$B=\begin{bmatrix} -1&2&1\\ -2& 2& 3 \end{bmatrix}$

$C=\begin{bmatrix} 2&5\\ 1&2 \end{bmatrix}$

Solve the matrix equation $CX=B$. Can someone show me how to solve this please..


Hint: Assuming that you know how to solve a system of the form $A\cdot x = b$, where $A$ is a square matrix and $x,b$ are column vectors, then it is sufficient to solve the $3$ systems: $$C\cdot x_1 = \begin{bmatrix} -1 \\-2 \end{bmatrix}\\[2ex] C\cdot x_2 = \begin{bmatrix} 2 \\2 \end{bmatrix}\\[2ex] C\cdot x_3 = \begin{bmatrix} 1 \\3 \end{bmatrix} $$ The solution $\mathbf{x}$, which is a $2\times 3$ matrix will be the horizontal concatenation of the $2\times 1$ column vectors $x_1,x_2,x_3$, i.e. $\mathbf{x} = \begin{bmatrix} x_1 & x_2 & x_3\end{bmatrix}.$

You may check that $C\cdot \mathbf{x} = B$.


As C is non singular ($|C|=-1$) you can do the following:

$$CX=B \implies X =C^{-1}B$$

  • 1
    $\begingroup$ Both answers are good, but I think this is the more "canonical" way to do it. As an analogy consider $ax = b$, where you solve by dividing both sides by $a$. Same thing happens here, only that computing a matrix inverse is somewhat more complicated (not too complicated for a 2x2 matrix, though). $\endgroup$ – Fryie Oct 20 '15 at 13:24

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