Tag: properties of multiplication of matrix

Questions Related to properties of multiplication of matrix

If $\left[\begin{array}{ll}
\mathrm{x} & \mathrm{y}^{3}\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 8\
2 & 0
\end{array}\right]$, then  $\left[\begin{array}{ll}
\mathrm{x} & \mathrm{y}\
2 & 0
\end{array}\right]^{-1}$ is equal to

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $-\dfrac{1}{4}$$\left[\begin{array}{ll}

    0 &-2\

    -2 & 1

    \end{array}\right]$

  2. $\dfrac{2}{4}$$\left[\begin{array}{ll}

    1 & 0\

    0 & 1

    \end{array}\right]$

  3. $\dfrac{1}{4}$$\left[\begin{array}{ll}

    0 & -8\

    -2 & 1

    \end{array}\right]$

  4. $\dfrac{1}{4}\left[\begin{array} \ 1&4 \7 &2 \end{array}\right]$

Show answer
Correct Option: A

$p=$ $\begin{bmatrix}
0 & x &0 \
 0& 0 & 1
\end{bmatrix}$, then $p^{-1}$=


business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. Not possible to get an inverse

  2. $\begin{bmatrix}

    x & -a &-bx \

    0&1 &0 \

    0&0 &x

    \end{bmatrix}$

  3. $\mathrm{x}$ $\begin{bmatrix}

    x & -a &-bx \

    0&1 &0 \

    0&0 &x

    \end{bmatrix}$

  4. $x^{2} \begin{bmatrix}

    x & -a &-bx \

    0&1 &0 \

    0&0 &x

    \end{bmatrix}$

Show answer
Correct Option: A
Explanation:
Requirements to have an Inverse

1. The matrix must be square (same number of rows and columns).
2. The determinant of the matrix must not be zero.
It doesn't satisfy the first condition  so inverse of the given matrice $\left[ \begin{matrix} 0 & x & 0 \\ 0 & 0 & 1 \end{matrix} \right] $ is not possible as it is not a square matrice

option A is correct.

A= $\begin{bmatrix}
cos\alpha  & -sin\alpha \
sin\alpha  & cos\alpha
\end{bmatrix}$ ,then find which of the following are correct 
I) A is singular matrix
II) $A^{-1}$=$A^{T}$
III) A is symmetric matrix
IV) $A^{-1}= -A$

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. only I and II

  2. only II and III

  3. only II

  4. only IV

Show answer
Correct Option: C
Explanation:

$\left | A \right |= \cos ^{2}\alpha + \sin ^{2}\alpha = 1$
$A^{T}= \begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq A$
$A^{-1}= \frac{1}{1}\begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq -A$
$\begin{bmatrix}A^{T}= A^{-1}\end{bmatrix}$

If AB=KI where $\displaystyle K\in R$ then $\displaystyle A^{-1}$= _____

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. B

  2. KB

  3. $\displaystyle \frac{1}{K}B$

  4. $\displaystyle \frac{1}{K^{2}}B$

Show answer
Correct Option: C
Explanation:
Given $AB=KI\quad K\epsilon R$
i.e., K is constant
Now ${ A }^{ -1 }=\cfrac { I }{ A } $
I is identity matrix
$AB=KI$
$\Rightarrow \cfrac { 1 }{ K } B=\cfrac { I }{ A } \Rightarrow { A }^{ -1 }=\cfrac { 1 }{ K } B$
OPTION C

If A=$\displaystyle \begin{vmatrix} 5 & -3   \ 4 & 2   \end{vmatrix}$ then find $\displaystyle AA^{-1}$

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

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

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

  4. Does not exist

Show answer
Correct Option: C
Explanation:

For any given square matrix , $ A{A}^{-1} $ is always equal to Identity Matrix $ I $

So, $ A{A}^{-1} = \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} $

If $\displaystyle A=\left[ \begin{matrix} \cos { \theta  }  & \sin { \theta  }  \ -\sin { \theta  }  & \cos { \theta  }  \end{matrix} \right] $, then $\displaystyle \underset { n\rightarrow \infty  }{ \lim } \frac { 1 }{ n } { A }^{ n }$ is?

