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I've solved the following problem by calculating matrix A's entries as variables and then evaluating them using the system of linear equations. As I'm a bit new to matrices, I wanted to know whether there are shorter ways to solving this problem cause solving it from the way I did seems so long...

If $$ \begin{bmatrix} 2 & 1 \\ \end{bmatrix} . A =\begin{bmatrix} 3 & 5 \\ \end{bmatrix} $$

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

what's the value of $$ \begin{bmatrix} 8 & 9 \\ \end{bmatrix} . A $$

$A$ is a matrix

JOUA
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2 Answers2

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You can use that matrix multiplication is linear: $(av+bw)A=a(vA)+b(wA)$, where $v,w$ are vectors and $a,b$ are numbers. With $[8~9]=[2~1]+2\cdot[3~4]$, you can find $[8~9]A$ quickly.

Vercassivelaunos
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You can notice that

$$ \begin{bmatrix} 8 & 9 \\ \end{bmatrix} = 2\begin{bmatrix} 3 & 4 \\ \end{bmatrix}+\begin{bmatrix} 2 & 1 \\ \end{bmatrix}$$

Hence $$ \begin{bmatrix} 8 & 9 \\ \end{bmatrix} A= 2\begin{bmatrix} 3 & 4 \\ \end{bmatrix} A+\begin{bmatrix} 2 & 1 \\ \end{bmatrix} A = 2\begin{bmatrix} -1 & 2 \\ \end{bmatrix}+\begin{bmatrix} 3 & 5 \\ \end{bmatrix} = \begin{bmatrix} 1 & 9 \\ \end{bmatrix}$$ by linearity.

mathcounterexamples.net
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