We know that if $A$ is a square of order $m$, and if there exists another square matrix $B$ of the same order $m$, such that $AB=I$, then $B$ is said to be the inverse of $A$. 

Given $A^{-1}=A$

We know that if $A^{-1}$ exist then $(cA)^{-1}=\dfrac{1}{c}A^{-1}$ where c is  a constant
Hence $(-A)^{-1}=\dfrac{1}{-1}A^{-1}=-A^{-1}$

We know that $|A^{n}|=|A|^{n}$  n be any integer
$\Rightarrow |A^{-1}|=|A|^{-1}=\displaystyle \frac{1}{|A|}$

We know, $(A^{n})^{-1}=(A^{-1})^{n}$
So, $(A^{2})^{-1}=(A^{-1})^{2}$
Hence, option A is correct.

We know that inverse of transpose of matrix is equal to transpose of inverse of matrix
$(A^{-1})' =(A')^{-1}$
Hence, option B is correct

For option C,
In the LHS, there is a matrix and in RHS , its a determinant i.e. a single value.
So, option C is incorrect.

Inverse of unit matrix also unit matrix.

Ans: B

It is fundamental concept that, $(AB)^{-1}=B^{-1}A^{-1}$

We know, $A^{-1}=\dfrac{adjA}{|A|}$

We know that 

Given, $\displaystyle A^{2}-A+2I=0$

Since, A satisfies the equation
$\displaystyle x^{3}-5x^{2}+4x+\lambda =0$
$\Rightarrow A^{3}-5A^{2}+4A+\lambda=O$
$\Rightarrow A^{3}A^{-1}-5A^{2}A^{-1}+4AA^{-1}+\lambda A^{-1}=O$
$\Rightarrow A^{2}-5A+4I+\lambda A^{-1}=O$
So, $A^{-1}$ exists if $\lambda\ne 0$

We have $I + p + p^2 + ..... + p^n = O$  ----------$(1)$
Since p is a nonsingular matrix, p is invertible.
Multiplying both sides of (1) by $p^{-1}$, we get
$p^{-1} + I + Ip + ..... + p^{n - 1} I = O. p^{-1}$
or $ p^{-1} + I (1 + p + ...... + p^{n-1}) = O$
or $ p^{-1} = - I (I + p + p^2 + ..... + p^{n - 1}) = - I(-p^n) = p^n$

Given, $A^{-1} XA = B$

Since A satisfies the given equation, therefore ${ A }^{ 3 }-5{ A }^{ 2 }+4A+kI=0$

Given, $A^{3} = 0$
$\Rightarrow I-A^{3} = I$

We have $AB=BA=B'A'=\left( AB \right) '$

$A^2 + A - I = 0$

using the property $(AB)^{-1}=B^{-1}A^{-1}$

$det{\left( {A}^{-1} \right)} = \left( 1 \times 2 \right) - \left( -2 \right) \times \left( -2 \right) = 2 - 4 = -2$

Given, $A^{2}-A+2I = 0$

Given : $\displaystyle A=A^{2}+I$

Consider $n=2$

Assume that $A$ is non-singular, then $A^{-1}$ exists. Thus
$AB =0    \Rightarrow A^{-1}(AB) =(A^{-1}:A)B = 0$
$ \Rightarrow   IB =0$
$\therefore  B =0$. A contradiction.
$\Rightarrow$ A is singular, similarly B is also singular.
Hence, both A and B must be singular.

 Let $A$ be an invertible symmetric matrix.
 $\therefore AA^{-1}=A^{-1}:A=I _n$
$\Rightarrow  ( AA^{-1})'=(A^{-1}:A)'=(I _n)'$
$\Rightarrow   ( A^{-1})'A'=A'(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'A=A(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'=A^{-1}$    [inverse of a matrix is unique]
i.e $A^{-1}$ is symmetric.
Hence, option A.

$A^{-1}=\begin{bmatrix}1 &0 \-1 &1 \end{bmatrix}$
$A^2=\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}=\begin{bmatrix}1 &0 \2 &1 \end{bmatrix}$
$A^{-2}=\begin{bmatrix}1 &0 \-2 &1 \end{bmatrix}$
$\Rightarrow A^{-n}=\begin{bmatrix}1 &0 \-n &1 \end{bmatrix}$
$\displaystyle \frac{1}{n} A^{-n}=\begin{bmatrix}1/n &0 \-1 &1/n \end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n}A^{-n}=\begin{bmatrix} 0&0 \-1 &0 \end{bmatrix}$
and $\displaystyle \frac{1}{n^2} A^{-n}=\begin{bmatrix}1/n^2 &0 \-1/n &1/n ^2\end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n^2}A^{-n}=\begin{bmatrix} 0&0 \0 &0 \end{bmatrix}$
Hence, options A,B and C.

$A$ and $B$ are $3\times 3$ matrices and $|A|\neq 0$
1.$|AB|=|A||B|=0$ $\Rightarrow |B|=0$
2.$|AB|=|A||B|\neq 0$ $\Rightarrow |B|\neq 0$
3.$AA^{-1}=I$ $\Rightarrow |A||A^{-1}|=1$
$\therefore |A^{-1}|=|A|^{-1}$
4.$|2A|= 8|A|$    ($\because  |kA|=k^n|A|$)
Hence, options A,B and C.

 $A =\begin{bmatrix}a &b \c &d \end{bmatrix}$ such that $A$ satisfies the relation $A^2- (a + d)A = 0$
$\Rightarrow A^2-(a+d)A=0$
$\Rightarrow \begin{bmatrix}a &b \c &d \end{bmatrix}\begin{bmatrix}a &b \c &d \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}a^2+bc &ab+bd \ac+cd &bc+d^2 \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow a^2+bc-a^2-ad=0$
$\Rightarrow ad-bc=\begin{vmatrix}a &b \c &d \end{vmatrix}=0$
$\therefore$ Inverse of A doesnot exist.
Hence, option D.

$A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$

$A=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ A }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},$

$AA^{-1}=I$
So, $|AA^{-1}=|I|=1$
$|A||A^{-1}=1|$
So, $det A^{-1}=\dfrac{1}{|A|}=\dfrac{1}{2}$

$det(A _{3 \times 3})=11$
$AA^{-1}=I$
$\therefore det(A)=\frac{1}{det(A^{-1})}$
$\therefore det(A^{-1})=\frac{1}{11}$

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

$A^{-1}B^{-1}=(BA)^{-1}=(AB)^{-1}$ since $A,B$ commute
$\Rightarrow A^{-1}B^{-1}=B^{-1}A^{-1}$
Hence $A^{-1}, B^{-1}$ also commute

$ABA^{-1}=X$
$\Rightarrow (ABA^{-1})(ABA^{-1})=XX=X^2$
$\Rightarrow ABAA^{-1}BA^{-1} =X^2$
$\Rightarrow ABBA^{-1} =X^2[\because AA^{-1}=I]$
$\Rightarrow AB^2A^{-1} =X^2$
Pre and post multiplying both sides with $A^{-1}$ and $A$ respectively we get,
$\Rightarrow B^2 =A^{-1}X^2A$

${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$

Given $A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}$
The characteristic equation of $A$ is given by 
$|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda  & 0 & 0 \ 0 & 1-\lambda  & 1 \ 0 & -2 & 4-\lambda  \end{vmatrix}=0$