Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

If $A^{-1}=\begin{bmatrix} 1 & -2 \ -2 & 2 \end{bmatrix}$, then what is $det(A)$ equal to ?

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $2$

  2. $-2$

  3. $1/2$

  4. $-1/2$

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Correct Option: D
Explanation:

$det{\left( {A}^{-1} \right)} = \left( 1 \times 2 \right) - \left( -2 \right) \times \left( -2 \right) = 2 - 4 = -2$

As we know that,
$det{\left( {A}^{-1} \right)} = \cfrac{1}{det{\left( A \right)}}$
$\Rightarrow det{\left( A \right)} = \cfrac{1}{det{\left( {A}^{-1} \right)}} = \cfrac{1}{-2} = - \cfrac{1}{2}$

A square, non-singular matrix $A$ satifies $A^2 - A + 2I = 0$, then $A^{-1} = $

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $I - A$

  2. $\dfrac {(I - A) }{2}$

  3. $I + A$

  4. $\dfrac {(I + A)}{2}$

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Correct Option: B
Explanation:

Given, $A^{2}-A+2I = 0$

$\Rightarrow 2I = A-A^{2}$

$\Rightarrow 2A^{-1}I = A^{-1}A-A^{-1}A^{2}$

$\therefore A^{-1} = \displaystyle\frac{I-A}{2}$

If matrix $A=\left| \begin{matrix} sin\theta  & cosec\theta  & 1 \ cosec\theta  & 1 & sin\theta  \ 1 & sin\theta  & cosec\theta  \end{matrix} \right| $ a non invertible matrix. then possible value of $\theta$ is-

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $n\pi+(-1)^n\dfrac{\pi}{4}$

  2. $n\pi+(-1)^n\dfrac{\pi}{3}$

  3. $n\pi+(-1)^n\dfrac{\pi}{6}$

  4. $2n\pi+\dfrac{\pi}{2}$

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Correct Option: A

If $A$ be a $3\times 3$ matrix and $I$ be the unit matrix of that order such that $\displaystyle A=A^{2}+I$ then $A^{-1}$ is equal to

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $A$

  2. $A+I$

  3. $I-A$

  4. $A-I$

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Correct Option: C
Explanation:

Given : $\displaystyle A=A^{2}+I$

$\Rightarrow A^{2}-A+I=O$

$\Rightarrow A^{2}A^{-1}-AA^{-1}+IA^{-1}=O$

$\Rightarrow A-I+A^{-1}=O$

$\Rightarrow A^{-1}=I-A$

If $A$ is a square matrix, $B$ is a singular matrix of same order, then for a positive integer $n,(A^{-1}BA)^n$ equals

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $A^{-n}B^nA^n$

  2. $A^nB^nA^{-n}$

  3. $A^{-1}B^nA$

  4. $n(A^{-1}BA)$

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Correct Option: C
Explanation:

Consider $n=2$


${ \left( { A }^{ -1 }BA \right)  }^{ 2 }=\left( { A }^{ -1 }BA \right) \left( { A }^{ -1 }BA \right) =\left( { A }^{ -1 }B \right) \left( { A }^{ -1 }A \right) \left( BA \right) ={ A }^{ -1 }{ B }^{ 2 }A$

Again for $n=3$ we have ${ \left( { A }^{ -1 }BA \right)  }^{ 3 }=\left( { A }^{ -1 }{ B }^{ 2 }A \right) \left( { A }^{ -1 }BA \right) ={ A }^{ -1 }{ B }^{ 3 }A$

Thus generalizing the case 

${ \left( { A }^{ -1 }BA \right)  }^{ n }={ A }^{ -1 }{ B }^{ n }A$

If $A$ is a scalar matrix with scalar $k \neq 0$, of order $3$, then $kA^{-1}$ is:

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $\dfrac{1}{k}I$

  2. $\dfrac{1}{k^2}I$

  3. ${k^2}I$

  4. $\dfrac{1}{k^3}I$

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Correct Option: C
Explanation:
It is well-known that if you find an inverse for a matrix, that inverse matrix will be unique.
So what we have to do is to show that $\left(k{A}^{−1}\right)⋅\left(kA\right)=Id=\left(kA\right)⋅\left(k{A}^{−1}\right)$.
We have $\left(k{A}^{−1}\right).\left(kA\right)=\left(kk\right)⋅\left({A}^{−1}.A\right)=Id=Id$ and $\left(kA\right)⋅\left(k{A}^{−1}\right)=\left(kk\right)⋅\left(A{A}^{−1}\right)=Id$.
So by the uniqueness of the inverse matrix we have that $k{A}^{−1}$ is the inverse of the matrix $kA$.

