Q.If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$, then verify that $A\,\operatorname{adj} A = |A|\,I$. Also find $A^{-1}$.

This checks the identity $A\,(\operatorname{adj}A)=|A|\,I$ and uses it to invert $A$.

Determinant. Expanding along column 1,

Adjoint. The cofactor matrix is $\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}$, so its transpose is

Verify. $A\,(\operatorname{adj}A)=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=1\cdot I=|A|\,I.$ ✓

Inverse. Since $|A|=1$, $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A=\operatorname{adj}A$.

Why the identity holds

For any square matrix, $A\,(\operatorname{adj}A)=|A|\,I$. Each diagonal entry of the product is the expansion of $|A|$ along a row, while each off-diagonal entry is the expansion of a determinant with two equal rows, which is $0$. When $|A|\neq0$ this gives $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$.

Step 1 — Determinant

Expanding along column 1,

Step 2 — Cofactors

So the cofactor matrix is $\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}$.

Step 3 — Adjoint (transpose the cofactors)

Step 4 — Verify $A\,(\operatorname{adj}A)=|A|\,I$

Multiplying row by column, for example row 1: $1(7)+3(-1)+3(-1)=1$, $\ 1(-3)+3(1)+3(0)=0$, $\ 1(-3)+3(0)+3(1)=0$. Carrying this through all rows,

The identity is verified.

Step 5 — Inverse

Since $|A|=1\neq0$,

Method: Verifying $A\cdot\operatorname{adj}(A)=|A|\,I$ and Extracting the Inverse

This method both proves the central adjoint identity for a specific matrix and uses it to find the matrix's inverse — the standard "adjoint method" for a $3\times3$ (or larger) matrix.

Steps

Step 1: Compute the determinant $|A|$

Expand along whichever row or column is most convenient. If $|A|=0$, stop here — the matrix has no inverse and $A\cdot\operatorname{adj}(A)$ will equal the zero matrix instead.

Step 2: Compute every cofactor $C_{ij}$

For each of the nine positions, delete the row and column, evaluate the $2\times2$ minor, and attach the checkerboard sign.

Step 3: Transpose the cofactor matrix to get $\operatorname{adj}(A)$

This transpose step is easy to forget — double check that the off-diagonal cofactors have been swapped, not left in place.

Step 4: Multiply $A\cdot\operatorname{adj}(A)$ and confirm it equals $|A|\,I$

Carry out the full $3\times3$ matrix multiplication. Every diagonal entry of the product should come out equal to $|A|$, and every off-diagonal entry should come out exactly $0$ — that's the identity being verified.

Step 5: Extract the inverse

Once verified,

If $|A|=1$, the inverse is simply the adjoint itself, with no further scaling needed.

This method is the general-purpose route to inverting any $3\times3$ matrix with a nonzero determinant, and doubles as the standard "prove the identity" exam question when the verification itself is asked for.

Common Mistakes

Mistake 1: A sign error in one of the nine cofactors, going undetected until the verification fails

Why it's wrong: with nine separate $2\times2$ minors and their signs to track, a single slip throws off both the adjoint and the final inverse. Correct approach: use the identity $A\cdot\operatorname{adj}(A)=|A|\,I$ itself as a check — if the off-diagonal entries of the product aren't exactly zero, a cofactor was computed incorrectly and needs to be re-derived.

Mistake 2: Forgetting to divide by $|A|$ when forming $A^{-1}$ (or not realizing division is still a required step when $|A|=1$)

Why it's wrong: the formula $A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)$ always needs that division step written explicitly — skipping it because $|A|$ happens to equal $1$ here can build a bad habit that produces wrong answers whenever $|A|\neq1$ in a later problem. Correct approach: always write the division step explicitly, even when it doesn't change any numbers.