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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 

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

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

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a diagonal matrix D = diag(d1, d2, ..., dn), the inverse is D^-1 = diag(1/d1, 1/d2, ..., 1/dn), provided all di are non-zero.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $det A = 0= 20 \lambda -60$
$\lambda =3$

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) $