Q.Find the inverse of the following matrix, if it exists: $A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$

Use $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$ for $A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$.

Determinant (expand along row 1): $2(-1-0) - 1(4-0) + 3(8-7) = -2-4+3 = -3 \neq 0$, so $A^{-1}$ exists.

Cofactors: $C_{11}=-1,\ C_{12}=-4,\ C_{13}=1$; $C_{21}=5,\ C_{22}=23,\ C_{23}=-11$; $C_{31}=3,\ C_{32}=12,\ C_{33}=-6.$

Adjoint (transpose of the cofactor matrix): $\operatorname{adj}(A) = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}.$

Divide by $\det A = -3$: …

The inverse of a $3\times3$ matrix is $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$, where $\operatorname{adj}(A)$ is the transpose of the cofactor matrix. It exists only when $\det A \neq 0$.

1. Determinant

Expanding along row 1:

Since $-3 \neq 0$, $A$ is invertible.

2. Cofactors $C_{ij}=(-1)^{i+j}M_{ij}$

3. Adjoint

Transpose the cofactor matrix: …

Method: Inverse of a 3×3 Matrix via the Adjoint

This method finds the inverse of any square matrix whose determinant is non-zero, using cofactors and the adjoint — the standard NCERT route for a $3\times3$ inverse.

Steps

Step 1: Compute $\det A$ and confirm it is non-zero

The inverse of a matrix exists only when $\det A \neq 0$ (a singular matrix has no inverse). Expand the determinant along any convenient row or column — pick one with the most zeros to minimise the arithmetic.

Step 2: Compute all nine cofactors

For each entry $a_{ij}$, delete its row and column to get the minor $M_{ij}$, then apply the checkerboard sign:

Work systematically row by row ($C_{11}, C_{12}, C_{13}$, then $C_{21}, \dots$) so no entry is skipped, keeping the alternating $+,-,+$ sign pattern in front of you rather than re-deriving it each time.

Step 3: Form the adjoint by transposing the cofactor matrix …

Common Mistakes

Mistake 1: Forgetting to transpose the cofactor matrix into the adjoint

Why it's wrong: the adjoint is defined as the transpose of the cofactor matrix, $\operatorname{adj}(A)=[C_{ij}]^T$ — using the cofactor matrix unchanged silently swaps every off-diagonal pair (e.g. $C_{12}$ and $C_{21}$), giving a wrong inverse that still "looks" plausible. Correct approach: write the cofactor matrix out fully first, then explicitly transpose it as its own step before dividing by $\det A$.

Mistake 2: A sign slip in the checkerboard pattern when computing cofactors …