Solve the following matrix equation for $X$.
$$\left[\begin{array}{cc} 5 &-8\cr 8 &1 \end{array}\right] X + \left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right] = \left[\begin{array}{cc} -1 &4\cr -3 &-1 \end{array}\right] X$$
Please give me some hint to do this question. Thanks.
On
This is a linear equation in four variables.
Let $X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}$.
Consider each entry of the RHS and LHS as a seperate equation:
$$\begin{align} 5x_{11} - 8x_{21} + 6 &= -1x_{11}+4x_{12} & \text{(top left entry)}\\ 5x_{12} - 8x_{22} + 6 &= -1x_{12}+4x_{22} & \text{(top right entry)}\\ 8x_{11} - 1x_{21} + 3 &= -3x_{11}-1x_{12} & \text{(bottom left entry)}\\ 8x_{12} - 1x_{22} + 5 &= -3x_{12}-1x_{22} & \text{(bottom right entry)}\\ \end{align}$$
Now you get a linear system of four equations wich can be solved using the usual methods.
As an alternative you can directly use matrix operations and solve
$$\left[\begin{array}{cc} 5 &-8\cr 8 &1 \end{array}\right] X + \left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right] = \left[\begin{array}{cc} -1 &4\cr -3 &-1 \end{array}\right] X$$
iff
$$\left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right] = \left[\begin{array}{cc} -1 &4\cr -3 &-1 \end{array}\right] X - \left[\begin{array}{cc} 5 &-8\cr 8 &1 \end{array}\right] X =\left[\begin{array}{cc} -6 &12\cr -11 &-2 \end{array}\right]X $$
iff
$$\left[\begin{array}{cc} -6 &12\cr -11 &-2 \end{array}\right]^{-1}\left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right] =X $$
(Note that the matrix that we are trying to invert is actually invertible since its determinant is $(-6)(-2)-(-11)12 = 144 \neq 0$.)
iff
$$X = \frac{1}{12}\left[\begin{array}{cc} -4 & -6\cr 4 &3 \end{array}\right] $$
Your matrix equation is $$ AX+B=CX$$
You solve for X.
$$(A-C)X=-B$$
$$X=-(A-C)^{-1}B$$ $$X= \left[\begin{array}{cc} -6 &12\cr -11 &-2 \end{array}\right]^{-1}\left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right]$$
$$ X = \frac{1}{12}\left[\begin{array}{cc} -4 & -6\cr 4 &3 \end{array}\right]$$