Class 12 Maths, Worked in Full Past papersGet updates
CBSE Sample Paper 2023-24 · Section A · 1 mark

Q2: Which statement about ∣AB−1∣|AB^{-1}|⁠, (AB)−1(AB)^{-1}⁠, (A+B)−1(A + B)^{-1} is not correct

Sample 2023-24Ch 4 · Determinants
The question

If AA and BB are invertible square matrices of the same order, then which of the following is not correct?

(a) ∣AB−1∣=∣A∣∣B∣\left|AB^{-1}\right| = \frac{|A|}{|B|}

(b) ∣(AB)−1∣=1∣A∣ ∣B∣\left|(AB)^{-1}\right| = \frac{1}{|A|\,|B|}

(c) (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

(d) (A+B)−1=B−1+A−1(A + B)^{-1} = B^{-1} + A^{-1}

Exam answer

(a) ∣AB−1∣=∣A∣ ∣B−1∣=∣A∣∣B∣\left|AB^{-1}\right| = |A|\,\left|B^{-1}\right| = \frac{|A|}{|B|}⁠. Correct.

Why?

Two standard results are used:

∣AB∣=∣A∣ ∣B∣∣B−1∣=1∣B∣\begin{aligned} |AB| &= |A|\,|B| \\ \left|B^{-1}\right| &= \frac{1}{|B|} \end{aligned}

The second follows from the first: BB−1=IBB^{-1} = I gives ∣B∣ ∣B−1∣=∣I∣=1|B|\,\left|B^{-1}\right| = |I| = 1⁠.

(b) ∣(AB)−1∣=1∣AB∣=1∣A∣ ∣B∣\left|(AB)^{-1}\right| = \frac{1}{|AB|} = \frac{1}{|A|\,|B|}⁠. Correct.

Why?

Use the same two results, with the matrix ABAB in place of BB⁠:

∣(AB)−1∣=1∣AB∣=1∣A∣ ∣B∣(∣AB∣=∣A∣ ∣B∣)\begin{aligned} \left|(AB)^{-1}\right| &= \frac{1}{|AB|} \\ &= \frac{1}{|A|\,|B|} \quad (|AB| = |A|\,|B|) \end{aligned}

(c) Multiplying ABAB by B−1A−1B^{-1}A^{-1}⁠,

(AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I\begin{aligned} &(AB)\left(B^{-1}A^{-1}\right) \\ &= A\left(BB^{-1}\right)A^{-1} \\ &= AIA^{-1} = AA^{-1} = I \end{aligned}

So (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠, the reversal law. Correct.

Why?

Matrix multiplication is associative, so the brackets can be moved. The middle pair BB−1BB^{-1} becomes II⁠.

A matrix that multiplies ABAB to give II is its inverse. Watch out: the order reverses. A−1B−1A^{-1}B^{-1} is wrong in general, because matrix products need not commute.

(d) For A=B=IA = B = I⁠: (A+B)−1=(2I)−1=12I(A + B)^{-1} = (2I)^{-1} = \frac{1}{2}I⁠, but B−1+A−1=2IB^{-1} + A^{-1} = 2I⁠. Not correct.

Why?

(2I)−1=12I(2I)^{-1} = \frac{1}{2}I because (2I)(12I)=I(2I)\left(\frac{1}{2}I\right) = I⁠. Also I−1=II^{-1} = I⁠, so B−1+A−1=I+IB^{-1} + A^{-1} = I + I⁠.

Since 12I≠2I\frac{1}{2}I \ne 2I⁠, the rule fails. One counterexample is enough to show that a statement is not always true.

Answer: (d) (A+B)−1=B−1+A−1(A + B)^{-1} = B^{-1} + A^{-1}

Where this comes from

Chapter 4, Determinants: Properties of determinants and inverses.

Question type: One-mark questions on ∣kA∣|kA|⁠, ∣adj⁡A∣|\operatorname{adj}A| and ∣A−1∣|A^{-1}|

How to spot it: click to see

Type 10. One-mark questions on ∣kA∣|kA|⁠, ∣adj⁡A∣|\operatorname{adj}A| and ∣A−1∣|A^{-1}|

Spot it: an MCQ about a matrix you are not given, only its order.

Here: No matrix is given, so test each option against the cards: ∣AB−1∣=∣A∣∣B∣|AB^{-1}| = \frac{|A|}{|B|}⁠, ∣(AB)−1∣=1∣A∣∣B∣|(AB)^{-1}| = \frac{1}{|A||B|} and (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} all hold. (d) is no card, and A=B=IA = B = I breaks it.

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 23 of 34 · see them all