Can't solve matrix equation.

Question

$$X\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 1 & -1 \\ \end{pmatrix}$$

I know a way to solve with inverse of matrix. But in this case it doesn't finds. Pls, help me.

Answer 1

Worst case scenario if you get so horribly stuck you can use dummy variables and solve from there:

Let $X = \begin{bmatrix}a&b\\c&d\end{bmatrix}$

Then if $X\cdot \begin{bmatrix}1&1\\1&1\end{bmatrix} = \begin{bmatrix}1&1\\-1&-1\end{bmatrix}$ we have

$\begin{bmatrix}a+b&a+b\\c+d&c+d\end{bmatrix} = \begin{bmatrix}1&1\\-1&-1\end{bmatrix}$

In other words, by making a system of equations setting each entry of the matrix equal between the left matrix and the right we have:

$$\begin{cases}a+b=1\\c+d=-1\end{cases}$$

So, all matrices $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ such that $a+b=1$ and $c+d=-1$ will satisfy your equation. For example $\begin{bmatrix}0.5&0.5\\-0.5&-0.5\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}500&-499\\\pi&-\pi-1\end{bmatrix}$ and many more. These will form a two dimension subspace of $M_{2,2}(\Bbb R)$

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For the new problem after it was made sideways:

We would have $\begin{bmatrix}a+b&a+b\\c+d&c+d\end{bmatrix}=\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$ which would have the system of equations:

$$\begin{cases}a+b=1\\a+b=-1\\c+d=1\\c+d=-1\end{cases}$$

Obviously, this system cannot be satisfied since $a+b$ cannot simultaneously equal $1$ and $-1$.