Tag: multiplication of matrices

Questions Related to multiplication of matrices

If $ \begin{bmatrix}
              2 & 1 \[0.3em]
              3 & 2
              \end{bmatrix} \ A \begin{bmatrix}
              -3 & 2 \[0.3em]
              5 & -3
              \end{bmatrix} = \begin{bmatrix}
              1 & 0 \[0.3em]
              0 & 1
              \end{bmatrix}$ then the matrix A is equal to

  1. $\begin{bmatrix}

    1 & 1 \[0.3em]

    1 & 0

    \end{bmatrix}$

  2. $\begin{bmatrix}

    1 & 1 \[0.3em]

    0 & 1

    \end{bmatrix}$

  3. $\begin{bmatrix}

    1 & 0 \[0.3em]

    1 & 1

    \end{bmatrix}$

  4. $\begin{bmatrix}

    0 & 1 \[0.3em]

    1 & 1

    \end{bmatrix}$


Correct Option: A
Explanation:

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

Pre multiplying by inverse of $1^{st}$ matrix on $LHS$
$A \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix} \, \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$
$A \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$
Post multiplying by $2^{nd}$ matrix.
$A = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix} \begin{bmatrix} -3 & -5 \ -2 & -3 \end{bmatrix}$
$= \begin{bmatrix} 1 &  1 \ 1 & 0 \end{bmatrix}$

Find the number of all possible ordered sets of two $(n\times n)$ matrices A and B for which $AB-BA=$$I$.

  1. Infinite

  2. $n^2$

  3. $n!$

  4. Zero


Correct Option: D
Explanation:
$AB-BA=I$

Let a $n\times n$ matrix. Take trace of both sides. trace $(AB-BA)=trace \ (I)$

trace $(AB)-$trace $(BA)=n$

$\Rightarrow \ n=0$

so zero possible ordered let of two $(n\times n)$ matrix.

If $A=|a _{ij}| _{2\times 2}$, where $a _{ij}=\left{\begin{matrix} i+j, & if & i\neq j\ i^2-2j, & if & i=j\end{matrix}\right.$, then $A^{-1}=?$

  1. $\dfrac{1}{9}\begin{bmatrix} 0 & 3\ 3 & 1\end{bmatrix}$

  2. $\dfrac{1}{9}\begin{bmatrix} 0 & -3 \ -3 & -1\end{bmatrix}$

  3. $\dfrac{1}{9}\begin{bmatrix} 4 & 1\ -1 & 2\end{bmatrix}$

  4. $\dfrac{-1}{9}\begin{bmatrix} 4 & 1\ -1 & 2\end{bmatrix}$


Correct Option: A
Explanation:

$\begin{array}{l} A={ \left| { aij } \right| _{ 2\times 2 } } \ { a _{ 11 } }=1-2=-1 \ { a _{ 12 } }=3 \ { a _{ 21 } }=3 \ { a _{ 22 } }=4-2\times 2=0 \ A=\left[ \begin{array}{l} -1 & 3 \ 3 & D \end{array} \right]  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA \ =\frac { { -1 } }{ { 09 } } \left[ \begin{array}{l} 0 & -3 \ -3 & -1 \end{array} \right]  \ =\frac { 1 }{ 9 } \left[ \begin{array}{l} 0 & 3 \ 3 & 1 \end{array} \right]  \end{array}$

If $\omega$ is the complex cube root of unity, then inverse of $\begin{bmatrix} \omega  & 0 & 0 \ 0 & { \omega  }^{ 2 } & 0 \ 0 & 0 & { \omega  }^{ 2 } \end{bmatrix}$ is

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

  2. $\begin{bmatrix} \omega^{2} & 0 & 0 \ 0 & { \omega }& 0 \ 0 & 0 & 1 \end{bmatrix}$

  3. $\begin{bmatrix} \omega^{3} & 0 & 0 \ 0 & { \omega } & 0 \ 0 & 0 & 1 \end{bmatrix}$

  4. $\begin{bmatrix} \omega & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & { \omega }^{ 2 } \end{bmatrix}$


