Linear Algebra · Question of the Day

Linear Algebra Question of the Day

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Tuesday, September 8, 2026

Let $\mathbf{u} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ , $\mathbf{v} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix}$ , and $\mathbf{w} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$ . If $A$ is a $2 \times 3$ matrix such that $A\mathbf{u} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$ and $A\mathbf{v} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ , what is $A(2\mathbf{u} - \mathbf{v})$ ?

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Question of the Day

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$\begin{pmatrix} 4 \\ 4 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 10 \\ 2 \end{pmatrix}$
$\begin{pmatrix} 16 \\ -2 \end{pmatrix}$
$\begin{pmatrix} -4 \\ 6 \end{pmatrix}$

Explanation: Using linearity of matrix multiplication: $A(2\mathbf{u} - \mathbf{v}) = 2A\mathbf{u} - A\mathbf{v} = 2\begin{pmatrix} 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 10 \\ 2 \end{pmatrix} - \begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$ . Choice B forgets to subtract $A\mathbf{v}$ . Choice C incorrectly adds instead of subtracting. Choice D uses incorrect signs throughout.