Q.$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$ Verify $A (\text{adj } A) = (\text{adj } A) A = |A| I$ in Exercises 3 and 4

Concept: Adjoint Matrix Property — For any square matrix $A$, $A(\text{adj } A) = (\text{adj } A)A = |A|I$.

Step 1: Compute $|A|$

Expanding along $R_1$:

$|A| = 1(3\cdot1 - 5\cdot0) - (-1)(2\cdot1 - 5\cdot(-2)) + 2(2\cdot0 - 3\cdot(-2))$

$= 1(3) + 1(2 + 10) + 2(0 + 6) = 3 + 12 + 12 = 27$.

Step 2: Find $\text{adj } A$

Cofactors:

$C_{11} = 3$, $C_{12} = -12$, $C_{13} = 6$

$C_{21} = 1$, $C_{22} = 5$, $C_{23} = 2$

$C_{31} = -11$, $C_{32} = -1$, $C_{33} = 5$

So $\text{adj } A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$.

Step 3: Verify $A(\text{adj } A)$

$A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$

$= \begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix} = 27I$.

Similarly, $(\text{adj } A)A$ gives the same result.

The property $A(\text{adj }A) = (\text{adj }A)A = |A|I$ is one of the most elegant results in matrix algebra. It tells us that the adjoint (or adjugate) of a matrix is essentially a "scaled inverse" — when $A$ is invertible, dividing the adjoint by the determinant gives the inverse. But the identity holds for any square matrix, invertible or not.

Why does this work? Each entry of $\text{adj }A$ is a cofactor (signed minor) of $A$. When you multiply $A$ by $\text{adj }A$, the $(i,j)$ entry becomes the sum of products of row $i$ of $A$ with column $j$ of cofactors. For $i=j$, this sum is exactly the Laplace expansion of $|A|$ along row $i$. For $i\neq j$, it's like expanding a matrix with two identical rows — which gives zero. So the product is diagonal, with $|A|$ on every diagonal entry.

Let's verify this concretely.

Step 1: Compute the determinant $|A|$.

We have

Expand along the third row (it has a zero, which saves work):

The first term: $(-2) \cdot (+1) \cdot [(-1)(5) - (2)(3)] = (-2)[-5 - 6] = (-2)(-11) = 22$.

The third term: $1 \cdot (+1) \cdot [(1)(3) - (-1)(2)] = 3 + 2 = 5$.

So $|A| = 22 + 5 = 27$.

Step 2: Find all cofactors $C_{ij}$.

We need nine cofactors. Let's compute them systematically.

• $C_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3\cdot 1 - 5\cdot 0) = 3$.
• $C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -[2\cdot 1 - 5\cdot(-2)] = -[2 + 10] = -12$.
• $C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = (2\cdot 0 - 3\cdot(-2)) = 0 + 6 = 6$.
• $C_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -[(-1)\cdot 1 - 2\cdot 0] = -[-1 - 0] = 1$.
• $C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = (1\cdot 1 - 2\cdot(-2)) = 1 + 4 = 5$.
• $C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -[1\cdot 0 - (-1)\cdot(-2)] = -[0 - 2] = 2$.
• $C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = [(-1)\cdot 5 - 2\cdot 3] = -5 - 6 = -11$.
• $C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -[1\cdot 5 - 2\cdot 2] = -[5 - 4] = -1$.
• $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1\cdot 3 - (-1)\cdot 2) = 3 + 2 = 5$.

Step 3: Write the adjoint matrix.

The adjoint is the transpose of the cofactor matrix:

Step 4: Compute $A(\text{adj }A)$.

Multiply $A$ (on the left) by $\text{adj }A$ (on the right). Let's do it entry by entry.

Row 1 of $A$: $[1, -1, 2]$.

• $(1,1)$ entry: $1\cdot 3 + (-1)\cdot(-12) + 2\cdot 6 = 3 + 12 + 12 = 27$.
• $(1,2)$ entry: $1\cdot 1 + (-1)\cdot 5 + 2\cdot 2 = 1 - 5 + 4 = 0$.
• $(1,3)$ entry: $1\cdot(-11) + (-1)\cdot(-1) + 2\cdot 5 = -11 + 1 + 10 = 0$.

Row 2 of $A$: $[2, 3, 5]$.

• $(2,1)$ entry: $2\cdot 3 + 3\cdot(-12) + 5\cdot 6 = 6 - 36 + 30 = 0$.
• $(2,2)$ entry: $2\cdot 1 + 3\cdot 5 + 5\cdot 2 = 2 + 15 + 10 = 27$.
• $(2,3)$ entry: $2\cdot(-11) + 3\cdot(-1) + 5\cdot 5 = -22 - 3 + 25 = 0$.

