MCQ
If $A$ is a square matrix such that $A^2 = I,$ then $A^{-1}$ is equal to:
  • A
    $A + I$
  • $A$
  • C
    $0$
  • D
    $2A$

Answer

Correct option: B.
$A$

$A^2 = I$
$A^{-1}A^2 = A^{-1}I$
$A = A^{-1}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:
  1. $\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  2. $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  3. ${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
  4. $2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$
If the vectors $4 \hat{i}+11 \hat{j}+m \hat{k}, 7 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\hat{i}+5 \hat{j}+4 \hat{k}$ are coplanar, then $m =$
The system of vectors i, j, k is:
  1. Orthogonal
  2. Collinear
  3. Coplana
  4. None of these
A fair die is tossed eight times. The probability that a third six is observed in the eight throw is:
The value of $\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{ dx}$ is:
  1. 1
  2. e - 1
  3. 0
  4. 1
The point on the curve $y = x^2 - 3x + 2$ where tangent is perpendicular to $y = x$ is:
 The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=0$ with y(1) = 1 is given by.
  1. $\text{y}=\frac{1}{\text{x}^{2}}$
  2. $\text{x}=\frac{1}{\text{y}^{2}}$
  3. $\text{x}=\frac{1}{\text{y}}$
  4. $\text{y}=\frac{1}{\text{x}}$ 
Choose the correct answer from the given four option.
The solution of $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}},\text{ y}(0)$is:
  1. $\text{y}=\text{e}^{\text{x}}(\text{x}-1)$
  2. $\text{y}=\text{x}\text{e}^{-\text{x}}$
  3. $\text{y}=\text{x}\text{e}^{\text{x}}+1$
  4. $\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
Choose the correct answer
The planes: $2x – y + 4z = 5$ and $5x – 2.5y + 10z = 6$ are:
Derivative of $\cos ^{-1}\left(4 x^3-3 x\right)$ :