Tag: business maths

Questions Related to business maths

The inverse of a skew-symmetric matrix of an odd order is

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagonal matrix

  4. does not exists


Correct Option: D
Explanation:

Let A be a skew-symmeteic matric of order $n.$

By definition $\displaystyle { A }^{ T }=-A$ 
$\displaystyle\Rightarrow \left| { A }^{ T } \right| =\left| -A \right| \Rightarrow \left| A \right| ={ \left( -1 \right)  }^{ n }\left| A \right| \$
$\displaystyle \Rightarrow \left| A \right| =-\left| A \right|\quad\quad[\because $ n is odd $]$
$\displaystyle \Rightarrow 2\left| A \right| =0\Rightarrow\left| A \right| =0$
$\therefore{ A }^{ -1 }$ does not exist. 

If $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$, $B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$ then ${(A+B)}^{2}$ equals

  1. ${A}^{2}+{B}^{2}$

  2. ${A}^{2}+{B}^{2}+2AB$

  3. ${A}^{2}+{B}^{2}+AB-BA$

  4. none of these


Correct Option: A
Explanation:

Given, $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix},B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$


$ A+B=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}+\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}$

$ { \left( A+B \right)  }^{ 2 }=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}=\begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$

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

$ { B }^{ 2 }=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

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

If $D=diag({d} _{1}, {d} _{2}, {d} _{3}........{d} _{n})$, where ${d} _{1}\ne 0$ for all $i=1, 2,.....n$, then ${D}^{-1}$ is equal to

  1. $D$

  2. ${I} _{n}$

  3. diag $({d} _{1}^{-1}, {d} _{2}^{-1}, ........{d} _{n}^{-1})$

  4. None of these


Correct Option: C

lf $\mathrm{A}$ is $\left{\begin{array}{lll}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{array}\right}$  is a singular matrix then  $\lambda =$ 

  1. 3

  2. 4

  3. 2

  4. 5


Correct Option: A
Explanation:

Given, $A=\begin{pmatrix}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{pmatrix}$ is a singular matrix
So, det A=0
$\therefore $ BY operation of matrix (s),
$det A=8(7 \lambda-16)+6[-6 \lambda + 8]+2[24-14]$
$=56 \lambda - 128 -36 \lambda +48 +20$
$=20 \lambda - 60$
So, $det A = 0= 20 \lambda -60$
$\lambda =3$

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


Correct Option: A

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


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

  1. only I and II

  2. only II and III

  3. only II

  4. only IV


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 $\displaystyle A^{-1}$= _____

  1. B

  2. KB

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

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


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 $\displaystyle AA^{-1}$

  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


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?

  1. A null matrix

  2. An identity matrix

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

  4. None of these


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) $