CBSE Sample Paper 2023-24 · Section A · 1 mark

Q13: $|\operatorname{adj}(2A)|$ when $A$ is of order 3 and $|A| = -2$

Sample 2023-24Ch 4 · Determinants

The question

Given that $A$ is a square matrix of order 3 and $|A| = -2$ ⁠, then $|\operatorname{adj}(2A)|$ is equal to

(a) $-2^6$

(b) $+4$

(d) $2^8$

Exam answer

For a square matrix $B$ of order $n$ ⁠, $|\operatorname{adj} B| = |B|^{n - 1}$ ⁠, and $|kA| = k^n|A|$ ⁠. Here $n = 3$ ⁠.

Why?

$|\operatorname{adj} B| = |B|^{n - 1}$ comes from $B(\operatorname{adj} B) = |B|\,I$ ⁠. Taking determinants, $|B|\,|\operatorname{adj} B| = |B|^n$ ⁠. Then divide by $|B| \ne 0$ ⁠.

$|kA| = k^n|A|$ because each of the $n$ rows of $kA$ carries a factor $k$ ⁠.

\begin{aligned} |\operatorname{adj}(2A)| &= |2A|^2 \\ &= \left(2^3|A|\right)^2 \\ &= 2^6 \times (-2)^2 = 2^8 \end{aligned}

Why?

Use $B = 2A$ with $n - 1 = 2$ ⁠. Then $|2A| = 2^3|A| = 8(-2) = -16$ ⁠, which is not 0.

So the value is $(-16)^2 = 256 = 2^8$ ⁠.

In powers of 2, step by step:

\begin{aligned} \left(2^3|A|\right)^2 &= \left(2^3\right)^2|A|^2 \\ &= 2^{3 \times 2} \times (-2)^2 \\ &= 2^6 \times 4 \\ &= 2^6 \times 2^2 \\ &= 2^{6 + 2} = 2^8 \end{aligned}

Watch out: $|2A|$ is not $2|A|$ ⁠. The factor 2 comes out once for each of the 3 rows. Squaring also removes the minus sign, so (a) and (c) cannot be right.

Answer: (d) $2^8$

Where this comes from

Chapter 4, Determinants: Determinant of the adjoint of kA.

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 3 and $|A| = -2$ are given: $|\operatorname{adj}(2A)| = |2A|^2 = (2^3|A|)^2 = 2^8$ ⁠.

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Determinants in past papers: question 26 of 34 · see them all

Q13: ∣adj⁡(2A)∣|\operatorname{adj}(2A)|∣adj(2A)∣ when AAA is of order 3 and ∣A∣=−2|A| = -2∣A∣=−2

Q13: $|\operatorname{adj}(2A)|$ when $A$ is of order 3 and $|A| = -2$