Q.Solve the following equation by the method of inversion.
$2x - y + z = 1$,
$x + 2y + 3z = 8$,
$3x + y - 4z = 1$

With $A=\begin{bmatrix}2&-1&1\\1&2&3\\3&1&-4\end{bmatrix}$, $B=\begin{bmatrix}1\\8\\1\end{bmatrix}$: $|A|=-40$, $\text{adj}(A)=\begin{bmatrix}-11&-3&-5\\13&-11&-5\\-5&-5&5\end{bmatrix}$, and $X=A^{-1}B$ gives $x=1,\ y=2,\ z=1$.

Step 1 — matrix form $AX=B$:

$A=\begin{bmatrix}2&-1&1\\1&2&3\\3&1&-4\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}1\\8\\1\end{bmatrix}.$

Step 2 — determinant.

$|A|=2(2\cdot(-4)-3\cdot1)-(-1)(1\cdot(-4)-3\cdot3)+1(1\cdot1-2\cdot3)$

$=2(-11)+1(-13)+1(-5)=-22-13-5=-40\ ( eq0),$

so $A^{-1}$ exists.

Step 3 — cofactors and adjoint. The cofactor matrix is

$\begin{bmatrix}-11&13&-5\\-3&-11&-5\\-5&-5&5\end{bmatrix},\qquad \text{adj}(A)=\begin{bmatrix}-11&-3&-5\\13&-11&-5\\-5&-5&5\end{bmatrix}\ (\text{its transpose}).$

Step 4 — inverse.

$A^{-1}=\frac{1}{|A|}\,\text{adj}(A)=-\frac{1}{40}\begin{bmatrix}-11&-3&-5\\13&-11&-5\\-5&-5&5\end{bmatrix}.$

…