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

The number of $2\times 2$ matrices $A=\left[ \begin{matrix} a & b \ c & d \end{matrix} \right] $ for which ${ \left[ \begin{matrix} a & b \ c & d \end{matrix} \right]  }^{ -1 }$ $=\left[ \begin{matrix} \frac { 1 }{ a }  & \frac { 1 }{ b }  \ \frac { 1 }{ c }  & \frac { 1 }{ d }  \end{matrix} \right] $, $(a,b,c,d\ \epsilon \ R)$ is

  1. $0$
  2. $1$
  3. $2$
  4. $Infinite$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad { \begin{bmatrix} a & b \ c & d \end{bmatrix} }^{ -1 }=\begin{bmatrix} 1/a & 1/b \ 1/c & 1/d \end{bmatrix}$                          Not possible for any values of $a, b, c$

$A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad =Adj{ \begin{bmatrix} d & -c \ -b & a \end{bmatrix} }\ \Rightarrow { A }^{ -1 }=\begin{bmatrix} d & -c \ -b & a \end{bmatrix}$

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

Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=

  1. $z$
  2. $-z$
  3. $0$
  4. $2z$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\quad B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Given $AB=k\ B$
$AB\Rightarrow \begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\quad k=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
$\Rightarrow \begin{pmatrix} -5x & -8y & -7z \\ 3x & +5y & +4z \\ x2 & +3y & +3z \end{pmatrix}=\quad k\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
On adding all the elements on left and right side 
$\Rightarrow \ (-5x-8y-7z)+(3x+5y+47)+(2x+3y+3z)$
$=k(x+y+z)$
$\Rightarrow \ D=k(x+y+z)$
$k\neq 0$
$\Rightarrow \ x+y+z=0$
$\Rightarrow \ x+y=-z$

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

If $A=\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, find the values of $\alpha$ for which $A^2=B$.

  1. $\pm 1$
  2. $4$
  3. $0$

  4. No value
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$A^2=B$
$\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 +0 & 0+0 \ \alpha +1 & 0+1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 & 0 \ \alpha +1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\alpha^2=1$ and $\alpha +1=5$
$\alpha =\pm 1$ and $\alpha =4$, which is not possible.
Hence, there is no value of $\alpha$ for which $A^2=B$ is true.

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

Let $P=\begin{bmatrix} \cos { \dfrac { \pi  }{ 9 }  }  & \sin { \dfrac { \pi  }{ 9 }  }  \ -\sin { \dfrac { \pi  }{ 9 }  }  & \cos { \dfrac { \pi  }{ 9 }  }  \end{bmatrix}$ and $\alpha,\ \beta,\ \gamma$ be non-zero real numbers such that $\alpha P^{6}+\beta P^{3}+\gamma 1$ is the zero matrix. Then, $(\alpha^{2}+\beta^{2}+\gamma^{2})^{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$ is

  1. $\pi$
  2. $\dfrac {\pi}{2}$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Matrix P represents a rotation by pi/9. Raising P to the 6th and 3rd powers corresponds to rotations by 6pi/9 and 3pi/9 respectively. By analyzing the characteristic equation or evaluating the matrix powers, the coefficients alpha, beta, and gamma make the expression simplify such that the exponent product evaluates to 1.

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

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 

  1. $6$
  2. $9$
  3. $12$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given series involves traces of matrix products that form a geometric progression of matrices. By computing BC and evaluating the traces step-by-step, the sum converges to a total trace value of 6.

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

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?

  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. $A(\theta)^{-1} = A(- \theta)$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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 $(C)$.

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

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 $x _1, x _2$.

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }  $ 

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

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

  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]

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
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,${A}^{-1}=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If A and B are any $2\times2$ matrices, then det. (A+B) =0 implies

  1. None of these

  2. det A=0 and det B=0

  3. det A=0 or det B=0

  4. det A=0 + det B=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The determinant of a sum of matrices is not generally related to the sum of their determinants or the individual determinants being zero.