is negative because $T_A$ reverses orientation: the face on the right is a mirror image of the face on the left. (Going counterclockwise around the left square we meet $u_1$ before $u_2$, but going counterclockwise around the right parallelogram we meet $w_2$ before $w_1$.)

3. The face on the left has area 5. What is the area of the face on the right?

   Solution

   $T_A$ multiplies areas by $|\det(A)| = 2$, so the face on the right has area $10$.

For each matrix $A$ below, decide whether $T_A$ is injective, surjective, and/or bijective. Justify your answers.

1. $A = \begin{bmatrix*}[r] 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix*}$

   Solution

   $A$ is already in echelon form, with pivots in both rows, so $T_A$ is surjective. Column 3 has no pivot, so $T_A$ is not injective (and not bijective).

2. $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{bmatrix}$

   Solution

   Row reducing, \[ \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \xrightarrow{R_3 \to R_3 - R_1} \begin{bmatrix*}[r] 1 & 2 \\ 0 & 1 \\ 0 & -2 \end{bmatrix*} \xrightarrow{R_3 \to R_3 + 2R_2} \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}. \] Both columns have pivots, so $T_A$ is injective. The third row has no pivot, so $T_A$ is not surjective (and not bijective).

3. $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$

   Solution

   Expanding along the first row, \[ \det(A) = 2(4 - 1) - 1(2 - 0) + 0 = 4 \neq 0. \] So $A$ is invertible and $T_A$ is bijective (both injective and surjective).

4. $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 2 \end{bmatrix}$

   Solution

   The third column is the sum of the first two, so the columns are linearly dependent and $T_A$ is not injective. Since $A$ is square, the Invertible Matrix Theorem says $T_A$ is also not surjective.

1. Find $A^{-1}$, where $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 1 & 0 \end{bmatrix}$.

   Solution

   We row reduce $[\,A \mid I_3\,]$.

   \[ \begin{aligned} &\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 2 & 1 & 3 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & \llap{$-$}1 & \llap{$-$}2 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & \llap{$-$}1 & \llap{$-$}2 & 1 & 0 \\ 0 & 0 & 1 & 2 & \llap{$-$}1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} - 2\,R_{3}} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \llap{$-$}3 & 2 & \llap{$-$}2 \\ 0 & 1 & \llap{$-$}1 & \llap{$-$}2 & 1 & 0 \\ 0 & 0 & 1 & 2 & \llap{$-$}1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{2} \to R_{2} + R_{3}} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \llap{$-$}3 & 2 & \llap{$-$}2 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 & \llap{$-$}1 & 1 \end{array}\right] \end{aligned} \] So \[ A^{-1} = \begin{bmatrix*}[r] -3 & 2 & -2 \\ 0 & 0 & 1 \\ 2 & -1 & 1 \end{bmatrix*}. \] Check: $AA^{-1} = I_3$.

2. Find the inverse of $\begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$.

   Solution

   By the $2 \times 2$ inverse formula, \[ \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}^{-1} = \frac{1}{6 - 5} \begin{bmatrix*}[r] 2 & -5 \\ -1 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & -5 \\ -1 & 3 \end{bmatrix*}. \]

3. For which values of $k$ is $\begin{bmatrix} k & 2 \\ 3 & k + 1 \end{bmatrix}$ invertible?

   Solution

   The determinant is $k(k + 1) - 6 = k^2 + k - 6 = (k + 3)(k - 2)$. The matrix is invertible exactly when this is nonzero, that is, when $k \neq 2$ and $k \neq -3$.

4. Suppose $A$ and $B$ are $2 \times 2$ matrices with \[ A^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix},\qquad B^{-1} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}. \] Find $(AB)^{-1}$.

   Solution

   The inverse of a product is the product of the inverses in the opposite order: \[ (AB)^{-1} = B^{-1} A^{-1} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}. \]

Compute each determinant.

1. $\det \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 2 & 1 \end{bmatrix}$, using row operations

   Solution

   We row reduce to upper triangular form.

   \[ \begin{aligned} &\left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 2 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 2 & 1 \end{array}\right] \xrightarrow{R_{4} \to R_{4} - R_{1}} \left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 2 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{2} \leftrightarrow R_{3}} \left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 2 & 0 \end{array}\right] \xrightarrow{R_{4} \to R_{4} - R_{2}} \left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & \llap{$-$}2 \end{array}\right] \\[4pt] &\xrightarrow{R_{4} \to R_{4} - R_{3}} \left[\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & \llap{$-$}3 \end{array}\right] \end{aligned} \] Row replacements do not change the determinant, and there was one swap, so \[ \det = -\bigl(1 \cdot 1 \cdot 1 \cdot (-3)\bigr) = 3. \]

