$\begin{vmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}-0\begin{vmatrix} 2 & 0 \ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \ 3 & 1 \end{vmatrix}$

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
Therefore, it is a singular matrix.

$\begin{vmatrix} 1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6  \end{vmatrix}=0$

$\Rightarrow 1\begin{vmatrix} -1 & 7 \ 4 & -6 \end{vmatrix}-2\begin{vmatrix} 4 & 7 \ 2 & -6 \end{vmatrix}+x\begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix}=0$

$\Rightarrow (6-28)-2(-24-14)+x(16+2)=0$
$\Rightarrow -22+76+18x=0$
$\Rightarrow 18x=-54$

Given $AB =0$

For a matrix to be not invertible, determinant of the matrix should be $0$
So, $\left| \begin{matrix} \alpha  & 2 & 2 \ -3 & 0 & 4 \ 1 & -1 & 1 \end{matrix} \right| =0$

For matrix

$AB=O$
taking determinant on both sides
$|AB|=|A||B|=0$
$\Rightarrow$ Either $A$ or $B$ should be a singular matrix.
Hence, option D.

${ A }^{ -1 }=\frac { adj\quad A }{ \left| A \right|  } $

$\begin{array}{l} { \lambda _{ 1 } }=3,\, \, { \lambda _{ 2 } }=-2 \ \left| A \right| =4 \ adj\left( A \right) =\left( A \right) \cdot { A^{ -1 } } \ Ax=\lambda x \ \frac { 1 }{ \lambda  } x={ A^{ -1 } }x \ \left( { \lambda I\cdot { A^{ -1 } } } \right) =\frac { { \left( A \right) \cdot \left( { \lambda I-{ A^{ -1 } } } \right)  } }{ { \left( A \right)  } }  \ exigent\, value\, of\, adj\left( A \right) \, is\, \frac { { \left( A \right)  } }{ { exigent\, value\, of\, A } }  \ =\frac { 4 }{ 3 } ,\frac { 4 }{ { -8 } }  \ =\left( { \frac { 4 }{ 3 } ,-2 } \right)  \ Hence,\, option\, B\, is\, correct\, answer. \end{array}$

Matrix A is invertible if $|A| \neq 0$, i.e.,
$\begin{bmatrix}1 & 0 & -K\ 2 & 1 & 3\ K & 0 & 1\end{bmatrix} \neq 0$
or $1(1) - K (-K) \neq 0$
Expanding along second column
$|A| =-0+1(1-(-K)(K))=1+K^2 \neq 0$ which is true for all real K.
Hence, A is invertible for all real values of K.

The inverse of a matrix exists if its determinant is not equal $0$.
For option A, 
Let $A=\begin{bmatrix}1 & {\omega} \ {\omega} & {\omega}^2\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option B, 
Let $A=\begin{bmatrix}{\omega}^{2} & 1 \ 1 & {\omega}\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

For option C, 
Let $A=\begin{bmatrix}{\omega} & {\omega}^{2} \ {\omega}^{2} & 1\end{bmatrix}$
Here, $|A|=0$
Hence, inverse does not exists.

Hence, the inverse does not exist for any of the given matrices

It is a $3 \times 3$ so it is a square matrix,

Given, the matrix $A$ is singular.

Let A be a skew-symmetric matrix of order n.

Given,  $(A-3I)(A-5I)=O$ for a $3\times 3$ non-singular matrix.

Given A is a square matrix with all entries as integer
(a) Now $\displaystyle \left|A\right|\neq 1,-1$ mean it may be zero also $\displaystyle \left|A\right| =0,$then $\displaystyle A^{-1}$ does'not exist $\displaystyle \therefore $ choice (a) is false.

(b) If $\displaystyle \left|A\right| =1,-1$ then $\displaystyle A^{-1}$ certainly exist but A is a square matrix with all integral entries so all cofactors are integers .So adj. A matrix has all integral entries .
$\displaystyle A^{-1}=\frac{1}{\left|A\right| }(adj A)=\pm(adj A) $
$\therefore $ choice (b) is correct.

(c) $\displaystyle \left|A\right| =1,-1 $ $\displaystyle \therefore $ $\displaystyle A^{-1}$ must exist but given $\displaystyle A^{-1}$ does not exist which is false result $\displaystyle \therefore $  choice (c) is incorrect

(d) $\displaystyle \left|A\right| =1,-1$ it is true that $\displaystyle A^{-1}$ exist but we have to follow that A has all integral entries but choice (d) says that it is not necessarily that entries are integer which mean adj. A, may or may not be with integral entries. Hence choice (d) is false.

Let the characteristic root of $A$ be $\lambda $.

$AB=AC\Rightarrow B=C$

Let $B=A+{ I } _{ n }$.

$\Delta =\begin{vmatrix} a & b & (a\alpha -b) \ b & c & (b\alpha -c) \ 2 & 1 & 0 \end{vmatrix}\ =2\left( b(b\alpha -c)-c(a\alpha -b) \right) -1\left( a(b\alpha -c)-b(a\alpha -b) \right)$

Solution:

Given $A=\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is a singular matrix.

$\begin{array}{l} { { Sin } }gular\, \, means \ \left| A \right| =0 \ \left( { \begin{array} { *{ 20 }{ c } }0 & x & { 16 } \ x & 5 & 7 \ 0 & 9 & x \end{array} } \right) =0 \ 0\left( { 51-63 } \right) -x\left( { { x^{ 2 } } } \right) +16\left( { 9x } \right) =0 \ -{ x^{ 3 } }+144x=0 \ { x^{ 3` } }-144x=0 \ x\left( { { x^{ 2 } }-144 } \right) =0 \ x=0 \ { x^{ 2 } }-144=0 \ { x^{ 2 } }=144 \ x=\pm 12 \ x=0,\pm 12 \end{array}$

$\displaystyle A=\begin{bmatrix} \frac{1}{2}\left ( e^{ix}+ e^{-ix}\right )&\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) \\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) &\frac{1}{2}\left ( e^{ix}+ e^{-ix}\right ) \end{bmatrix}$

Let $B$ be non-singular, then ${ B }^{ -1 }$ exists.

$A=\left[ \begin{matrix} x+\lambda  & x & x \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right] $
${ A }^{ -1 }$ exists if $\left| A \right| \neq 0$
$\left| A \right| =\left| \begin{matrix} x+\lambda  & x & x \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right| \neq 0$
${ R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }+{ R } _{ 3 }$
$\left( 3x+\lambda  \right) \left| \begin{matrix} 1 & 1 & 1 \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right| \neq 0$
${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 }$ and ${ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$
$\left( 3x+\lambda  \right) \left| \begin{matrix} 1 & 0 & 0 \ x & \lambda  & 0 \ x & 0 & \lambda  \end{matrix} \right| \neq 0$
$\left( 3x+\lambda  \right) { \left[ { \lambda  }^{ 2 } \right]  }\neq 0$
$3x+\lambda \neq 0,\lambda \neq 0$

$adj\left| { A }^{ -1 } \right| =\frac { A }{ \left| A \right|  } $