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. A null matrix

  2. An identity matrix

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

  4. None of these

Show answer
Correct Option: A
Explanation:
$A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix}\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n } { A }^{ n } } $
${ A }^{ n }={ \begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix} }^{ n }$
${ A }^{ n }={ \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }$
Now,
$\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n } { A }^{ n } } =\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n }  } { \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }$
$=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}=$Null matrix
Proof for ${ A }^{ n }={ \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }=P\left( n \right) $
$P\left( n \right) $is true for $n=1$
For $n=k,k\ge 1$
${ A }^{ k }={ \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }$
${ A }^{ k+1 }={ A }^{ k }A$
$={ \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }{ \begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix} }$
$\Rightarrow { \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }$
$\therefore P\left( n \right) $ is true for $n=k+1\left( k\ge 1 \right) $

If A is invertible, then which of the following is not true?

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $\displaystyle { A }^{ -1 }={ \left| A \right| }^{ -1 }$

  2. $\displaystyle { \left( { A }^{ 2 } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ 2 }$

  3. $\displaystyle { \left( { A }^{ ' } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ ' }$

  4. None of these

Show answer
Correct Option: A
Explanation:
A is invertible
$\Rightarrow { A }^{ -1 }$ exists
Option A: ${ A }^{ -1 }={ \left| A \right|  }^{ -1 }$
But we cannot write that a matrix and its determinant are both equal
$\therefore $ option A is not true
Option B: ${ \left( { A }^{ 2 } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
This option is true from the property
${ \left( { A }^{ n } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
Option C: ${ \left( { A }^{ -1 } \right)  }^{ 1 }={ \left( { A }^{ 1 } \right)  }^{ -1 }$
Consider $\left( { A }^{ T } \right) { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 }A \right)  }^{ T }={ I }^{ T }=I$
Similarly
${ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }={ \left( A{ A }^{ -1 } \right)  }^{ 1 }{ I }^{ 1 }=1$
From $1$ and $2$
${ A }^{ T }{ \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }=I$
$\Rightarrow { A }^{ 1 }$ is multiplicative inverse of ${ \left( { A }^{ -1 } \right)  }^{ 1 }$
$\Rightarrow { \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$

Which of the following matrices is not invertible?

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  2. $\displaystyle \left[ \begin{matrix} -1 & -1 \ -1 & 2 \end{matrix} \right] $

  3. $\displaystyle \left[ \begin{matrix} 2 & 3 \ 4 & 6 \end{matrix} \right] $

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

Show answer
Correct Option: C
Explanation:

A square matrix that is not invertible is called singular matrix in which its determinant is 0.
$\displaystyle \left[\begin{matrix} 2 & 3 \ 4 & 6 \end{matrix}  \right]$ is a non invertible matrix.
$\because$ its determinant is    $12-12 =0$
Option C is correct.

If the matrix $\displaystyle \left[ \begin{matrix} a \ c \end{matrix}\begin{matrix} b \ d \end{matrix} \right] $ is commutative with the matrix $\displaystyle \left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 1 \ 1 \end{matrix} \right] $, then

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $a=0, b=c$

  2. $b=0, c=d$

  3. $c=0, d=a$

  4. $d=0, a=b$

Show answer
Correct Option: C
Explanation:

Since matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is commutative with the matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$


So,

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}=\begin{bmatrix} a+c & b+d \\ c & d \end{bmatrix}$

$a=a+c,\quad a+b=b+d$
$c=0,\quad a=d$

Consider two matrix $A = \begin{bmatrix} 1 & 2\ 2 & 1\ 1 & 1 \end{bmatrix}$ and $ B = \begin{bmatrix} 1 & 2 & -4\  2 & 1 & -4 \end{bmatrix}$. Which one of the following is correct ?

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. B is the right inverse of A

  2. B is the left inverse of A

  3. B is the both sided inverse of A

  4. None of the above

Show answer
Correct Option: B
Explanation:

For two $m \times  n$ and $n \times m$ matrices $A$ and $B$

If $AB =I$, then $B$ is right inverse of $A$
If $BA=I$, then $B$ is left inverse of $A$
Note that if $AB=BA=I$, $B$ is both sided inverse of $A$
If $m=n$, which isn't true here
$AB=\begin{bmatrix} 1 & 2 \ 2 & 1 \ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & -4 \ 2 & 1 & -4 \end{bmatrix}=\begin{bmatrix} 5 & 4 & -4 \ 4 & 5 & -4 \ 3 & 3 & 0 \end{bmatrix}\neq I$
$BA=\begin{bmatrix} 1 & 2 & -4 \ 2 & 1 & -4 \end{bmatrix}\begin{bmatrix} 1 & 2 \ 2 & 1 \ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}=I$
$\therefore$ B is left inverse of A