Question

Which of the following statements is/are true about square matrix A of order n?

• A. $(-A{)}^{-1}$ is equal to $-A^{-1}$ when n is odd only.
• B. If $A^n=O$, then $I+A+A^2+.....+A^{n-1}=(I-A{)}^{-1}$.
• C. If A is skew-symmetric matrix of odd order, then its inverse does not exist.
• D. $(A^T{)}^{-1}=(A^{-1}{)}^T$ holds exist.

Answer 1

Answer: (B), (C)

If $A^n=O$, then $I+A+A^2+.....+A^{n-1}=(I-A{)}^{-1}$.
If A is skew-symmetric matrix of odd order, then its inverse does not exist.

The options are examined below:

$\left(A\right){(-A)}^{-1}=\frac{adj(-A)}{\left|A\right|}=\frac{{(-1)}^{n-1}adjA}{{(-1)}^n\left|A\right|}=\frac{adjA}{-\left|A\right|}={-A}^{-1}$

So ${(-A)}^{-1}={-A}^{-1}$ (For any value of $n$ , odd or even)

$\therefore$ Option $A$ is false

$\left(B\right)A^n=0\therefore (I-A)(I+A+A^2+......A^{n-1})=I-A^2=I-0\Rightarrow (I-A)(I+A+A^2......)=I\therefore {(I-A)}^{-1}=I+A+A^2.....A^{n-1}$

$\therefore$ Option $B$ is true.

$\left(C\right)$ If $A$ is skew-symmetric matrix of odd order, then $\left|A\right|=0$ and so inverse of $A$ does not exist

$\therefore$ Option $C$ is true.

$\left(D\right){\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T$ does not hold always. (Hold only when $A$ is non-singular)

$\therefore$ Option $D$ is false.