Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

Let $A$ be an $n\times n$ matrix such that $A^n=\alpha A,$ where $\alpha$ is a real number different from $1$ and $-1$. Then, the matrix $A+I _n$ is

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

  1. singular

  2. non-singular, i.e., invertible

  3. scalar

  4. None of these

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Correct Option: B
Explanation:

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


Since $A=B-{ I } _{ n }$, the condition ${ A }^{ n }=\alpha A$ can be written in the form $\displaystyle { \left( { B-I } _{ n } \right)  }^{ n }\alpha \left( B-{ I } _{ n } \right) $

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n }{ I } _{ n }=\alpha B-\alpha { I } _{ n }$

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ C } _{ 2 }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n-1 }B-\alpha B=-\alpha { I } _{ n }-{ \left( -1 \right)  }^{ n }{ I } _{ n },$

$\displaystyle \Rightarrow B\left( { B }^{ n-1 }-^{ n }{ { C } _{ 1 }{ B }^{ n-2 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-3 }+...+{ \left( -1 \right)  }^{ n-1 }{ I } _{ n }-\alpha { I } _{ n } } \right) =\left[ { \left( -1 \right)  }^{ n+1 }-\alpha  \right] { I } _{ n }.$

Since $\displaystyle { \left( -1 \right)  }^{ n+1 }-\alpha \neq 0,\left[ \because \alpha \neq \pm 1 \right] $

$\therefore$ $B$ is invertible 

Matrix $\begin{bmatrix}a & b &(a\alpha -b) \b  & c & (b\alpha -c)\2 & 1 & 0\end{bmatrix}$ is non invertible if 

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

  1. $\alpha = 1/2$

  2. a, b, c are in A.P.

  3. a, b, c are in G.P.

  4. a, b, c are in H.P.

Show answer
Correct Option: A,C
Explanation:

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

 
Expanding along $R _3$

$ =2{ b }^{ 2 }\alpha -2bc-2ac\alpha +2bc-ab\alpha +ca+ab\alpha -{ b }^{ 2 }\\ =\left( { b }^{ 2 }-ac \right) \left( 2\alpha -1 \right) $

Matrix is non invertible when $\Delta =0$
i.e $\displaystyle\alpha=\frac{1}{2}$ and $a,b,c$ are in $G.P$

If $\left |\begin{matrix}1 & -1 &x \ 1 & x & 1\ x & -1 & 1\end{matrix} \right|$ has no inverse, then the real value of $x$ can be is

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

  1. 2

  2. 3

  3. 0

  4. 1

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Correct Option: D
Explanation:
$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}$
No inverse
$\Rightarrow \left| A \right| =0$
$x+1+1(1-x)+x(-1-{ x }^{ 2 })=0$
$x+1+1-x-x-{ x }^{ 3 }=0$
${ x }^{ 3 }-x-2=0$
$\Rightarrow { x }^{ 3 }+2x-x-2=0$
$\left( { x }^{ 2 }-1 \right) \left( x+2 \right) =0$
$\Rightarrow x=\pm 1,x=-2$
Real value of $x=1$
Option D

If $A$ and $B$ are any two matrices such that $AB = 0$ and $A$ is non-singular, then

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

  1. $B = 0$

  2. $B$ is singular

  3. $B$ is non-singular

  4. $B = A$

Show answer
Correct Option: A
Explanation:
Given that $AB = 0$, taking determinants on both sides, we have
$|A\cdot B| = 0$
$\therefore |A| \times |B| = 0$
Product of two numbers is zero if one of them is zero.
Since $A$ is non-singular matrix, $B$ has to be singular, meaning that its determinant has to be zero.
The matrix need not be zero.

If the matrix $\begin{bmatrix} -1& 3 &2 \1&k&-3\1&4&5\end{bmatrix}$ has an inverse then the values of $k$.

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

  1. $k$ is any real number

  2. $k = -4$

  3. $k \neq -4$

  4. $k \neq 4$

Show answer
Correct Option: C
Explanation:
For a matrix to be invertible, its determinant must not be zero.
$\therefore -1(5k + 12) - 3(5 - (-3)) + 2(4 - k) \neq 0$
$\therefore -5k - 12 - 24 + 8 - 2k \neq 0$
$\therefore -7k \neq 28$
$\therefore k \neq -4$

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.