Tag: elementary transformations of a matrix

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Questions Related to elementary transformations of a matrix

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

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 =

  1. $Icos\theta +Jsin\theta $
  2. $Icos\theta -Jsin\theta $
  3. $Isin\theta +Jcos\theta $
  4. $-Icos\theta +Jsin\theta $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $I=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, J=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

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

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

$=I\cos\theta +J\sin\theta$.
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 business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

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

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

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given I + P + P^2 + ... + P^n = 0, we have P + P^2 + ... + P^n = -I. Factoring out P gives P(I + P + ... + P^(n-1)) = -I. This is a standard identity for geometric series of matrices.

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

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

  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] $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If AX = B, then A^-1 = XB^-1. Here X is the matrix on the left and B is the matrix on the right. Calculating B^-1 and multiplying by X yields the inverse.

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

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

  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. $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

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

A is an involuntary matrix given by $A=\begin{bmatrix} 0 & 1 & -1\ 4 & -3 & 4\ 3 & -3 & 4\end{bmatrix}$ then the inverse of $\dfrac{A}{2}$ will be?

  1. $2A$
  2. $\dfrac{A^{-1}}{2}$
  3. $\dfrac{A}{2}$
  4. $A^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An involuntary matrix satisfies A^2 = I, which means A = A^-1. If we want the inverse of (A/2), we use the property (kA)^-1 = (1/k)A^-1. Thus, (A/2)^-1 = 2A^-1 = 2A.

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

If $A\begin{bmatrix} 1 & 1\ 2 & 0\end{bmatrix}=\begin{bmatrix} 3 & 2\ 1 & 1\end{bmatrix}$, then $A^{-1}$ is given by?

  1. $\begin{bmatrix} 0 & -1\\ 2 & -4\end{bmatrix}$
  2. $\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$
  3. $\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let X = [[1, 1], [2, 0]] and B = [[3, 2], [1, 1]]. We have AX = B, so A = BX^-1. Then A^-1 = (BX^-1)^-1 = XB^-1. Calculating B^-1 and multiplying by X gives the result.