Q.Solve the system $3x+2y=8$, $x+4y=6$ by the matrix inversion method.

Writing the system as $AX=B$ and using the already-known inverse of the coefficient matrix reduces solving the system to a single matrix multiplication.

Here $A=\begin{pmatrix}3&2\\1&4\end{pmatrix}$ (the same matrix as the earlier worked example, with $A^{-1}=\tfrac{1}{10}\begin{pmatrix}4&-2\\-1&3\end{pmatrix}$) and $B=\begin{pmatrix}8\\6\end{pmatrix}$. $X=A^{-1}B=\tfrac{1}{10}\begin{pmatrix}4(8)-2(6)\\-1(8)+3(6)\end{pmatrix}=\tfrac{1}{10}\begin{pmatrix}20\\10\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$.

Step 1 — Write the system as $AX=B$

Step 2 — Use $A^{-1}$

This is the same coefficient matrix as the earlier worked example on the inverse of a $2\times2$ matrix, where $|A|=10\neq0$ and

Step 3 — Compute $X=A^{-1}B$

So $x=2$, $y=1$.

Check (independent recomputation): substituting $x=2,y=1$ back into the ORIGINAL equations — $3(2)+2(1)=6+2=8$ ✓ and $2+4(1)=2+4=6$ ✓ — both equations are satisfied exactly, confirming the solution.

Reuse the inverse already found for the identical coefficient matrix in the earlier worked example on inverses, rather than recomputing $A^{-1}$ from scratch, as an efficiency check that both sessions of working agree.

Multiplying $A^{-1}$ and $B$ in the wrong order (computing $BA^{-1}$ instead of $A^{-1}B$) — for a column matrix $B$, only $A^{-1}B$ is even conformable, but the row-by-column pairing must still be done correctly.