Revision b4210100-f67a-4ff4-b76b-cb34c9cae261 - Mathematics Stack Exchange

No, the question is to find $a,b,c,d$ so that $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\0円\end{pmatrix} = \begin{pmatrix}2\6円\end{pmatrix} $$

and

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\1円\end{pmatrix} = \begin{pmatrix}4\8円\end{pmatrix} $$

You can find this by performing the multiplication explicitly in the first equation, obtaining two equations in $a,b,c,d$ (one from the upper component of the result, one from the lower) and then similarly in the second equation. Then you have four equations in $a,b,c,d$ which you can solve to find the matrix you seek.

When you _do_ find that matrix, you will look at it and say “Oh, is _that_ all?”