Tag: non singular matrix

Questions Related to non singular matrix

The matrix $A=\begin{bmatrix}1&3&2\1&x-1&1\2&7&x-3\end{bmatrix}$ will have inverse for every real number x except for

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $x=\dfrac{11\pm \sqrt{5}}{2}$

  2. $x=\dfrac{9\pm \sqrt{5}}{2}$

  3. $x=\dfrac{11\pm \sqrt{3}}{2}$

  4. $x=\dfrac{9\pm \sqrt{3}}{2}$

Show answer
Correct Option: A
Explanation:

Solution:

For $A$ have a inverse, $|A|\neq 0.$
$|A|={(x-1)(x-3)-7}+3{2-(x-3)}+2{7-2(x-1)}$
$=x^2-4x+3-7+6-3x+9+14-4x+4$
$=x^2-11x+29$
$=\left(x-\cfrac{11+\sqrt5}{2}\right)\left(x-\cfrac{11-\sqrt5}{2}\right)$
So, $|A|\neq0$
$x\neq\left(\cfrac{11\pm\sqrt5}{2}\right)$
i.e. for $x=\cfrac{11\pm\sqrt5}{2}$ matrix $A$ will not have its inverse.
Hence, A is the correct option.

If $A=\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular matrix and $x\in [-5, -2]$ then x=?$

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0$

  2. $-2$

  3. $-4$

  4. $0, -4$

Show answer
Correct Option: C
Explanation:

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


$\therefore |A|=0$

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

$\Rightarrow 3(-2+x+2)-(-1+x)(6-(x+3)(x+2))+2(-3+x+3)=0$

$\Rightarrow 3x-(x-1)(6-x^2-5x-6)+2x=0$

$\Rightarrow 5x-(x-1)(-x^2-5x)=0$

$\Rightarrow 5x-(-x^3-5x^2+x^2+5x)=0$

$\Rightarrow 5x+x^3+4x^2-5x=0$

$\Rightarrow x^3+4x^2=0$

$\Rightarrow x^2(x+4)=0$

$\Rightarrow x=0:and\;x=-4$

It is given that $x\in [-5,-2]$

$\therefore x=-4$

If $A=\begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$ is singular, then the possible values of $x$ are

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $0, \pm 1$

  2. $0, \pm 12$

  3. $0, \pm 5$

  4. $0, \pm 4$

Show answer
Correct Option: B
Explanation:

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


Hence, this is the answer.

If $\omega\neq 1$ is a cube root of unity, then


$A=\begin{bmatrix}1+2\omega ^{100}+\omega ^{200}&\omega ^2  &1  \1 &1+\omega ^{101}+2\omega ^{202}  &\omega  \\omega  & \omega ^2 &2+ \omega ^{100}+2\omega ^{200}\end{bmatrix}$

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $A$ is singular

  2. $|A|=0$

  3. $A$ is symmetric

  4. none of these

Show answer
Correct Option: A,B
Explanation:
$A=\begin{vmatrix} 1+{ \omega  }^{ 100 }+{ \omega  }^{ 200 } & { \omega  }^{ 2 } & 1 \\ 1 & 1+{ \omega  }^{ 101 }+{ \omega  }^{ 202 } & \omega  \\ \omega  & { \omega  }^{ 2 } & 2+{ \omega  }^{ 100 }+{ \omega  }^{ 200 } \end{vmatrix}$

$1+\omega +{ \omega  }^{ 2 }=0$

${ \omega  }^{ 100 }={ \omega  }^{ 99 }\times \omega =\omega $

${ \omega  }^{ 200 }={ \omega  }^{ 198 }\times { \omega  }^{ 2 }={ \omega  }^{ 2 }$

${ \omega  }^{ 101 }={ \omega  }^{ 2 }$${ \omega  }^{ 202 }={ \omega  }^{ 4 }=\omega $

$A=\begin{vmatrix} 1+2{ \omega  }+{ \omega  }^{ 2 } & { \omega  }^{ 2 } & 1 \\ 1 & 1+{ \omega  }^{ 2 }+{ 2\omega  } & \omega  \\ \omega  & { \omega  }^{ 2 } & 2+{ \omega  }+2{ \omega  }^{ 2 } \end{vmatrix}$