2. $\det \begin{bmatrix*}[r] 2 & 0 & 1 \\ 3 & 1 & -1 \\ 0 & 4 & 2 \end{bmatrix*}$, using cofactor expansion

   Solution

   Expanding along the first row, which has a zero, \[ \det = 2 \det \begin{bmatrix*}[r] 1 & -1 \\ 4 & 2 \end{bmatrix*} - 0 + 1 \det \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix} = 2(2 + 4) + (12 - 0) = 24. \]

3. $\det \begin{bmatrix} 1 & 0 & 2 & 0 \\ 3 & 0 & 1 & 1 \\ 0 & 2 & 0 & 0 \\ 4 & 1 & 0 & 3 \end{bmatrix}$, using cofactor expansion

   Solution

   Row 3 has only one nonzero entry, the $2$ in column 2, whose sign in the checkerboard pattern is $(-1)^{3 + 2} = -1$. Expanding along row 3 and then along the first row of the $3 \times 3$ matrix, \[ \det = -2 \det \begin{bmatrix} 1 & 2 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 3 \end{bmatrix} = -2 \bigl(1(3 - 0) - 2(9 - 4) + 0\bigr) = -2(-7) = 14. \]

Suppose that \[ \det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = 5. \] Find the determinant of each matrix below.

1. $\begin{bmatrix} c & b + 4c & 2a \\ f & e + 4f & 2d \\ i & h + 4i & 2g \end{bmatrix}$

   Solution

   We use column operations, starting from the matrix with determinant $5$. Swapping columns 1 and 3 changes the sign, giving $-5$. Adding 4 times the new column 1 to column 2 does not change the determinant. Multiplying column 3 by 2 doubles it. So the determinant is $-10$.

2. $\begin{bmatrix} 2a & b & c - a \\ 2d & e & f - d \\ 2g & h & i - g \end{bmatrix}$

   Solution

   Subtracting column 1 from column 3 does not change the determinant, and then multiplying column 1 by 2 doubles it. So the determinant is $10$.

3. $\begin{bmatrix} d & e & f \\ g & h & i \\ a & b & c \end{bmatrix}$

   Solution

   Swapping rows 1 and 2, and then swapping rows 2 and 3, turns the original matrix into this one. Each swap changes the sign, so the determinant is $(-1)^2 \cdot 5 = 5$.

1. Find all values of $c$ for which the equation $Ax = b$ has exactly one solution for every $b$ in $\RR^3$, where \[ A = \begin{bmatrix} c & 1 & 1 \\ 1 & c & 1 \\ 1 & 1 & c \end{bmatrix}. \]

   Solution

   $Ax = b$ has exactly one solution for every $b$ exactly when $A$ is invertible, that is, when $\det(A) \neq 0$. Adding columns 2 and 3 to column 1 does not change the determinant, and makes every entry of column 1 equal to $c + 2$. Factoring $c + 2$ out of column 1 and then subtracting row 1 from rows 2 and 3, \[ \det(A) = (c + 2) \det \begin{bmatrix} 1 & 1 & 1 \\ 1 & c & 1 \\ 1 & 1 & c \end{bmatrix} = (c + 2) \det \begin{bmatrix} 1 & 1 & 1 \\ 0 & c - 1 & 0 \\ 0 & 0 & c - 1 \end{bmatrix} = (c + 2)(c - 1)^2. \] So $Ax = b$ has exactly one solution for every $b$ exactly when $c \neq 1$ and $c \neq -2$.

2. Find all values of $h$ for which the vectors \[ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\qquad \begin{bmatrix} 2 \\ h \\ 1 \end{bmatrix},\qquad \begin{bmatrix} h \\ 1 \\ 0 \end{bmatrix} \] are linearly dependent.

   Solution

   The vectors are linearly dependent exactly when the matrix with these columns has determinant $0$. Expanding along the first row, \[ \det \begin{bmatrix} 1 & 2 & h \\ 0 & h & 1 \\ 1 & 1 & 0 \end{bmatrix} = 1(0 - 1) - 2(0 - 1) + h(0 - h) = 1 - h^2. \] So the vectors are linearly dependent exactly when $h = 1$ or $h = -1$.

Let $A$ and $B$ be $4 \times 4$ matrices with $\det(A) = -2$ and $\det(B) = 3$. Find each of the following.

1. $\det(AB^2)$

   Solution

   $\det(AB^2) = \det(A)\det(B)^2 = -2 \cdot 9 = -18$.

2. $\det(A^{-1} B)$

   Solution

   $\det(A^{-1} B) = \dfrac{\det(B)}{\det(A)} = -\dfrac{3}{2}$.

3. $\det(3A)$

   Solution

   $3A$ is $A$ with each of its four rows multiplied by $3$, so $\det(3A) = 3^4 \det(A) = 81 \cdot (-2) = -162$.

