Non singular matrix - class-XII

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

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
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

For a matrix A with integer entries, the inverse is given by (1/det(A)) * adj(A). If det(A) = +/- 1, then 1/det(A) is +/- 1. Since the adjugate matrix consists of determinants of integer submatrices, it also contains only integers. Thus, A^-1 has only integer entries.

$AB=O$
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|  } $

Given AB = BA + A, we can rewrite this as AB - BA = A. This is a commutation relation. Using properties of determinants and the given condition, it can be shown that |B| = |I + B|.

If AB = A and BA = B, then A^2 = A(BA) = (AB)A = AA = A^2. Also, A = AB implies |A| = |A||B|. If |A| is not 0, then |B| = 1. However, if A and B are singular, |A| = 0 and |B| = 0, which satisfies the equations.

A 3x3 matrix with four 1s and five 0s has a determinant that can be calculated by expanding along rows or columns. By testing various configurations, one finds that many such matrices are singular, but there are indeed more than 7 non-singular configurations possible.

For square matrices A and B, if AB = I, then A and B are inverses of each other. By the properties of inverse matrices, if AB = I, then BA = I must also hold.

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

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.

A matrix is singular if its determinant is 0. Calculating the determinant: 1(k*-4 - 3*-2) - k(3*-4 - 2*-2) + 3(33 - 2k) = 0. This simplifies to 1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0, which is -4k + 6 + 8k + 27 - 6k = 0, or -2k + 33 = 0, so k = 33/2.

A matrix is invertible if its determinant is non-zero. Determinant = 4(x+1) - (x+2)(2x-3) = 4x + 4 - (2x^2 + x - 6) = -2x^2 + 3x + 10. Setting this to 0: 2x^2 - 3x - 10 = 0. Using the quadratic formula, x = (3 +/- sqrt(9 - 42-10)) / 4 = (3 +/- sqrt(89)) / 4. Since these values are not in the options, 'none of these' is the correct choice.

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|  } $