Q.Find the inverse of $A=\begin{pmatrix}3&2\\1&4\end{pmatrix}$ by the adjoint method, and verify that $AA^{-1}=I$.

For a $2\times2$ matrix, the adjoint is found by swapping the diagonal entries and negating the off-diagonal entries, after first confirming the determinant is non-zero.

Step 1 — Check non-singularity

Since $|A|\neq0$, $A$ is non-singular and an inverse exists.

Step 2 — Find the adjoint

For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\text{adj}(A)=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$:

Step 3 — Divide by the determinant

Step 4 — Verify $AA^{-1}=I$

$AA^{-1}=\frac{1}{10}\begin{pmatrix}3&2\\1&4\end{pmatrix}\begin{pmatrix}4&-2\\-1&3\end{pmatrix}=\frac{1}{10}\begin{pmatrix}3(4)+2(-1)&3(-2)+2(3)\\1(4)+4(-1)&1(-2)+4(3)\end{pmatrix}=\frac{1}{10}\begin{pmatrix}10&0\\0&10\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ …

Recognise the $2\times2$ shortcut directly: swap $a$ and $d$, negate $b$ and $c$, then divide the whole matrix by $|A|$ — no co …

Forgetting to divide by the determinant (leaving the answer as just the adjoint), or swapping which entries get negated (nega …