CBSE Sample Paper 2025-26 · Section A · 1 mark

Q4: $A(\operatorname{adj} A)$ when $|\operatorname{adj} A| = 27$ ⁠, order 4

The question

If $A$ is a square matrix of order 4 and $|\operatorname{adj} A| = 27$ ⁠, then $A\,(\operatorname{adj} A)$ is equal to

(A) $3$

(B) $9$

(D) $9\,I$

Exam answer

For order $n = 4$ ⁠, $|\operatorname{adj} A| = |A|^{n - 1} = |A|^3$ ⁠.

Why?

Standard result for a square matrix $A$ of order $n$ ⁠:

|\operatorname{adj} A| = |A|^{n - 1}

Here $n = 4$ ⁠, so the power is $4 - 1 = 3$ ⁠. Watch out: the power is $n - 1$ ⁠, not $n$ ⁠.

\begin{aligned} &|A|^3 = 27 = 3^3 \\ \Rightarrow\ &|A| = 3 \end{aligned}

Why?

$27 = 3 \times 3 \times 3$ ⁠. A real number has only one real cube root, so $|A| = 3$ ⁠. There is no $\pm$ here, unlike $x^2 = 9$ ⁠.

A\,(\operatorname{adj} A) = |A|\,I = 3\,I

Why?

Standard result, where $I$ is the identity matrix of the same order:

A\,(\operatorname{adj} A) = |A|\,I

Each diagonal entry of the product is $|A|$ ⁠, and every other entry is 0.

$A\,(\operatorname{adj} A)$ is a matrix, so the numbers in (A) and (B) are ruled out, and $9\,I$ would need $|A| = 9$ ⁠.

Answer: (C) $3\,I$

Where this comes from

Chapter 4, Determinants: Adjoint: |adj A| = |A|ⁿ⁻¹ and A(adj A) = |A|I.

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 the order 4 and $|\operatorname{adj}A| = 27$ are given: $|A|^3 = 27$ ⁠, so $|A| = 3$ ⁠, and $A(\operatorname{adj}A) = |A|I = 3I$ ⁠.

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Determinants in past papers: question 16 of 34 · see them all

Q4: A(adj⁡A)A(\operatorname{adj} A)A(adjA) when ∣adj⁡A∣=27|\operatorname{adj} A| = 27∣adjA∣=27⁠, order 4