Sample 2023-24 Class 12 Maths Q1 Solution: A² when aᵢⱼ = 1 for i ≠ j and 0 for i = j

The question

If $A = \left[a_{ij}\right]$ is a square matrix of order 2 such that

a_{ij} = \begin{cases} 1, & \text{when } i \ne j \\ 0, & \text{when } i = j \end{cases},

then $A^2$ is

(a) $\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}_{2 \times 2}$

(b) $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}_{2 \times 2}$

(c) $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}_{2 \times 2}$

(d) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}_{2 \times 2}$

Exam answer

Here $a_{11} = a_{22} = 0$ and $a_{12} = a_{21} = 1$ ⁠.

Why?

$a_{ij}$ is the entry in row $i$ and column $j$ ⁠.

On the diagonal the row and column numbers are equal, so $a_{11}$ and $a_{22}$ follow the rule for $i = j$ and are 0. The other two entries have $i \ne j$ ⁠, so they are 1.

A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

\begin{aligned} A^2 &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{aligned}

Why?

Each entry of the product is a row of the first matrix times a column of the second: multiply term by term and add.

\begin{aligned} (1, 1)&: 0 \cdot 0 + 1 \cdot 1 = 1 \\ (1, 2)&: 0 \cdot 1 + 1 \cdot 0 = 0 \\ (2, 1)&: 1 \cdot 0 + 0 \cdot 1 = 0 \\ (2, 2)&: 1 \cdot 1 + 0 \cdot 0 = 1 \end{aligned}

So $A^2 = I$ ⁠, the identity matrix of order 2. Options (a), (b) and (c) each have a 1 off the diagonal, so they are ruled out.

Why?

Watch out: $A^2$ means the matrix product $AA$ ⁠, not the square of each entry. Squaring each entry would give $A$ back.

Answer: (d) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}_{2 \times 2}$

Where this comes from

Chapter 3, Matrices: Constructing a matrix from a rule; A².

Question type: Build a matrix from a rule for $a_{ij}$ + Multiply two matrices (combines 2 types)

How to spot it: click to see

Type 1. Build a matrix from a rule for $a_{ij}$

Spot it: "Construct a $2\times2$ matrix whose elements are given by" a rule for $a_{ij}$ ⁠.

Here: First, the rule gives 0 where $i = j$ and 1 where $i \ne j$ ⁠, so row 1 of $A$ is $0, 1$ and row 2 is $1, 0$ ⁠.

Type 5. Multiply two matrices: row times column

Spot it: two matrices written side by side, or $AB$ ⁠, $BA$ or $A^2$ ⁠.

Here: Then $A^2 = AA$ ⁠, row times column, gives 1 on the diagonal and 0 elsewhere, so $A^2 = I$ ⁠, option (d).

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