CBSE Board Examination 2025 · Section A · 1 mark

Q6: $\det(4\,\text{adj}\,A)$ when $\det A = 4$ ⁠, order 2

The question

If A is a square matrix of order 2 such that $\det(A) = 4$ ⁠, then $\det(4\,\text{adj}\,A)$ is equal to :

(A) $16$

(B) $64$

(D) $512$

Exam answer

For order n: $|kB| = k^n|B|$ and $|\text{adj}\,A| = |A|^{n-1}$ ⁠. Here $n = 2$ ⁠.

Why?

$|kB| = k^n|B|$ ⁠: the matrix kB has every one of its n rows multiplied by k, and each such row multiplies the determinant by k.

$|\text{adj}\,A| = |A|^{n-1}$ ⁠: take determinants in $A\,(\text{adj}\,A) = |A|\,I$ ⁠:

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

Then divide by $|A| \ne 0$ ⁠.

\begin{aligned} |4\,\text{adj}\,A| &= 4^2\,|\text{adj}\,A| \\ &= 16\,|A|^{2-1} \\ &= 16 \times 4 = 64 \end{aligned}

Why?

Here the matrix is $\text{adj}\,A$ and $k = 4$ ⁠. Its order is 2, so $k^2 = 16$ ⁠. Also $|A|^{2-1} = |A| = 4$ ⁠.

Watch out: $|4\,\text{adj}\,A|$ is not $4\,|\text{adj}\,A|$ ⁠. That slip gives 16, option (A).

Check with an example of order 2 with $|A| = 4$ ⁠:

\begin{aligned} A &= \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \\ |A| &= (2)(2) - (1)(0) = 4 \end{aligned}

For order 2, adj A swaps the two diagonal entries and changes the sign of the other two:

\begin{aligned} \text{adj}\,A &= \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix} \\ 4\,\text{adj}\,A &= \begin{bmatrix} 8 & -4 \\ 0 & 8 \end{bmatrix} \end{aligned}

\begin{aligned} \begin{vmatrix} 8 & -4 \\ 0 & 8 \end{vmatrix} &= (8)(8) - (-4)(0) \\ &= 64 - 0 = 64 \end{aligned}

Answer: (B) $64$

Where this comes from

Chapter 4, Determinants: Determinant of k adj A.

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 $\det A = 4$ and the order 2 are given: $|4\operatorname{adj}A| = 4^2|\operatorname{adj}A| = 16|A|^{2 - 1} = 64$ ⁠.

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Determinants in past papers: question 4 of 34 · see them all

Q6: det⁡(4 adj A)\det(4\,\text{adj}\,A)det(4adjA) when det⁡A=4\det A = 4detA=4⁠, order 2