Properties of inverses of matrices - class-XII

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 $A^{-1}=A$

Given A is non-singular and AA' = A'A, we have B = A^-1 A'. Then BB' = (A^-1 A')(A^-1 A')'. Since (A^-1 A')' = (A')'(A^-1)' = A(A^-1)' = A(A')^-1, the expression simplifies based on the commutativity of A and A'.

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

The matrix A(alpha, beta) is a rotation matrix combined with a scaling factor e^beta. The inverse of a rotation matrix is obtained by replacing alpha with -alpha, and the inverse of the scaling factor e^beta is e^-beta.

The matrix is non-invertible if its determinant is zero. Given the structure and the roots of unity, the determinant evaluates to zero only under specific conditions related to the values of a, b, and c.

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 (A-B)C = BA^-1, we can multiply both sides by A on the right to get (A-B)CA = B. Rearranging or manipulating the equation leads to C(A-B) = A^-1 B.

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 $\lambda\ne 0$

A matrix has no inverse if its determinant is zero. Setting the determinant of the given matrix to zero results in an equation that solves for x = 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, $A^{2}-A+2I = 0$

The matrix is non-invertible if its determinant is zero. Calculating the determinant of the given matrix and setting it to zero leads to a trigonometric equation that yields the specified value for theta.

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

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

Given ABC = D, then C = (AB)^-1 * D. Calculate AB first, then find its inverse, and multiply by the matrix D.

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

The property (Adj A)^-1 = Adj(A^-1) is a standard identity in matrix algebra.

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

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

$A^{-1}B^{-1}=(BA)^{-1}=(AB)^{-1}$ 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 
$|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda  & 0 & 0 \ 0 & 1-\lambda  & 1 \ 0 & -2 & 4-\lambda  \end{vmatrix}=0$