Matrix Operations - GRE Subject Test: Math

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Question

Perform the following operation.

$\begin{bmatrix} 2& 4& 1\ 5& 3& -2\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\ 4\ 1 \end{bmatrix}$ $+\begin{bmatrix} -7\ 4\ 11 \end{bmatrix}$

Answer

The first step to solving this operation is to do the multiplication:

$\begin{bmatrix} 2& 4& 1\ 5& 3& -2\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\ 4\ 1 \end{bmatrix}$ $=\begin{bmatrix} 2(2)& 4(4)& 1(1)\ 5(2)& 3(4)& -2(1)\ -3(2)& -1(4)&0(1) \end{bmatrix}$ $=\begin{bmatrix} 4+16+1\ 10+12-2\ -6-4+0 \end{bmatrix}$ $=\begin{bmatrix} 21\ 20\ -10 \end{bmatrix}$

Once we have multiplied the matrices, we can perform the addition portion:

$\begin{bmatrix} 21\ 20\ -10 \end{bmatrix} + \begin{bmatrix} -7\ 4\ 11 \end{bmatrix} = \begin{bmatrix} 14\ 24\ 1 \end{bmatrix}$

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Question

Perform the following operation.

$3\left(\begin{bmatrix} 7\ -3\ 2 \end{bmatrix}+\begin{bmatrix} -1\ 4\ 6 \end{bmatrix}\right)$

Answer

The first step is to solve whatever is in the parentheses, in this case it is addition:

$\begin{bmatrix} 7\ -3\ 2 \end{bmatrix}+\begin{bmatrix} -1\ 4\ 6 \end{bmatrix}=\begin{bmatrix} 6\ 1\ 8 \end{bmatrix}$

We then substitute our solution into the parentheses:

$3\left(\begin{bmatrix} 6\ 1\ 8 \end{bmatrix}\right)$

Our next, and final step in this problem, is to carry out the multiplication:

$3\left(\begin{bmatrix} 6\ 1\ 8 \end{bmatrix}\right)=\begin{bmatrix} 18\ 3\ 24 \end{bmatrix}$

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