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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.

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closed as off-topic by Saad, user99914, Xander Henderson, Parcly Taxel, Brian Borchers Mar 12 '18 at 2:52

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  • 3
    $\begingroup$ How would you solve the scalar equation $ax+b=cx$ for $x$? The same ideas apply here. $\endgroup$ – amd Mar 11 '18 at 15:58
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Your matrix equation is $$ AX+B=CX$$

You solve for X.

$$(A-C)X=-B$$

$$X=-(A-C)^{-1}B$$ $$X= \left[\begin{array}{cc} -6 &12\cr -11 &-2 \end{array}\right]^{-1}\left[\begin{array}{cc} 6 &6\cr 3 &5 \end{array}\right]$$

$$ X = \frac{1}{12}\left[\begin{array}{cc} -4 & -6\cr 4 &3 \end{array}\right]$$

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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

$$X = \frac{1}{12}\left[\begin{array}{cc} -4 & -6\cr 4 &3 \end{array}\right] $$

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  • $\begingroup$ it should be $$ X = \frac{1}{12}\left[\begin{array}{cc} -4 & -6\cr 4 &3 \end{array}\right]$$, thanks. $\endgroup$ – kdhug886 Mar 11 '18 at 9:53

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