Let $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ and let $B = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}$ Solve $A X = B$ for a matrix $X$
My guess is that i:
let $X = \begin{bmatrix} x \\ y \end{bmatrix}$
Then solve it using a linear equation but i'm not too sure. Any help would be appreciated.
On
It's not possible to multiply $2\times 2$ matrix with $2\times 1$ matrix and give a $2\times 2$ matrix.
But if in your question $B$ is a $2\times 1$ matrix, you can solve this system of equation by elimination:
$$\begin{cases}2x+3y=a\\4x+5y=b\end{cases}$$
On
Let
$$X=\left[\begin{matrix} a &b\\ c &d\end{matrix}\right].$$
Expanding the matrix product, you need to solve
$$\begin{cases}2a+3c=3\\4a+5c=5\\2b+3d=4\\4b+5d=6\end{cases}$$
Hint: $$\begin{cases}2a+3c=3\\4a+5c=5\end{cases}$$ $$\begin{cases}2b+3d=4\\4b+5d=6\end{cases}$$
The real, efficient method is by solving the system (by Gaussian elimination) for all right-hand sides simultaneously.
$$\begin{cases}\begin{align}2x+3y=3|4\\4x+5y=5|6\end{align}\end{cases}$$
$$\begin{cases}2x+3y=\ \ \ 3|\ \ \ \ \ 4\\0x-1y=-1|-2\end{cases}$$
$$\begin{cases}2x+0y=\ \ \ 0|-2\\0x-1y=-1|-2\end{cases}$$
$$\begin{cases}x=0|-1\\y=1|\ \ \ \ \ 2\end{cases}$$
Hint: Think about inverting Matrix A and rewriting it as $X=A^{-1}B$. The inverse of a 2 by 2 matrix is simple to generate by hand. Your approach is the right one but you will have to introduce a 2 by 2 matrix for $X$ instead of a vector.