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We have matrices. (Same matrices as in post earlier). You can find earlier post from here. $$A=\begin{bmatrix} -2 & 0 \\ -5 & 6 \\ \end{bmatrix}\\ B^{-1}=\begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix}\\ C=\begin{bmatrix} -15 & -2 \\ -8 & -14 \\ \end{bmatrix} $$

I would like to solve $X$ from this equation $$CX=-6X+I$$ $$CX+6X=I$$ $$X(C+6)=I$$ Multiplying both sides with $(C+6)^{-1}$ $$XI=I(C+6)^{-1}$$ $$X=(C+6)^{-1}$$ Is this correct ? $$X=(\begin{bmatrix} -15 & -2 \\ -8 & -14 \\ \end{bmatrix}+6)^{-1}$$ $$X=(\begin{bmatrix} -9 & -2 \\ -2 & -8 \\ \end{bmatrix})^-1$$ $$X=\begin{bmatrix} -\frac{2}{17} & \frac{1}{34} \\ \frac{1}{34} & -\frac{9}{68} \end{bmatrix}$$

Arthur
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Tuki
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  • In general, one way of seeing if you are correct is plugging the matrix back into the equation to see if it works. – alphacapture Sep 30 '17 at 19:23
  • Matrix multiplication is not commutative. Do not interchange $CX$ and $XC$. – Donald Splutterwit Sep 30 '17 at 19:25
  • Yep i tried plugging in numbers and it seems this is not correct solution indeed. – Tuki Sep 30 '17 at 19:27
  • Just a silly mistake in adding $6$ $$\left( \begin{array}{ll} -9 & +4 \ -2 & -8 \ \end{array} \right)^{-1}=\left( \begin{array}{cc} -\frac{1}{10} & -\frac{1}{20} \ \frac{1}{40} & -\frac{9}{80} \ \end{array} \right)$$ Is wrong anyway. The actual result is $$\left( \begin{array}{cc} -\frac{1}{7} & \frac{1}{28} \ \frac{1}{7} & -\frac{9}{56} \ \end{array} \right)$$ – Raffaele Sep 30 '17 at 19:58
  • But is the equation correct ? $X=(C+6)^{-1}$ – Tuki Sep 30 '17 at 20:28
  • It appears that this is correct solution – Tuki Sep 30 '17 at 21:18

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