Need help understanding solution to a matrix multiplication I'm given the following matrix equation:
\begin{align}
\begin{pmatrix}
2&4\\4&8
\end{pmatrix}X=
\begin{pmatrix}
0&0\\0&0
\end{pmatrix}
\end{align}
The solution is given as follows - let $X$ be of the form $$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ Then, the matrix equation becomes the following system:
\begin{align}
\begin{cases}
2a+4c=0\\
2b+4d=0\\
4a+8c=0\\
4b+8d=0
\end{cases}
\end{align}
So, the matrix is of the form $$X=\begin{pmatrix}-2c&-2d\\c&d\end{pmatrix}$$
I don't follow the above explanation, can someone help?
 A: You need to take into account several things:


*

*First of all, if you multiply an $n\times m$ matrix by a $m\times p$ matrix, you obtain a $n\times p$ matrix. In your case, $n=m=2$ due to the matrix on the LHS and $n=p=2$ because of the RHS one. Therefore, $X$ has $m=2$ rows and $p=2$ columns; that is, it is of the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

*I assume you already know how matrix multiplication works. If you compute the product on the left, you obtain
$$\begin{pmatrix}2a+4c&2b+4d\\4a+8c&4b+8d\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$$
Now, since two matrices are equal if and only if each of their elements are equal, you can deduce that $2a+4c=0$ and so on. The system you have, precisely.

*Finally, you just need to solve the system. Note that the third and fourth equations are the same as the first and the second one, respectively (divide them by 2), so just focus on the latters:
$$2a+4c=0 \Rightarrow a=-2c$$
$$2b+4d=0 \Rightarrow b=-2d$$
Substitute that in your initial guess for $X$ and you finally get
$$X=\begin{pmatrix}-2c&-2d\\c&d\end{pmatrix}$$
Note that any value of $c$ and $d$ satisfies the initial matrix equation, so that is the final answer indeed.
