CBSE Sample Paper 2022-23 · Section A · 1 mark

Q13: $|\operatorname{adj} A|$ when $|A| = 5$ ⁠, order 3

The question

If A is a square matrix of order 3 and |A| = 5, then $|\operatorname{adj} A| =$

(a) $5$

(b) $25$

(d) $\frac{1}{5}$

Exam answer

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

Why?

This comes from $A(\operatorname{adj} A) = |A|\,I$ ⁠. Take determinants of both sides, using $|kI| = k^n$ for order n:

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

Dividing by $|A| \ne 0$ gives $|\operatorname{adj} A| = |A|^{n - 1}$ ⁠.

|\operatorname{adj} A| = 5^{3 - 1} = 25

Why?

Here $n = 3$ and $|A| = 5$ ⁠, so the power is $3 - 1 = 2$ and $5^2 = 25$ ⁠.

Watch out: $5^3 = 125$ ⁠, option (c), is $|A|^n$ ⁠, which equals $|A|\,|\operatorname{adj} A|$ ⁠, not $|\operatorname{adj} A|$ itself.

Answer: (b) $25$

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 $|A| = 5$ and the order 3 are given: $|\operatorname{adj}A| = 5^{3 - 1} = 25$ ⁠.

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Determinants in past papers: question 30 of 34 · see them all

Q13: ∣adj⁡A∣|\operatorname{adj} A|∣adjA∣ when ∣A∣=5|A| = 5∣A∣=5⁠, order 3