The Timpson Lyceum

1. (2 pts) Compute $A\mathbf{x}$ for $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \\ 0 & 4 \end{bmatrix}$ and $\mathbf{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ , writing $A\mathbf{x}$ as a linear combination of the columns of $A$ .

Solution

1. $A\mathbf{x} = 2\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix} + 1\begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 - 2 \\ 6 + 0 \\ 0 + 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ 4 \end{bmatrix}.$

2. (3 pts) Solve the matrix equation $A\mathbf{x} = \mathbf{b}$ for $A = \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$ . Also write the equation in the form of a vector equation.

Solution

2. The system $A\mathbf{x} = \mathbf{b}$ is $\begin{aligned} x_1 - 2x_2 &= 1 \\ 2x_1 - 3x_2 &= 4. \end{aligned}$ Applying $R_2 - 2R_1$ : $x_2 = 2$ , then $x_1 = 1 + 2(2) = 5$ . So $\mathbf{x} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$ . As a vector equation:

$x_1\begin{bmatrix} 1 \\ 2 \end{bmatrix} + x_2\begin{bmatrix} -2 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}.$

3. (3 pts) Do the columns of $A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 5 & 1 \\ 1 & 4 & -4 \end{bmatrix}$ span $\mathbb{R}^3$ ? Justify using pivot positions.

Solution

3. Row reduce $A$ :

$\begin{bmatrix} 1 & 3 & -1 \\ 2 & 5 & 1 \\ 1 & 4 & -4 \end{bmatrix} \xrightarrow[R_3 - R_1]{R_2 - 2R_1} \begin{bmatrix} 1 & 3 & -1 \\ 0 & -1 & 3 \\ 0 & 1 & -3 \end{bmatrix} \xrightarrow{R_3 + R_2} \begin{bmatrix} 1 & 3 & -1 \\ 0 & -1 & 3 \\ 0 & 0 & 0 \end{bmatrix}.$

There are only two pivots (rows 1 and 2); row 3 has no pivot. Since $A$ does not have a pivot in every row, its columns do not span $\mathbb{R}^3$ (they span a plane).

4. (3 pts) Does $A\mathbf{x} = \mathbf{b}$ have a solution for every $\mathbf{b}$ in $\mathbb{R}^3$ , where $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 3 \end{bmatrix}$ ? Justify using pivot positions.

Solution

4. Row reduce $A$ :

$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 3 \end{bmatrix} \xrightarrow{R_3 - R_1} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{bmatrix} \xrightarrow{R_3 - R_2} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}.$

There is a pivot in each of the three rows, so by the equivalent conditions of Theorem 4 the columns of $A$ span $\mathbb{R}^3$ and $A\mathbf{x} = \mathbf{b}$ has a solution for every $\mathbf{b} \in \mathbb{R}^3$ .

5. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) If $A$ is $m \times n$ and $A\mathbf{x} = \mathbf{b}$ is consistent for every $\mathbf{b}$ in $\mathbb{R}^m$ , then $A$ has a pivot position in every row.

(b) A $3 \times 4$ matrix can have columns that span $\mathbb{R}^3$ .

(c) The product $A\mathbf{x}$ is defined only when the number of columns of $A$ equals the number of entries in $\mathbf{x}$ .

Solution

5.

(a) True. This is one of the equivalent statements of Theorem 4: $A\mathbf{x}=\mathbf{b}$ is consistent for all $\mathbf{b}$ iff the columns span $\mathbb{R}^m$ iff $A$ has a pivot in every row.

(b) True. A $3 \times 4$ matrix can have three pivots (one per row), e.g. $A$ in Problem 4 padded with a fourth column; three pivots means the columns span $\mathbb{R}^3$ .

(c) True. $A\mathbf{x}$ is defined as $x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ , which requires exactly as many entries in $\mathbf{x}$ as columns of $A$ .

6. (3 pts) Suppose $A$ is a matrix with $A\mathbf{u} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $A\mathbf{v} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ . Without knowing $A$ , compute $A(2\mathbf{u} - 3\mathbf{v})$ , and state which properties of the matrix–vector product you used.

Solution

6. Using $A(c\mathbf{u} + d\mathbf{v}) = cA\mathbf{u} + dA\mathbf{v}$ (linearity of the matrix–vector product):

$A(2\mathbf{u} - 3\mathbf{v}) = 2A\mathbf{u} - 3A\mathbf{v} = 2\begin{bmatrix} 1 \\ 2 \end{bmatrix} - 3\begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} - \begin{bmatrix} 9 \\ -3 \end{bmatrix} = \begin{bmatrix} -7 \\ 7 \end{bmatrix}.$

The properties used are $A(\mathbf{u}+\mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ and $A(c\mathbf{u}) = c(A\mathbf{u})$ .

7. (3 pts) Let $A$ be an $m \times n$ matrix, $\mathbf{u}, \mathbf{v}$ vectors in $\mathbb{R}^n$ , and $c$ a scalar. Prove that $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ and $A(c\mathbf{u}) = c(A\mathbf{u})$ , using the definition of $A\mathbf{x}$ as a linear combination of the columns of $A$ .

Solution

7. Let the columns of $A$ be $\mathbf{a}_1, \dots, \mathbf{a}_n$ , and write $\mathbf{u} = (u_1, \dots, u_n)$ , $\mathbf{v} = (v_1, \dots, v_n)$ . By definition $A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ . Then

$A(\mathbf{u}+\mathbf{v}) = \sum_{j=1}^n (u_j + v_j)\mathbf{a}_j = \sum_{j=1}^n u_j\mathbf{a}_j + \sum_{j=1}^n v_j\mathbf{a}_j = A\mathbf{u} + A\mathbf{v},$

using distributivity of scalar multiplication over vector addition. Similarly

$A(c\mathbf{u}) = \sum_{j=1}^n (cu_j)\mathbf{a}_j = c\sum_{j=1}^n u_j\mathbf{a}_j = c(A\mathbf{u}).$

$\blacksquare$

8. (4 pts ••) Prove that the equation $A\mathbf{x} = \mathbf{b}$ has a solution if and only if $\mathbf{b}$ is a linear combination of the columns of $A$ .

Solution

8. Let $\mathbf{a}_1, \dots, \mathbf{a}_n$ be the columns of $A$ . By definition, for any $\mathbf{x} = (x_1, \dots, x_n)$ ,

$A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n.$

( $\Rightarrow$ ) If $A\mathbf{x} = \mathbf{b}$ has a solution $\mathbf{x}$ , then $\mathbf{b} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ is a linear combination of the columns.

( $\Leftarrow$ ) Conversely, if $\mathbf{b} = c_1\mathbf{a}_1 + \cdots + c_n\mathbf{a}_n$ for some scalars $c_j$ , then $\mathbf{x} = (c_1, \dots, c_n)$ satisfies $A\mathbf{x} = \mathbf{b}$ , so the equation is solvable.

Hence $A\mathbf{x} = \mathbf{b}$ has a solution if and only if $\mathbf{b}$ is a linear combination of the columns of $A$ . $\blacksquare$