Tag: inverse of a matrix and linear equations

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

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.

Let $A=$ $\begin{bmatrix}1&3&0\ 4&0&-2\ 2&6&0\end{bmatrix}$