Maths. Section AMatrices (Section A)

Let A and B be two non-singular, square matrices of same order, and A…

Let A and B be two non-singular, square matrices of same order, and

A. (AB)⁻¹ = B⁻¹.A⁻¹

B. (A + B)⁻¹ = B⁻¹ + A⁻¹

C. adj. A = |A|.A⁻¹

D. det(A⁻¹) = [det A]⁻¹

Choose the correct answer from the options given below

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → A, C and D only

The standard results for non-singular matrices of the same order:

\((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I\)

\(A(\mathrm{adj}\,A) = |A|\,I\)

\(|A|\,|A^{-1}| = |AA^{-1}| = |I| = 1\)

So \((AB)^{-1} = B^{-1}A^{-1}\), \(\mathrm{adj}\,A = |A|\,A^{-1}\) and \(\det(A^{-1}) = \frac{1}{\det A}\), while there is no such rule for the inverse of a sum.

Checking the statements

A. (AB)⁻¹ = B⁻¹.A⁻¹ — Correct. The reversal law: \(B^{-1}A^{-1}\) multiplied by AB gives I.

B. (A + B)⁻¹ = B⁻¹ + A⁻¹ — Incorrect. With A = B = I: \((2I)^{-1} = \frac{1}{2}I\), but \(B^{-1} + A^{-1} = 2I\).

C. adj. A = |A|.A⁻¹ — Correct. Multiply \(A(\mathrm{adj}\,A) = |A|\,I\) on the left by \(A^{-1}\).

D. det(A⁻¹) = [det A]⁻¹ — Correct. \(|A|\,|A^{-1}| = |I| = 1\).