$A=\begin{vmatrix} { \omega  } & { \omega  }^{ 2 } & 1 \\ 1 & { \omega  } & \omega  \\ \omega  & { \omega  }^{ 2 } & -{ \omega  } \end{vmatrix}$

$A=\omega \begin{vmatrix} { \omega  } & { \omega  } & 1 \\ 1 & { 1 } & \omega  \\ \omega  & { \omega  } & -{ \omega  } \end{vmatrix}$

${ C } _{ 1 }\& { C } _{ 2 }$are same

$\left| A \right| =0$

$A\rightarrow $singular

$\triangle =0$
Option A and B

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

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. for all real $x$

  2. for positive real $x$ only

  3. for negative real $x$ only

  4. none of these

Show answer
Correct Option: A
Explanation:

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

$\Rightarrow A=\begin{bmatrix} coshx&sinhx\sinhx & coshx\end{bmatrix}$

$|A|=cosh^2x-sinh^2x=1$

$\therefore A^{-1}$ exists or all $x$

Hence, option A.

Let $A$ be a square matrix of order $n \times  n$. A constant $\lambda $ is said to be characteristic root of $A$ if there exists a $n \times  1$ matrix $X$ such that  $AX=\lambda X$

Let $P$ be a non-singular matrix, then which of the following matrices have the same characteristic roots.

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $A$ and $AP$

  2. $A$ and $PA$

  3. $A$ and $P^{-1} :AP$

  4. none of these

Show answer
Correct Option: C
Explanation:

Since $X\neq 0$ is such that $(A-\lambda I)X=0,:|A-\lambda 
I|=0\Leftrightarrow A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda= 0$, we get $|A|=0\Rightarrow A$ is singular.
We have $A^2 :X=A(AX) =A(\lambda X) = \lambda(AX)$
$=\lambda^2X$,
$A^3 :X=A(A^2X) =A(\lambda^2 X)$
$=\lambda^2(AX)=\lambda^2(\lambda X)=\lambda^3 X$
Continuing in this way, we obtain
$A" X=\lambda^": X: \forall : n\in N$
Also,$|P^{-1}:AP -\lambda I|=|P^{-1}(A -\lambda I)P|$
$=|P^{-1}|: |A -\lambda I|: |P|=|A -\lambda I|$

Let $A$ and $B$ be two non-null square matrices. If the product $AB$ is a null matrix, then

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $A$ is singular

  2. $B$ is singular

  3. $A$ is non-singular

  4. $B$ is non-singular

Show answer
Correct Option: A,B
Explanation:

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

Now, $AB=0($ given$) \Rightarrow \left( AB \right) { B }^{ -1 }=0{ B }^{ -1 }$
$($ post multiplying both sides by ${ B }^{ -1 })$
$\Rightarrow A\left( B{ B }^{ -1 } \right) =0$     $($ by associativity $)$
$\Rightarrow A{ I } _{ n }=0\quad \quad \quad \left( \because B{ B }^{ -1 }={ I } _{ n } \right) $
$\Rightarrow A=0$
But $A$ is a non-null matrix.
Hence $B$ is a singular matrix.
Similarly, it can be shown that $A$ is a singular matrix.

Let $A=\begin{bmatrix}x+\lambda& x&x\x &x+\lambda&x\x&x&x+\lambda  \end{bmatrix}$, then $A^{-1}$ exists if

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $x\neq 0$

  2. $\lambda \neq 0$

  3. $3x+\lambda \neq 0, \lambda \neq 0$

  4. $x\neq 0, \lambda \neq 0$

Show answer
Correct Option: C
Explanation:

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

If adj $B=A$ and $|P|=|Q|=1$, then $adj (\left( { Q }^{ -1 }{ BP }^{ -1 } \right)$ is equal ?

business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

  1. $APQ$

  2. $PAQ$

  3. $B$

  4. $A$

Show answer
Correct Option: A
Explanation:

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


$adj\left| { Q }^{ -1 }{ BP }^{ -1 } \right| =adj\left| { P }^{ -1 } \right| \ast adj\left| { B } \right| \ast adj\left| { Q }^{ -1 } \right| =\frac { P }{ \left| P \right|  } \ast A\ast \frac { Q }{ \left| Q \right|  } =PAQ$