Jednadžba s kvadratnim matricama

Riješite matričnu jednadžbu

$\displaystyle (AX)^{-1}+X^{-1}=B,$

gdje je

$\displaystyle A=\begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \quad\textrm{ i }\quad B=\displaystyle \begin{bmatrix} 3 & 4 \\ 1 & -3 \end{bmatrix}.$

Rješenje. Prema formulama iz [M1, poglavlje 2.8] slijedi

$\displaystyle (AX)^{-1}+X^{-1}$	$\displaystyle =B,$
$\displaystyle X^{-1}A^{-1}+X^{-1}$	$\displaystyle =B,$
$\displaystyle X^{-1}(A^{-1}+I)$	$\displaystyle =B.$

Množenjem jednadžbe s lijeve strane matricom

i s desne strane matricom $B^{-1}$ dobivamo

$\displaystyle X\cdot X^{-1}(A^{-1}+I)\cdot B^{-1}$	$\displaystyle =X\cdot B\cdot B^{-1},$
$\displaystyle I\cdot(A^{-1}+I)\cdot B^{-1}$	$\displaystyle =X\cdot I,$
$\displaystyle (A^{-1}+I)\cdot B^{-1}$	$\displaystyle =X,$

odnosno

$\displaystyle X=(A^{-1}+I)\cdot B^{-1}.$

Matrice

imaju inverze jer je

$\displaystyle \det A=2\cdot 4-(-1)\cdot3=11\neq0 \quad\textrm{ i }\quad \det B=3\cdot(-3)-4\cdot 1=-13\neq0$

Prema zadatku 2.23 vrijedi

$\displaystyle A^{-1}=\frac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}\qu... ...\ -1 & 3\end{bmatrix}= \frac{1}{13}\begin{bmatrix}3 & 4 \\ 1 & -3\end{bmatrix}.$

Stoga je

$\displaystyle X$	$\displaystyle =\left(\frac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}+ \... ...nd{bmatrix}\right)\cdot \frac{1}{13}\begin{bmatrix}3 & 4 \\ 1 & -3\end{bmatrix}$
	$\displaystyle =\left(\frac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}+ \... ...nd{bmatrix}\right)\cdot \frac{1}{13}\begin{bmatrix}3 & 4 \\ 1 & -3\end{bmatrix}$
	$\displaystyle =\frac{1}{11\cdot 13}\begin{bmatrix}15 & 1 \\ -3 & 13\end{bmatrix... ...& -3\end{bmatrix}= \frac{1}{143}\begin{bmatrix}46 & 57 \\ 4 & -51\end{bmatrix}.$

Matrična jednadžba LINEARNA ALGEBRA Rješavanje matrične jednadžbe invertiranjem