Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

If the matrix $\begin{bmatrix} 0 & 2\beta & \Upsilon \ \alpha & \beta & -\Upsilon \ \alpha & -\beta & \Upsilon \end{bmatrix}$is orthogonal, then

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

  1. $\alpha = \pm\dfrac{1}{\sqrt{2}}$

  2. $\beta = \pm\dfrac{1}{\sqrt{6}}$

  3. $\gamma = \pm\dfrac{1}{\sqrt{3}}$

  4. all of these

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Correct Option: D
Explanation:
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}$
for orthogonal matrix we have 
$A.A^{T}=I$
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}\begin{bmatrix} 0 & \alpha  & \alpha  \\ 2\beta  & \beta  & -\beta  \\ \gamma  & -\gamma  & \gamma  \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 0+4{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \\ 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } \\ 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$, 
$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$        $\beta^{2}=\dfrac{r^{2}}{2}$
$r^{2}[3]=1$
$r=\pm \dfrac{1}{\sqrt{3}}$
$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$
$\beta^{2}=\dfrac{\gamma^{2}}{2},  \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$
$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$
$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$
$\beta=\pm \dfrac{1}{\sqrt{6}}$            $\alpha=\pm \dfrac{1}{\sqrt{2}}$

The inverse of the $\begin{bmatrix}- 1 & 5\ - 3 & 2\end{bmatrix}$ is

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

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

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

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

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

Show answer
Correct Option: A
Explanation:
$A=\left[\begin{matrix} -1 & 5 \\ -3  & 2 \end{matrix} \right]$

$\left|A\right|=-2+15=13\neq 0$

Hence ${A}^{-1}$ exists.

${C} _{ij}={\left(-1\right)}^{i+j}{M} _{ij}$

${C} _{11}=2,\,{C} _{12}=3,\,{C} _{21}=-5$  and ${C} _{22}=-1$

${C} _{ij}=\left[\begin{matrix} 2 & 3 \\ -5  & -1 \end{matrix} \right]$

Adj${A}={{C} _{ij}}^{T}=\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$

${A}^{-1}=\dfrac{Adj{\left(A\right)}}{\left|A\right|}=\dfrac{1}{13}\left[\begin{matrix} 2 & -5 \\ 3  & -1 \end{matrix} \right]$

The inverse of the matrix $\begin{bmatrix} 5 & -2 \ 3 & 1 \end{bmatrix}$ is 

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

  1. $\dfrac { 1 }{ 11 } \begin{bmatrix} 1 & 2 \ -3 & 5 \end{bmatrix}$

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

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

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

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} A=\left[ \begin{array}{l} 5\, \, \, \, \, -2 \ 3\, \, \, \, \, \, \, \, \, 1 \end{array} \right]  \ \left| A \right| =5+6=11\ne 0 \ so,\, A\, \, is\, \, \, non-\sin  gular\, ,\, { A^{ -1 } }\, \, is\, \, exist \ so,m\, { A _{ 11 } }=1,\, \, \, \, \, { A _{ 12 } }=-3,\, \, \, \, { A _{ 21 } }=2,\, \, \, \, \, \, { A _{ 22 } }=5 \ A=\left( { \begin{array} { *{ 20 }{ c } }1 & { -3 } \ 2 & 5 \end{array} } \right) \Rightarrow AdjA=\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA\, \, \, \, \Rightarrow \, \, \, \, \frac { 1 }{ { 11 } } \left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \end{array}$


Hence, this is the answer.

Let $P(x, y)$ be any given point and $\displaystyle P(x _{1},y _{1})$ be the image of $P(x,y)$ after reflection.
The matrix of reflection of point $P$ through the line $\displaystyle y = x $ is given by

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

  1. $\displaystyle \begin{bmatrix}1 &0 \0 &1\end{bmatrix}$

  2. $\displaystyle \begin{bmatrix}-1 &0 \0 &-1\end{bmatrix}$

  3. $\displaystyle \begin{bmatrix}-1 &0 \0 &1\end{bmatrix}$

  4. $\displaystyle \begin{bmatrix}0 &1 \1&0\end{bmatrix}$

Show answer
Correct Option: D
Explanation:

Reflection of matrix $P(x,y)$ through the line $y=mx$ making an angle $\theta$ with $x-$axis is

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

Given line is $y=x$ which makes an angle of $45^{\circ}$ with $x-$axis

Hence, the transformation matrix is $\begin{bmatrix} 0&1\1&0\end{bmatrix}$

What is the inverse of the matrix
$A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  & 0 \ -\sin { \theta  }  & \cos { \theta  }  & 0 \ 0 & 0 & 1 \end{bmatrix}$ ?

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

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

  2. $\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \ 0 & 1 & 0 \ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$

  3. $\begin{bmatrix} 1 & 0 & 0 \ 0 & \cos { \theta } & -\sin { \theta } \ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$

  4. $\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \ -\sin { \theta } & \cos { \theta } & 0 \ 0 & 0 & 1 \end{bmatrix}$

Show answer
Correct Option: A
Explanation:
$A = \begin{bmatrix} \cos \theta &  \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Calculate first minors.
$M _{11} = \cos \theta , M _{13} = 0, M _{22} = \cos \theta$
$M _{12} = -\sin \theta, M _{21} = \sin \theta, M _{23} = 0$
$M _{31} = 0, M _{32} = 0, M _{33} = \cos^{2}\theta + \sin^{2}\theta = 1$
Cofactor Matrix $= \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{bmatrix} = C$
$det|A| = \cos^{2}\theta + \sin^{2}\theta = 1$

$adj (A) = C^{T} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{bmatrix}$

$A^{-1} = \dfrac {adj(A)}{(A)} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0 & 1\end{bmatrix}$.