CBSE Sample Paper 2024-25 · Section A · 1 mark

Q4: $\det(B^{-1}AB)^2$ when $\det(A) = 5$

Sample 2024-25Ch 4 · Determinants

The question

If $A$ and $B$ are non-singular matrices of same order with $\det(A) = 5$ ⁠, then $\det(B^{-1}AB)^2$ is equal to

(A) $5$

(B) $5^2$

(D) $5^5$

Exam answer

Using $|XY| = |X|\,|Y|$ and $|B^{-1}| = \frac{1}{|B|}$ ⁠:

Why?

For square matrices of the same order, the determinant of a product is the product of the determinants.

$B$ is non-singular, so $|B| \ne 0$ and $B^{-1}$ exists. Taking determinants of $BB^{-1} = I$ ⁠:

|B|\,|B^{-1}| = |I| = 1

\begin{aligned} \det(B^{-1}AB) &= |B^{-1}|\,|A|\,|B| \\ &= \frac{1}{|B|} \cdot 5 \cdot |B| = 5 \end{aligned}

Why?

The matrix $B^{-1}AB$ need not equal $A$ ⁠, because matrix products do not commute. But the determinants are numbers, and numbers can be reordered, so $|B|$ cancels against $\frac{1}{|B|}$ ⁠.

\begin{aligned} |B^{-1}|\,|A|\,|B| &= \frac{1}{|B|} \cdot 5 \cdot |B| \\ &= 5 \cdot \frac{|B|}{|B|} \\ &= 5 \cdot 1 = 5 \end{aligned}

\therefore\ \det(B^{-1}AB)^2 = 5^2

(Reading it as $\det((B^{-1}AB)^2)$ gives the same value.)

Why?

With $M = B^{-1}AB$ ⁠, the product rule gives $\det(M^2) = \det M \cdot \det M = 5 \times 5$ ⁠. So both readings give $5^2 = 25$ ⁠.

Watch out: option (A), 5, is $\det(B^{-1}AB)$ before squaring.

Answer: (B) $5^2$

Where this comes from

Chapter 4, Determinants: Determinant of a product of matrices.

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 $\det A = 5$ is given: $|B^{-1}AB| = \frac{1}{|B|} \cdot 5 \cdot |B| = 5$ ⁠, so its square is $5^2$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Determinants in past papers: question 20 of 34 · see them all

Q4: det⁡(B−1AB)2\det(B^{-1}AB)^2det(B−1AB)2 when det⁡(A)=5\det(A) = 5det(A)=5

Q4: $\det(B^{-1}AB)^2$ when $\det(A) = 5$