Calculate A⁻¹ using adjoint method. When does inverse not exist?

Find Inverse of a 2×2 Matrix

Question

Given matrix $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$ , find $A^{-1}$ . Also state the condition under which the inverse does not exist.

Solution — Step by Step

Before anything else, we calculate the determinant. If $|A| = 0$ , we stop — the inverse simply doesn’t exist.

|A| = (3)(2) - (1)(5) = 6 - 5 = 1

Since $|A| = 1 \neq 0$ , we’re good to proceed.

For a 2×2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the cofactors follow a fixed sign pattern $\begin{pmatrix} + & - \\ - & + \end{pmatrix}$ .

C_{11} = +2, \quad C_{12} = -5, \quad C_{21} = -1, \quad C_{22} = +3

Cofactor matrix $= \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}$

The adjoint (or adjugate) is the transpose of the cofactor matrix — rows become columns.

\text{adj}(A) = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

\boxed{A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}}

\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 6-5 & -3+3 \\ 10-10 & -5+6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ✓

Always spend 20 seconds on this verification in board exams — it’s a free error check.

Why This Works

The formula $A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$ comes from a deeper identity: $A \cdot \text{adj}(A) = |A| \cdot I$ . Dividing both sides by $|A|$ gives us the inverse. This is why $|A| = 0$ breaks everything — we’d be dividing by zero.

For a 2×2 matrix, there’s actually a pattern worth memorising: swap the diagonal elements, negate the off-diagonal elements, then divide by the determinant. So $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . This bypasses the cofactor step entirely for 2×2.

The condition for non-existence — $|A| = 0$ — means the matrix is singular. Geometrically, a singular matrix collapses 2D space onto a line, and you can’t “undo” that collapse.

Alternative Method

For 2×2 specifically, use the shortcut formula directly:

A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \implies A^{-1} = \frac{1}{(3)(2)-(1)(5)} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

Pattern: swap the main diagonal ( $3 \leftrightarrow 2$ ), negate the anti-diagonal ( $1 \to -1$ , $5 \to -5$ ), divide by determinant.

In CBSE 2024, a 2×2 inverse question appeared for 3 marks. Using this shortcut saves roughly 2 minutes versus the full cofactor method. For 3×3 matrices though, there’s no shortcut — you must go through all 9 cofactors.

Common Mistake

The most frequent error: forgetting to transpose when writing the adjoint. Students compute the cofactor matrix correctly, then write it directly as the adjoint. The adjoint is the transpose of the cofactor matrix — rows and columns must be swapped. In this problem, the cofactor matrix has $-5$ in position (1,2), but the adjoint has $-5$ in position (2,1). Mixing these up gives a wrong inverse that won’t satisfy $AA^{-1} = I$ , which is exactly why verification matters.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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