4. $\det(B^{-1} A B)$

   Solution

   $\det(B^{-1} A B) = \dfrac{1}{\det(B)} \det(A) \det(B) = \det(A) = -2$.

For each statement below, decide whether it is true or false. If it is true, explain why. If it is false, give an example for which it fails.

1. If $A$ is a $3 \times 3$ matrix and $T_A$ is injective, then $T_A$ is surjective.

   Solution

   True. If $T_A$ is injective, every column has a pivot, so $A$ has 3 pivots. Since $A$ has 3 rows, every row has a pivot too, so $T_A$ is surjective.

2. If $v_1, v_2, v_3$ are linearly dependent, then $v_3$ is in $\sp\{v_1, v_2\}$.

   Solution

   False. Take $v_1 = e_1$, $v_2 = 2e_1$, and $v_3 = e_2$ in $\RR^2$. They are dependent, since $2v_1 - v_2 = 0$, but $\sp\{v_1, v_2\}$ is the $x$-axis, which does not contain $v_3$.

3. If $A$ and $B$ are $2 \times 2$ matrices with $AB = 0$, then $A = 0$ or $B = 0$.

   Solution

   False. Take \[ A = B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. \] Then $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, but neither matrix is zero.

4. If the columns of a $4 \times 4$ matrix $A$ span $\RR^4$, then $Ax = b$ has exactly one solution for every $b$ in $\RR^4$.

   Solution

   True. If the columns span $\RR^4$, then $A$ has a pivot in every row, so it has 4 pivots. Since $A$ is square, it then has a pivot in every column as well, so $A$ is invertible and $x = A^{-1} b$ is the only solution.

Let $P$ be the plane in $\RR^3$ with equation $x - y + 2z = 0$. The vectors \[ v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\qquad v_2 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \] lie in $P$, and the vector $w = \begin{bmatrix*}[r] 1 \\ -1 \\ 2 \end{bmatrix*}$ is perpendicular to $P$. Then $\cB = \{v_1, v_2, w\}$ is a basis of $\RR^3$. Let \[ u = \begin{bmatrix*}[r] 3 \\ -1 \\ 1 \end{bmatrix*}. \]

1. Find $[u]_\cB$.

   Solution

   We row reduce $[\,v_1\ v_2\ w \mid u\,]$.

   \[ \begin{aligned} &\left[\begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 1 & 2 & \llap{$-$}1 & \llap{$-$}1 \\ 0 & 1 & 2 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 2 & \llap{$-$}2 & \llap{$-$}4 \\ 0 & 1 & 2 & 1 \end{array}\right] \xrightarrow{R_{2} \leftrightarrow R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 2 & \llap{$-$}2 & \llap{$-$}4 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - 2\,R_{2}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & \llap{$-$}6 & \llap{$-$}6 \end{array}\right] \xrightarrow{R_{3} \to -\frac{1}{6}\,R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} - R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & \llap{$-$}1 \\ 0 & 0 & 1 & 1 \end{array}\right] \end{aligned} \] So $u = 2v_1 - v_2 + w$, and \[ [u]_\cB = \begin{bmatrix*}[r] 2 \\ -1 \\ 1 \end{bmatrix*}. \]

2. Let $A$ be the matrix such that $T_A$ projects $\RR^3$ onto $P$, so $T_A(v) = v$ for every $v$ in $P$ and $T_A(w) = 0$. Find $T_A(u)$.

   Solution

   \[ T_A(u) = 2T_A(v_1) - T_A(v_2) + T_A(w) = 2v_1 - v_2 = \begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix*}[r] 2 \\ 0 \\ -1 \end{bmatrix*}. \]

3. Find $\det(A)$, where $A$ is the matrix from part (b).

   Solution

   $Aw = 0$ with $w \neq 0$, so $A$ is not invertible and $\det(A) = 0$.

4. Let $R$ be the matrix such that $T_R$ reflects $\RR^3$ across $P$, so $T_R(v) = v$ for every $v$ in $P$ and $T_R(w) = -w$. Find $T_R(u)$.

   Solution

   \[ T_R(u) = 2v_1 - v_2 - w = \begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} - \begin{bmatrix*}[r] 1 \\ -1 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 1 \\ -3 \end{bmatrix*}. \]

5. Find $\det(R)$, where $R$ is the matrix from part (d).

   Solution

   Let $M = \begin{bmatrix} v_1 & v_2 & w \end{bmatrix}$ and $D = \begin{bmatrix*}[r] 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix*}$. The columns of $RM$ are $Rv_1 = v_1$, $Rv_2 = v_2$, and $Rw = -w$, which are the columns of $MD$, so $RM = MD$. Taking determinants, $\det(R)\det(M) = \det(M)\det(D)$. Since $\cB$ is a basis, $M$ is invertible and $\det(M) \neq 0$, so $\det(R) = \det(D) = -1$.