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 perforingperforming 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?"
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 perforing 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?"
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?"
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 perforing 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?"