Matrices — types, operations, and application selection guide

Question

If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ , find $A^{-1}$ and verify that $AA^{-1} = I$ .

(CBSE 12 / JEE Main — Matrices)

Matrix Operations Classification

flowchart TD
    A["Matrix Problem"] --> B{What is asked?}
    B -->|Addition/Subtraction| C["Same order required"]
    B -->|Multiplication AB| D["Columns of A = Rows of B"]
    B -->|Transpose| E["Swap rows and columns"]
    B -->|Determinant| F["|A| for square matrices"]
    B -->|Inverse| G{"Is |A| != 0?"}
    G -->|Yes| H["A^-1 = adj(A) / |A|"]
    G -->|No| I["Singular: inverse does not exist"]
    B -->|Solve equations| J["AX = B => X = A^-1 B"]

Solution — Step by Step

|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1(4) - 2(3) = 4 - 6 = -2

Since $|A| \neq 0$ , the inverse exists.

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the adjoint is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

\text{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}

A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}

AA^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}

Row 1: $(1)(-2) + (2)(3/2) = -2 + 3 = 1$ , $(1)(1) + (2)(-1/2) = 1 - 1 = 0$

Row 2: $(3)(-2) + (4)(3/2) = -6 + 6 = 0$ , $(3)(1) + (4)(-1/2) = 3 - 2 = 1$

\boxed{AA^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \quad \checkmark}

Why This Works

The inverse matrix “undoes” the original transformation. If $A$ transforms a vector $\vec{x}$ to $\vec{b}$ (i.e., $A\vec{x} = \vec{b}$ ), then $A^{-1}$ recovers $\vec{x}$ from $\vec{b}$ (i.e., $\vec{x} = A^{-1}\vec{b}$ ). The condition $|A| \neq 0$ ensures the transformation is reversible — no information is lost.

For $2 \times 2$ matrices, the adjoint shortcut (swap diagonal, negate off-diagonal) is the fastest method. For $3 \times 3$ , you need cofactors.

Alternative Method — Row Reduction

Augment $A$ with the identity matrix $[A | I]$ and perform row operations until the left side becomes $I$ . The right side automatically becomes $A^{-1}$ :

\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array}\right] \xrightarrow{R_2 \to R_2 - 3R_1} \left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array}\right]

Continue row-reducing to get the same $A^{-1}$ .

For CBSE 12, the $2 \times 2$ inverse using the adjoint formula is enough for most problems. But JEE Main often gives $3 \times 3$ matrices — practice the cofactor method and row reduction for those. Also, $|AB| = |A||B|$ and $(AB)^{-1} = B^{-1}A^{-1}$ — these properties appear as 1-mark conceptual questions.

Common Mistake

In the $2 \times 2$ adjoint, students swap the wrong elements. The rule is: swap the main diagonal elements ( $a$ and $d$ ) and negate the off-diagonal elements ( $b$ and $c$ ). Writing $\begin{pmatrix} d & b \\ c & a \end{pmatrix}$ (forgetting to negate) gives a completely wrong inverse.