Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Inverse of $\begin{bmatrix} -1 & 5 \ -3 & 2 \end{bmatrix}$ is

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix} 2/13 & -5/13 \ 3/13 & -1/13 \end{bmatrix}$

  2. $\begin{bmatrix} -2/13 & 5/13 \ -3/13 & 1/13 \end{bmatrix}$

  3. $\begin{bmatrix} 2 & -5 \ 3 & -1 \end{bmatrix}$

  4. $Cannot\ be\ determined$

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Correct Option: A

Consider three matrices $A=\begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}, B=\begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A)+tr\left(\dfrac{ABC}{2}\right)+tr\left(\dfrac{A(BC)^{2}}{4}\right)+tr\left(\dfrac{A(BC)^{3}}{8}\right)+....+\infty$ is 

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $6$

  2. $9$

  3. $12$

  4. $3$

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Correct Option: B

If A is a 2 X 2 matrix such that $A^2009 + A^2008$= I, then : $(A^2008)^-1$= 

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $A^2008 + I$

  2. $A^2009 + 1$

  3. A + I

  4. A

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Correct Option: A

If $I=I=\left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 0 \ 1 \end{matrix} \right] ,j=\left[ \begin{matrix} 0 \ -1 \end{matrix}\begin{matrix} 1 \ 0 \end{matrix} \right] and B=\left[ \begin{matrix} cos\theta  \ -sin\theta  \end{matrix}\begin{matrix} sin\theta  \ cos\theta  \end{matrix} \right] ,$ then B =

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $Icos\theta +Jsin\theta $

  2. $Icos\theta -Jsin\theta $

  3. $Isin\theta +Jcos\theta $

  4. $-Icos\theta +Jsin\theta $

Show answer
Correct Option: A
Explanation:
Given, $I=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, J=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

and $B=\begin{bmatrix} \cos \theta &\sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$=\cos\theta\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} +\sin\theta \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

$=I\cos\theta +J\sin\theta$.

If $A(\theta) = \begin{bmatrix}\sin  \theta & i  \cos  \theta\ i  \cos  \theta & \sin  \theta\end{bmatrix}$, then which of the following is not true?

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $A(\theta)^{-1} = A(\pi - \theta)$

  2. $A(\theta) + A(\pi + \theta)$ is a null matrix

  3. $A(\theta)$ is invertible for all $\theta \in R$

  4. $A(\theta)^{-1} = A(- \theta)$

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Correct Option: A,B,C
Explanation:

Finding inverse of the matrix $A(\theta)= \begin{bmatrix} \sin\theta & i\cos\theta \ i\cos\theta & \sin\theta\end{bmatrix}$


Determinant of $A(\theta)$ is $|A(\theta)|=\sin^2\theta-i^2\cos^2\theta$
                                                    $= \sin^2\theta+\cos^2\theta$
                                                    $=  1$

Therefore $A(\theta)$ is a non-singular matrix. So , it is invertible of all $\theta \in R$

$A(\theta)^{-1} = \begin{bmatrix} \sin\theta & -i\cos\theta \-i\cos\theta & \sin\theta \end{bmatrix}$

Now. $A(\pi -\theta)=\begin{bmatrix} \sin(\pi-\theta) & i\cos(\pi-\theta) \i\cos(\pi-\theta) & \sin(\pi-\theta) \end{bmatrix}$
                          $=\begin{bmatrix} \sin\theta  & -i\cos\theta \ -i\cos\theta & \sin\theta \end{bmatrix}$
                          $= A(\theta)^{-1}$

Now, $A(\pi+\theta)= \begin{bmatrix} \sin(\pi+\theta) & i\cos(\pi+\theta) \i\cos(\pi+\theta) & \sin(\pi+\theta) \end{bmatrix} $
                          $= \begin{bmatrix} -\sin\theta & -i\cos\theta \-i\cos\theta & -\sin\theta \end{bmatrix}$
                          $= -A(\theta)$

Therefore, $A(\theta) + A(\pi+\theta)=0$.

Hence, the correct options are $(A), (B)$ and $(C)$.

Write the following transformation in matrix form
$\quad x _1 = \displaystyle\frac{\sqrt 3}{2}y _1 + \displaystyle\frac{1}{2}y _2; \quad x _2 = -\displaystyle\frac{1}{2}y _1 + \displaystyle\frac{\sqrt 3}{2}y _2$.
Hence find the transformation in matrix form which expresses $y _1, y _2$ in terms of $x _1, x _2$.

