Let A be a non-singular square matrix. Match each relation in List-I with the inverse \(A^{-1}\) it gives in List-II
| List-I (Relation) | List-II (\(A^{-1}\)) |
|---|---|
| (A) \(A^{2} = I\) | (I) \(A + I\) |
| (B) \(A^{2} - 3A + 2I = O\) | (II) A |
| (C) \(A^{3} = I\) | (III) \(A^{2}\) |
| (D) \(A^{2} + A - I = O\) | (IV) \(\displaystyle\frac{1}{2}(3I - A)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
In each case the relation can be written as \(A \cdot (\text{something}) = I\); that something is \(A^{-1}\), because the inverse is unique.
(A) \(A \cdot A = I\), so \(A^{-1} = A\) → (II)
(B) \(A^{2} - 3A = -2I\)
\(A(3I - A) = 2I\)
\(\displaystyle A \cdot \frac{1}{2}(3I - A) = I\), so \(\displaystyle A^{-1} = \frac{1}{2}(3I - A)\) → (IV)
(C) \(A \cdot A^{2} = I\), so \(A^{-1} = A^{2}\) → (III)
(D) \(A^{2} + A = I\)
\(A(A + I) = I\), so \(A^{-1} = A + I\) → (I)