Find the inverse of a 3×3 matrix using adjoint method

Question

Find the inverse of the matrix using the adjoint method:

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}

(NCERT Class 12, Exercise 4.5)

Solution — Step by Step

Expand along the first row:

|A| = 1(0 - 24) - 2(0 - 20) + 3(0 - 5)

= 1(-24) - 2(-20) + 3(-5)

= -24 + 40 - 15 = 1

Since $|A| = 1 \neq 0$ , the inverse exists. A determinant of 1 is great news — it means $A^{-1} = \text{adj}(A)$ directly.

The cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ , where $M_{ij}$ is the minor.

$C_{11} = +(0 - 24) = -24$ , $C_{12} = -(0 - 20) = 20$ , $C_{13} = +(0 - 5) = -5$

$C_{21} = -(0 - 18) = 18$ , $C_{22} = +(0 - 15) = -15$ , $C_{23} = -(6 - 10) = 4$

$C_{31} = +(8 - 3) = 5$ , $C_{32} = -(4 - 0) = -4$ , $C_{33} = +(1 - 0) = 1$

Cofactor matrix:

C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}

\text{adj}(A) = C^T = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

Since $|A| = 1$ :

\boxed{A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}}

Quick verification: Multiply $A \cdot A^{-1}$ and check if you get the identity matrix $I$ . In exams, verify at least the $(1,1)$ entry: $1(-24) + 2(20) + 3(-5) = -24 + 40 - 15 = 1$ ✓. This takes 10 seconds and catches sign errors.

Why This Works

The adjoint method relies on the fact that $A \cdot \text{adj}(A) = |A| \cdot I$ . This identity comes from cofactor expansion — when you multiply $A$ by the transpose of its cofactor matrix, the diagonal entries give $|A|$ and the off-diagonal entries give zero (because they represent expansion of a determinant with two identical rows).

Dividing both sides by $|A|$ gives us $A^{-1} = \frac{\text{adj}(A)}{|A|}$ .

Alternative Method — Row Reduction

Augment $A$ with $I$ and row-reduce $[A \mid I]$ to $[I \mid A^{-1}]$ :

Write $[A \mid I]$ and apply elementary row operations. This method avoids computing 9 cofactors and is often faster for numerical matrices. However, CBSE marking schemes typically expect the adjoint method when the question specifically says “using adjoint.”

In CBSE board exams, the adjoint method question carries 5-6 marks. The marking scheme awards marks stepwise: 1 mark for $|A|$ , 2 marks for cofactors, 1 mark for transpose (adj), 1 mark for the final inverse. Show every cofactor calculation — don’t skip steps.

Common Mistake

The biggest error: forgetting to transpose the cofactor matrix. Students compute all 9 cofactors correctly, arrange them in a matrix, and then directly divide by $|A|$ — missing the transpose step. Remember: adj $(A)$ is the transpose of the cofactor matrix, not the cofactor matrix itself. The cofactors go into their $(i,j)$ position first, then the whole matrix gets transposed.

Question

Solution — Step by Step

Why This Works

Alternative Method — Row Reduction

Common Mistake

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