If $A$ and $B$ are two non-zero square matrices of the same order such that the product $AB=0$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. both A and B must be singular

  2. exactly one of them must be singular

  3. atleast one of them must be non-singular

  4. none of these

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Correct Option: A
Explanation:

Assume that $A$ is non-singular, then $A^{-1}$ exists. Thus
$AB =0    \Rightarrow A^{-1}(AB) =(A^{-1}:A)B = 0$
$ \Rightarrow   IB =0$
$\therefore  B =0$. A contradiction.
$\Rightarrow$ A is singular, similarly B is also singular.
Hence, both A and B must be singular.


The inverse of a symmetric matrix (if it exists) is

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. a symmetric matrix

  2. a skew symmetric matrix

  3. a diagonal matrix

  4. none of these

Show answer
Correct Option: A
Explanation:

 Let $A$ be an invertible symmetric matrix.
 $\therefore AA^{-1}=A^{-1}:A=I _n$
$\Rightarrow  ( AA^{-1})'=(A^{-1}:A)'=(I _n)'$
$\Rightarrow   ( A^{-1})'A'=A'(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'A=A(A^{-1})'=(I _n)$
$\Rightarrow   ( A^{-1})'=A^{-1}$    [inverse of a matrix is unique]
i.e $A^{-1}$ is symmetric.
Hence, option A.

Let $A=\begin{bmatrix} 1&0 \1 &1 \end{bmatrix}$ then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $A^{-n}=\begin{bmatrix} 1&0 \-n &1 \end{bmatrix}\forall : n: \in: N$.

  2. $\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n}A^{-n}=\begin{bmatrix} 0&0 \-1 &0 \end{bmatrix}$

  3. $\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n^2}A^{-n}=\begin{bmatrix} 0&0 \0 &0 \end{bmatrix}$

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:

$A^{-1}=\begin{bmatrix}1 &0 \-1 &1 \end{bmatrix}$
$A^2=\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}=\begin{bmatrix}1 &0 \2 &1 \end{bmatrix}$
$A^{-2}=\begin{bmatrix}1 &0 \-2 &1 \end{bmatrix}$
$\Rightarrow A^{-n}=\begin{bmatrix}1 &0 \-n &1 \end{bmatrix}$
$\displaystyle \frac{1}{n} A^{-n}=\begin{bmatrix}1/n &0 \-1 &1/n \end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n}A^{-n}=\begin{bmatrix} 0&0 \-1 &0 \end{bmatrix}$
and $\displaystyle \frac{1}{n^2} A^{-n}=\begin{bmatrix}1/n^2 &0 \-1/n &1/n ^2\end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n^2}A^{-n}=\begin{bmatrix} 0&0 \0 &0 \end{bmatrix}$
Hence, options A,B and C.

If $A$ and $B$ are $3\times 3$ matrices and $|A|\neq 0$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $|AB|=0\Rightarrow |B|=0$

  2. $|AB|\neq 0\Rightarrow |B|\neq 0$

  3. $|A^{-1}|=|A|^{-1}$

  4. $|2A|=2|A|$

Show answer
Correct Option: A,B,C
Explanation:

$A$ and $B$ are $3\times 3$ matrices and $|A|\neq 0$
1.$|AB|=|A||B|=0$ $\Rightarrow |B|=0$
2.$|AB|=|A||B|\neq 0$ $\Rightarrow |B|\neq 0$
3.$AA^{-1}=I$ $\Rightarrow |A||A^{-1}|=1$
$\therefore |A^{-1}|=|A|^{-1}$
4.$|2A|= 8|A|$    ($\because  |kA|=k^n|A|$)
Hence, options A,B and C.