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 + \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 + \displaystyle\frac{\sqrt 3}{2}x _2$

  2. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 - \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 + \displaystyle\frac{\sqrt 3}{2}x _2$

  3. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 - \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 - \displaystyle\frac{\sqrt 3}{2}x _2$

  4. None of these

Show answer
Correct Option: B
Explanation:

$ \displaystyle  { x } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { y } _{ 1 }+\frac { 1 }{ 2 } { y } _{ 2 }  $ and $\displaystyle { x } _{ 2 }=\frac { -1 }{ 2 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { y } _{ 2 } $ 
We observe $ \displaystyle \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }=\frac { 3 }{ 4 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } .\frac { 1 }{ 2 } { y } _{ 2 }+\frac { 1 }{ 4 } { y } _{ 1 }-\frac { \sqrt { 3 }  }{ 2 } \frac { 1 }{ 2 } { y } _{ 2 } $
$ \displaystyle \Rightarrow \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }={ y } _{ 1 } $
Similarly $ \displaystyle \frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }=\frac { 1 }{ 4 } { y } _{ 2 }+\frac { 3 }{ 4 } { y } _{ 2 }={ y } _{ 2 } $
$ \displaystyle \therefore { y } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 };{ y } _{ 2 }=\frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }  $ 

Let p be a non-singular matrix, $1+p+p^{2}+....+p^{n}=0$ (0 denotes the null matrix) then $p^{-1}=$

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $p^{n}$

  2. -$p^{n}$

  3. -(1+p+...+$p^{n}$)

  4. none

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Correct Option: A

Let A be a $3 \times 3$  matrix such that is: $A\left[ \begin{matrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{matrix} \right]  $Then $A^{-1}$ is

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\left[ \begin{matrix} 0 & 1 & 3 \ 0 & 2 & 3 \ 1 & 1 & 1 \end{matrix} \right] $

  2. $\left[ \begin{matrix} 3 & 2 & 1 \ 3 & 2 & 0 \ 1 & 1 & 0 \end{matrix} \right] $

  3. $\left[ \begin{matrix} 1 & 2 & 3 \ 0 & 1 & 1 \ 0 & 2 & 3 \end{matrix} \right] $

  4. $\left[ \begin{matrix} 3 & 1 & 2 \ 3 & 0 & 2 \ 1 & 0 & 1 \end{matrix} \right] $

Show answer
Correct Option: A

Use the method of elementary row transformation to compute the inverse of 
$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

  2. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$

  3. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$

  4. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Show answer
Correct Option: A
Explanation:

Let $\quad A = \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

$\Rightarrow \quad Write \space A A^{-1}= I$

$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix} A^{-1}= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _{21}(-2)\ \mbox{~}\ R _{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ -2 & 1 & 0 \ 1 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _2(-1) \ \mbox{~} \ R _3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \ 0 & 1 & 9 \ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _{12}(-2) \ \mbox{~} \ R _{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _3(-1/7)\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

$\quad \begin{matrix}R _{13}(13) \ \mbox{~} \ R _{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Hence, $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

If $
A=\left[ \begin{array}{ll}{x} & {1} \ {1} & {0}\end{array}\right]
 $ and $
A^{2}=I
 $, $
A^{-1}
 $ is equal to ...............

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $

    \left[ \begin{array}{ll}{0} & {1} \ {1} & {0}\end{array}\right]

    $

  2. $

    \left[ \begin{array}{ll}{1} & {0} \ {0} & {1}\end{array}\right]

    $

  3. $

    \left[ \begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}\right]

    $

  4. $

    \left[ \begin{array}{ll}{0} & {0} \ {0} & {0}\end{array}\right]

    $

Show answer
Correct Option: A
Explanation:
$A=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
Given: ${A}^{2}=I$ where $I$ is $2\times 2$ identity matrix
Let us find ${A}^{2}$
$=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
$=\left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]$
Given ${A}^{2}=I$
$\Rightarrow \left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]=\left[\begin{matrix} 1 & 0 \\ 0 & 1  \end{matrix}\right]$
Equating,we get
${x}^{2}+x=1,x=0$
Put $x=0$ in $A$
$A=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
We have ${A}^{2}=I$
Pre-multiply ${A}^{-1}$ both sides,we get
${A}^{-1}{A}^{2}={A}^{-1}I$
$\Rightarrow A={A}^{-1}$
Hence,${A}^{-1}=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$