Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

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

Show answer
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

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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.

If $A =\begin{bmatrix}a &b \c &d \end{bmatrix}$ such that $A$ satisfies the relation $A^2- (a + d)A = 0$, then inverse of $A$ is

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

  1. $I$

  2. $A$

  3. $(a + d)A$

  4. none of these

Show answer
Correct Option: D
Explanation:

 $A =\begin{bmatrix}a &b \c &d \end{bmatrix}$ such that $A$ satisfies the relation $A^2- (a + d)A = 0$
$\Rightarrow A^2-(a+d)A=0$
$\Rightarrow \begin{bmatrix}a &b \c &d \end{bmatrix}\begin{bmatrix}a &b \c &d \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}a^2+bc &ab+bd \ac+cd &bc+d^2 \end{bmatrix}-(a+d)\begin{bmatrix}a &b \c &d \end{bmatrix}=\begin{bmatrix}0 &0 \0 &0 \end{bmatrix}$
$\Rightarrow a^2+bc-a^2-ad=0$
$\Rightarrow ad-bc=\begin{vmatrix}a &b \c &d \end{vmatrix}=0$
$\therefore$ Inverse of A doesnot exist.
Hence, option D.

Let the matrix A and B be defined as $A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$ then the value of Det.$(2A^9B^{-1})$, is 

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

  1. $2$

  2. $1$

  3. $-1$

  4. $-2$

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Correct Option: D
Explanation:

$A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$

$|A| = \begin{vmatrix} 3 & 2 \ 2 & 1 \end{vmatrix} = -1$

$|B| = \begin{vmatrix} 3 & 1 \ 7 & 3 \end{vmatrix} = 2$

$\displaystyle |2A^9 B^{-1}| = 2^2|A|^9\frac{1}{|B|}$

                     $\displaystyle= 4\times (-1)\times \frac{1}{2}$

$\therefore |2A^9 B^{-1}|=-2$

Hence, option D.

If $P$ is a two-rowed matrix satisfying $P^T = P^{-1}$, then $P$ can be

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

  1. $\begin{bmatrix}cos\, \theta & -sin\, \theta \ -sin\,\theta & cos\, \theta \end{bmatrix}$

  2. $\begin{bmatrix}cos\, \theta & sin\, \theta \ -sin\,\theta & cos\, \theta \end{bmatrix}$

  3. $\begin{bmatrix}-cos\, \theta & sin\, \theta \ sin\,\theta & -cos\, \theta \end{bmatrix}$

  4. none of these

Show answer
Correct Option: B
Explanation:

$A=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ A }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},$

${ A }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ B }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix},$
${ B }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ C=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},{ C }^{ T }=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},$
${ C }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix}$
hence $P=B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix}$

Let A be an invertible matrix then which of the following is/are true

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

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

  2. $(A^2)^{-1} = (A^{-1})^2$

  3. $(A^T)^{-1} = (A^{-1})^T$

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:
Option A
$\left| { A }^{ -1 } \right| ={ \left| A \right|  }^{ -1 }$
$det\left( A \right) (det\left( B \right) )$
$d\left( A{ A }^{ -1 } \right) =detAdet\left( { A }^{ -1 } \right) $
$det\left( I \right) =1$
$\Rightarrow det\left( A \right) \ast det\left( { A }^{ -1 } \right) =I$
$det\left( { A }^{ -1 } \right) ={ \left( detA \right)  }^{ -1 }$

Option B:
A is invertible $A{ A }^{ -1 }={ A }^{ -1 }A=I$
$\Rightarrow { A }^{ 2 }$ is also invertible
${ \left( A{ A }^{ -1 } \right)  }^{ 2 }={ I }^{ 2 }$
${ A }^{ 2 }{ \left( { A }^{ -1 } \right)  }^{ 2 }=I$
${ \left( { A }^{ -1 } \right)  }^{ 2 }={ ({ A }^{ 2 }) }^{ -1 }$

Option C:
${ \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$
$\left( { A }^{ T } \right) { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 }A \right)  }^{ T }={ I }^{ T }=I$
Also,
${ \left( { A }^{ -1 } \right)  }^{ T }\left( { A }^{ T } \right) ={ \left( A{ A }^{ -1 } \right)  }^{ T }={ I }^{ T }=I$
${ A }^{ 1 }{ \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 } \right)  }^{ T }\left( { A }^{ T } \right) =I$
$\Rightarrow { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ T } \right)  }^{ -1 }$

Option A,B,C are correct