CBSE Sample Paper 2026-27 · Section A · 1 mark

Q5: $|\text{adj}(PQ)|$ for $2 \times 2$ matrices $P$ and $Q$

The question

If $P$ and $Q$ are matrices of same order such that $P = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$ ⁠, $|Q| = 3$ ⁠, then find $|adj(PQ)|$

(A) $-3$

(B) $1$

(D) $36$

Exam answer

$|P| = 2 - 0 = 2$ ⁠. $P$ and $Q$ are of order $2 \times 2$ ⁠, so $PQ$ is of order 2.

Why?

For a $2 \times 2$ matrix the determinant is $ad - bc$ ⁠:

\begin{aligned} |P| &= \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} \\ &= (2)(1) - (1)(0) \\ &= 2 - 0 = 2 \end{aligned}

$Q$ has the same order as $P$ ⁠, so it is $2 \times 2$ ⁠, and the product of two $2 \times 2$ matrices is $2 \times 2$ ⁠.

Using $|\text{adj}\,A| = |A|^{n-1}$ ⁠, $n = 2$ ⁠:

Why?

For a square matrix of order $n$ ⁠, $A(\text{adj}\,A) = |A|\,I$ ⁠. Taking determinants gives $|A|\,|\text{adj}\,A| = |A|^n$ ⁠, so

|\text{adj}\,A| = |A|^{n-1}

For $n = 2$ the power is 1, so $|\text{adj}\,A| = |A|$ ⁠.

\begin{aligned} |\text{adj}(PQ)| &= |PQ| = |P||Q| \\ &= 2 \times 3 = 6 \end{aligned}

Why?

The determinant of a product is the product of the determinants: $|PQ| = |P|\,|Q|$ ⁠.

Watch out: $36 = 6^2$ in (D) uses the power $n - 1 = 2$ ⁠, which is right only for a $3 \times 3$ matrix.

Answer: (C) $6$

Where this comes from

Chapter 4, Determinants: Adjoint of a matrix: |adj A| = |A|ⁿ⁻¹.

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: $PQ$ is of order 2, so $|\operatorname{adj}(PQ)| = |PQ|^{2 - 1} = |P||Q|$ ⁠. With $|P| = 2$ and $|Q| = 3$ ⁠, that is 6.

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