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

Concept: Inverse via Adjoint — $A^{-1} = \frac{1}{\det A} \cdot \operatorname{adj} A$, so first check $\det A \neq 0$.

Step 1 – Compute determinant

Expand along first row:

$\det A = 1 \cdot \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} + 2 \cdot \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix}$

$= 1(8 - 6) + 1(0 + 9) + 2(0 - 6) = 2 + 9 - 12 = -1 \neq 0$, so inverse exists.

Step 2 – Find cofactor matrix

$C_{11} = +(8-6)=2$, $C_{12} = -(0+9)=-9$, $C_{13} = +(0-6)=-6$

$C_{21} = -((-1)(4) - (2)(-2)) = -(-4+4)=0$

$C_{22} = +((1)(4) - (2)(3)) = 4-6=-2$

$C_{23} = -((1)(-2) - (-1)(3)) = -(-2+3)=-1$

$C_{31} = +((-1)(-3) - (2)(2)) = 3-4=-1$

$C_{32} = -((1)(-3) - (2)(0)) = -(-3)=3$

$C_{33} = +((1)(2) - (-1)(0)) = 2$

Cofactor matrix: $\begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix}$ …

For a square matrix $A$, the inverse exists precisely when $\det(A) \neq 0$, and then $A^{-1} = \dfrac{1}{\det(A)}\,\text{adj}(A)$, where $\text{adj}(A)$ is the transpose of the cofactor matrix.

1. Compute the determinant. Expanding along the first row of

Since $\det(A) = -1 \neq 0$, the inverse exists.

2. Compute the cofactors $C_{ij} = (-1)^{i+j}M_{ij}$:

3. Form the adjoint (transpose of the cofactor matrix): …

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

Whenever a question asks for the inverse of a $3\times3$ matrix, the standard route is: determinant → cofactors → adjoint → divide.

Steps

Step 1: Find $\det A$ first, before anything else

Expand along the row or column with the most zeros to keep the arithmetic light. If $\det A = 0$, stop — the matrix has no inverse, and the question is really testing that you recognise this, not that you complete the adjoint calculation.

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

Each $M_{ij}$ is the $2\times2$ determinant left after crossing out row $i$ and column $j$. Compute all nine methodically — row 1, then row 2, then row 3 — applying the alternating sign as you go, rather than trying to remember signs at the end.

Step 3: Transpose the cofactor matrix to get the adjoint

$\operatorname{adj}(A) = [C_{ij}]^{T}$ …

Common Mistakes

Mistake 1: Mixing up signs when several matrix entries are negative

Why it's wrong: with negative entries like $-1$ or $-3$ in the matrix, it's easy to drop a sign while writing out a $2\times2$ minor determinant (e.g. writing $2\times4-(-3)\times(-2)$ as $8+6$ instead of $8-6$). Correct approach: write each minor's $2\times2$ determinant out fully in brackets before simplifying, rather than combining the subtraction mentally.

Mistake 2: Dividing by $\det A$ incorrectly when $\det A$ is negative …