Matrices: Speed-Solving Techniques (8)

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Question

If A=(2312)A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}, find A−1A^{-1} using the adjoint method, and verify that A⋅A−1=IA \cdot A^{-1} = I.

Solution — Step by Step

det⁡(A)=(2)(2)−(3)(1)=4−3=1\det(A) = (2)(2) - (3)(1) = 4 - 3 = 1

Since det⁡(A)≠0\det(A) \neq 0, the inverse exists.

For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the adjoint is (d−b−ca)\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} — swap diagonals, negate off-diagonals. So:

adj(A)=(2−3−12)\text{adj}(A) = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} A−1=1det⁡(A)⋅adj(A)=11(2−3−12)=(2−3−12)A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} A⋅A−1=(2312)(2−3−12)=(4−3−6+62−2−3+4)=(1001)=I✓A \cdot A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \checkmark

Final Answer: A−1=(2−3−12)A^{-1} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}.

Why This Works

For any invertible square matrix, A−1=1det⁡(A)adj(A)A^{-1} = \dfrac{1}{\det(A)} \text{adj}(A). The adjoint is the transpose of the cofactor matrix; for a 2×22 \times 2, the formula collapses to the swap-and-negate shortcut. For larger matrices, you compute cofactors explicitly.

If det⁡(A)=0\det(A) = 0, the matrix is singular and has no inverse. Always compute the determinant first as a quick sanity check.

Alternative Method

For 2×22 \times 2, use row-reduction on [A∣I][A | I] to get [I∣A−1][I | A^{-1}]. Slower than the adjoint shortcut for 2×22 \times 2 but more general (and faster for n≥3n \geq 3, where computing the full adjoint is tedious).

Forgetting to negate the off-diagonals in the adjoint shortcut is the most common error. The pattern is: swap a↔da \leftrightarrow d, negate bb and cc. Drill this until it’s reflex.

Once you have A−1A^{-1}, you can solve any system Ax⃗=b⃗A\vec{x} = \vec{b} as x⃗=A−1b⃗\vec{x} = A^{-1}\vec{b}. JEE Main loves to embed this inside a “find the value of x+yx + y” question — recognize the matrix structure first.

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