Correct Option: B
Explanation:
$M=\begin{bmatrix} w & 0 & 0 \\ 0 & { w }^{ 2 } & 0 \\ 0 & 0 & { w }^{ 2 } \end{bmatrix}$
$(M)=w3w^2.w^3$
$=w^3.w^2$
$=w^3(w^3=1)$
$=(1)$

$adjA\Rightarrow \begin{bmatrix} { w }^{ 5 } & 0 & 0 \\ 0 & { w }^{ 4 } & 0 \\ 0 & 0 & { w }^{ 3 } \end{bmatrix}$

${ A }^{ -1 }=\dfrac { 1 }{ 1 } \begin{bmatrix} { w }^{ 2 }.{ w }^{ 3 } & 0 & 0 \\ 0 & { w }.{ w }^{ 3 } & 0 \\ 0 & 0 & \left( { w }^{ 3 } \right)  \end{bmatrix}$

$=\dfrac { 1 }{ 1 } \begin{bmatrix} { w }^{ 2 }.{ w }^{ 3 } & 0 & 0 \\ 0 & { w }.{ w }^{ 3 } & 0 \\ 0 & 0 & \left( { w }^{ 3 } \right)  \end{bmatrix}$
option $B$ is correct

The inverse of the matrix $\begin{bmatrix}1 & 0 & 1\ 0 & 2 & 3\ 1 & 2& 1\end{bmatrix}$ is

  1. $\dfrac {-1}{6} \begin{bmatrix}-4 & 2 & -2\ 3 & 0 & -3\ -2 & -2& 2\end{bmatrix}$

  2. $\dfrac {1}{6} \begin{bmatrix}-4 & 2 & -2\ 3 & 0 & -3\ -2 & -2& 2\end{bmatrix}$

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

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


Correct Option: A

If A =$\left[ \begin{matrix} i \ 0 \end{matrix}\begin{matrix} 0 \ -1 \end{matrix} \right] $, than check whether: ${{\text{A}}^2} =  - {\text{I,(}}{{\text{i}}^2} =  - 1)$

  1. True

  2. False


Correct Option: B
Explanation:

Given $A=\begin{bmatrix} i & 0  \ 0 & -1 \end{bmatrix}$ where $i^{2}=-1$


Taking LHS

$\Rightarrow A^{2}=A.A$

$\begin{bmatrix} i & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} i & 0 \ 0 & -1 \end{bmatrix}$ (using matrix multiplication)

$\Rightarrow \begin{bmatrix} { i }^{ 2 }+0 & 0+0 \ 0+0 & 0+1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$ (wihch is not equal to $-I$)

as $I=\begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$

$\therefore A^{2}=-I$ is not a valid relation.

If $M = \left[ \begin{array}{l}0\,\,\,\,2\5\,\,\,\,\,0\end{array} \right]\,\,\,and\,\,N = \left[ \begin{array}{l}0\,\,\,\,5\2\,\,\,\,\,0\end{array} \right]$,then ${M^{2011}}$ is-

  1. ${10^{1005}}M$

  2. ${10^{1005}}N$

  3. ${10^{2010}}M$

  4. ${10^{2011}}M$


Correct Option: D

If $A = \left[ \begin{array}{l}\cos \theta \,\,\,\,\sin \theta \ - \sin \theta \,\,\,\cos \theta \end{array} \right]$ where $\theta  = \frac{{2\pi }}{{19}}$ then ${A^{2017}} = $

  1. $A$

  2. ${A^3}$

  3. ${A^5}$

  4. $i$


Correct Option: B

If A and B are matrices of the same order, then $\displaystyle :\left ( A+B \right )^{2}= A^{2}+2AB+B^{2}$ is possible, iff

  1. AB= I

  2. BA= I

  3. AB= BA

  4. none of these


Correct Option: C
Explanation:

$\displaystyle :\left ( A+B \right )^{2}=(A+B)(A+B)$
$= A^{2}+AB+BA+B^{2}$
If $AB=BA$ then 
$\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}$


If $A$ and $B$ are any two matices, then  

  1. $AB=BA$

  2. $AB=I$

  3. $AB=0$

  4. $AB$ may or may not be defined


Correct Option: A