Tag: properties of multiplication of matrix

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Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  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. $\dfrac{1}{4}\left[\begin{array} \ 1&4 \\7 &2 \end{array}\right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the matrix equality, x=1 and y^3=8, which implies y=2. The matrix to invert is [[1, 2], [2, 0]]. The determinant is (1*0) - (2*2) = -4. The inverse is (1/det) * [[0, -2], [-2, 1]], which simplifies to -1/4 * [[0, -2], [-2, 1]].

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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


  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}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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) $A^{-1}= -A$

  1. only I and II

  2. only II and III

  3. only II

  4. only IV
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. B

  2. KB
  3. $\displaystyle \frac{1}{K}B$
  4. $\displaystyle \frac{1}{K^{2}}B$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

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

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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?

  1. A null matrix

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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) $
Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ \left( { A }^{ -1 } \right)  }^{ 1 }$
$\Rightarrow { \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$
Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Which of the following matrices is not invertible?

  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] $
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. $a=0, b=c$
  2. $b=0, c=d$
  3. $c=0, d=a$
  4. $d=0, a=b$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,

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

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

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

Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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 ?

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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