Sample 2020-21 Class 12 Maths Q5 Solution: A² for aᵢⱼ = 1 (i ≠ j), 0 (i = j). Or: |adj A| when |A| = −4

The question

(a) Find the value of $A^2$ ⁠, where A is a 2×2 matrix whose elements are given by

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

OR

(b) Given that A is a square matrix of order 3×3 and $|A| = -4$ ⁠. Find $|\text{adj}\,A|$

Exam answer

(a)

Here $a_{11} = a_{22} = 0$ (as $i = j$ ⁠) and $a_{12} = a_{21} = 1$ (as $i \ne j$ ⁠).

Why?

$a_{ij}$ is the entry in row $i$ and column $j$ ⁠. The diagonal entries $a_{11}$ ⁠, $a_{22}$ have $i = j$ ⁠, so they are 0. The other two entries 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} 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:

\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}

$\therefore\ A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

(b)

For a square matrix of order n, $|\text{adj}\,A| = |A|^{n-1}$ ⁠.

Why?

This comes from $A(\text{adj}\,A) = |A|\,I$ ⁠. Taking determinants, with $|kI| = k^n$ for order $n$ ⁠:

|A|\,|\text{adj}\,A| = |A|^n

Dividing by $|A| \ne 0$ gives the result:

\begin{aligned} |\text{adj}\,A| &= \frac{|A|^n}{|A|} \\ &= |A|^{n-1} \end{aligned}

|\text{adj}\,A| = (-4)^{3-1} = 16

Why?

Here $n = 3$ ⁠, so the power is $3 - 1 = 2$ ⁠, and $(-4)^2 = 16$ ⁠.

\begin{aligned} (-4)^{3-1} &= (-4)^2 \\ &= (-4)(-4) \\ &= 16 \end{aligned}

Watch out: the even power makes the answer positive. $-16$ and $(-4)^3 = -64$ are common slips.

$\therefore\ |\text{adj}\,A| = 16$

Where this comes from

Chapter 3, Matrices: Constructing a matrix; |adj A| = |A|ⁿ⁻¹.

Question type: (a) Build a matrix from a rule for $a_{ij}$ + Multiply two matrices (combined); (b) One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

How to spot it: click to see

(a) Ch 3, 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 puts 0 on the diagonal ( $i = j$ ⁠) and 1 in the other two places, so row 1 of $A$ is $0, 1$ and row 2 is $1, 0$ ⁠.

(a) Ch 3, 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: $A^2 = I$ ⁠.

(b) Ch 4, type 10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Here: Only $|A| = -4$ and the order 3 are given: use $|\operatorname{adj}A| = |A|^{n - 1}$ with $n = 3$ ⁠, so $(-4)^2 = 16$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 4, Determinants.

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Q5: A2A^2A2 for aij=1a_{ij} = 1aij​=1 (i≠ji \ne ji=j⁠), 000 (i=ji = ji=j⁠). Or: ∣adj A∣|\text{adj}\,A|∣adjA∣ when ∣A∣=−4|A| = -4∣A∣=−4

Q5: $A^2$ for $a_{ij} = 1$ ( $i \ne j$ ⁠), $0$ ( $i = j$ ⁠). Or: $|\text{adj}\,A|$ when $|A| = -4$