Q.Solve using the matrix inversion method: $2x+3y=8$, $x-2y=-3$.

Writing the system as $AX=B$, the solution is found by computing $A^{-1}$ once and then multiplying it by the constants column $B$. …

Write the system as $AX=B$ with $A=\begin{pmatrix}2&3\\1&-2\end{pmatrix}$, $X=\begin{pmatrix}x\\y\end{pmatrix}$, $B=\begin{pmatrix}8\\-3\end{pmatrix}$.

Determinant: $|A|=(2)(-2)-(3)(1)=-4-3=-7$. Since $|A|\ne0$, $A$ is non-singular and $A^{-1}$ exists.

Adjoint (swap-and-negate shortcut): $\operatorname{adj}(A)=\begin{pmatrix}-2&-3\\-1&2\end{pmatrix}$

Inverse: $A^{-1}=\dfrac{1}{-7}\begin{pmatrix}-2&-3\\-1&2\end{pmatrix}=\begin{pmatrix}2/7&3/7\\1/7&-2/7\end{pmatrix}$

Solve $X=A^{-1}B$:

$x=\tfrac{2}{7}(8)+\tfrac{3}{7}(-3)=\tfrac{16}{7}-\tfrac{9}{7}=\tfrac{7}{7}=1$ …

Alternative cross-check using Cramer's Rule: $D=-7$, $D_x=\begin{vmatrix}8&3\\-3&-2\end{vmatrix}=-16+9=-7$ so $x=-7/-7=1$; $D_y=\begin{vmatrix}2&8\\1&-3\end{vmatrix}=-6-8=-14$ so $y=-14/-7= …

Multiplying $B$ by $A^{-1}$ in the wrong order (attempting $BA^{-1}$ instead of $A^{-1}B$ — matrix multiplication order matters, and $X=A^{-1}B$ is the only order that makes sense dimensionally an …