CBSE Board Examination 2023 · Section A · 1 mark

Q3: $|A|$ when $|\text{adj}\,A| = 64$ for a $3 \times 3$ matrix

Board 2023Ch 4 · Determinants

The question

Let A be a $3 \times 3$ matrix such that $|\text{adj}\,A| = 64$ ⁠. Then $|A|$ is equal to :

(a) 8 only

(b) $-8$ only

(d) 8 or $-8$

Exam answer

For a square matrix of order $n$ ⁠, $|\text{adj}\,A| = |A|^{n - 1}$ ⁠. Here $n = 3$ ⁠.

Why?

This standard result 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 $|A|$ (here $|A| \ne 0$ ⁠, because $|\text{adj}\,A| = 64 \ne 0$ ⁠) gives $|\text{adj}\,A| = |A|^{n - 1}$ ⁠.

\begin{aligned} &|A|^{3 - 1} = 64 \\ \Rightarrow\ &|A|^2 = 64 \\ \Rightarrow\ &|A| = 8 \text{ or } -8 \end{aligned}

Why?

First $3 - 1 = 2$ ⁠. A number whose square is 64 can be $8$ or $-8$ ⁠, since $(-8)^2 = 64$ too.

\begin{aligned} (8)^2 &= 8 \times 8 = 64 \\ (-8)^2 &= (-8)(-8) = 64 \end{aligned}

Watch out: a determinant can be negative, so do not drop $-8$ ⁠. Both really occur: the diagonal matrices with diagonal entries $8, 1, 1$ and $-8, 1, 1$ each have $|\text{adj}\,A| = 64$ ⁠. So "8 only" is wrong.

Answer: (d) 8 or $-8$

Where this comes from

Chapter 4, Determinants: Determinant of the adjoint.

Question type: One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

How to spot it: click to see

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 $|\operatorname{adj}A| = 64$ and the order 3 are given: $|A|^{3 - 1} = 64$ ⁠, so $|A|^2 = 64$ and $|A| = \pm 8$ ⁠.

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Determinants in past papers: question 12 of 34 · see them all

Q3: ∣A∣|A|∣A∣ when ∣adj A∣=64|\text{adj}\,A| = 64∣adjA∣=64 for a 3×33 \times 33×3 matrix

Q3: $|A|$ when $|\text{adj}\,A| = 64$ for a $3 \times 3$ matrix