The Timpson Lyceum · Linear Algebra (2026)

1.4 Exercises

Problem sheet · 20 points · 4 extra credit

The Matrix Equation Ax = b

1. 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 $A$.

2. 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.

3. 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.

4. 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.

5. 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 $\mathbf{x}$.

6. 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.

7. 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 $A$.

8. 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 $A$.