Row 3 of $A$: $[-2, 0, 1]$.

• $(3,1)$ entry: $(-2)\cdot 3 + 0\cdot(-12) + 1\cdot 6 = -6 + 0 + 6 = 0$.
• $(3,2)$ entry: $(-2)\cdot 1 + 0\cdot 5 + 1\cdot 2 = -2 + 0 + 2 = 0$.
• $(3,3)$ entry: $(-2)\cdot(-11) + 0\cdot(-1) + 1\cdot 5 = 22 + 0 + 5 = 27$.

So

Step 5: Compute $(\text{adj }A)A$.

Now multiply $\text{adj }A$ (on the left) by $A$ (on the right).

Row 1 of $\text{adj }A$: $[3, 1, -11]$.

• $(1,1)$ entry: $3\cdot 1 + 1\cdot 2 + (-11)\cdot(-2) = 3 + 2 + 22 = 27$.
• $(1,2)$ entry: $3\cdot(-1) + 1\cdot 3 + (-11)\cdot 0 = -3 + 3 + 0 = 0$.
• $(1,3)$ entry: $3\cdot 2 + 1\cdot 5 + (-11)\cdot 1 = 6 + 5 - 11 = 0$.

Row 2 of $\text{adj }A$: $[-12, 5, -1]$.

• $(2,1)$ entry: $(-12)\cdot 1 + 5\cdot 2 + (-1)\cdot(-2) = -12 + 10 + 2 = 0$.
• $(2,2)$ entry: $(-12)\cdot(-1) + 5\cdot 3 + (-1)\cdot 0 = 12 + 15 + 0 = 27$.
• $(2,3)$ entry: $(-12)\cdot 2 + 5\cdot 5 + (-1)\cdot 1 = -24 + 25 - 1 = 0$.

Row 3 of $\text{adj }A$: $[6, 2, 5]$.

• $(3,1)$ entry: $6\cdot 1 + 2\cdot 2 + 5\cdot(-2) = 6 + 4 - 10 = 0$.
• $(3,2)$ entry: $6\cdot(-1) + 2\cdot 3 + 5\cdot 0 = -6 + 6 + 0 = 0$.
• $(3,3)$ entry: $6\cdot 2 + 2\cdot 5 + 5\cdot 1 = 12 + 10 + 5 = 27$.

Thus

Both products give the same diagonal matrix, confirming the identity.

Method: Verifying $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\,I$

The standard procedure for any "verify the adjoint identity" question.

Steps

Step 1: Compute $|A|$

Expand along whichever row or column has the most zeros (or the first row if none do).

Step 2: Compute all nine cofactors $C_{ij}=(-1)^{i+j}M_{ij}$

Work systematically through every entry, deleting its row and column to form the $2\times2$ minor, then applying the correct sign.

Step 3: Assemble the adjoint as the TRANSPOSE of the cofactor matrix

Don't skip the transpose — the adjoint is not simply the cofactor matrix itself.

Step 4: Multiply $A\cdot\operatorname{adj}(A)$ entry by entry

Every diagonal entry of the product should come out equal to $|A|$ (it is the row-expansion of $|A|$ along that row); every off-diagonal entry should come out $0$ (it is the expansion of a determinant with a repeated row).

Step 5: Repeat for $(\operatorname{adj} A)\cdot A$ and confirm both equal $|A|\,I$

The product in the reverse order gives the identical diagonal matrix, confirming the identity.

Common Mistakes

Mistake 1: Forgetting the $(-1)^{i+j}$ sign when computing a cofactor

Why it's wrong: for example $C_{12}$'s minor evaluates to $12$, but the correct sign for position $(1,2)$ is negative (since $1+2=3$ is odd) — skipping the sign gives $C_{12}=12$ instead of the correct $-12$, which then corrupts every product using it. Correct approach: write out $(-1)^{i+j}$ explicitly for every one of the nine cofactors before substituting the minor.

Mistake 2: Writing the adjoint as the cofactor matrix itself, without transposing

Why it's wrong: the adjoint is defined as the transpose of the cofactor matrix — using the untransposed cofactor matrix directly gives $A(\operatorname{adj} A)$ a non-diagonal result instead of $|A|I$, since the off-diagonal entries won't cancel correctly. Correct approach: always swap rows and columns of the cofactor matrix (i.e. $(\operatorname{adj} A)_{ij}=C_{ji}$) before multiplying by $A$.