Tag: properties of multiplication of matrix

Questions Related to properties of multiplication of matrix

If $A=\begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$ then $A^n=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \ 3^{n-1} & 3^{n-1} & 3^{n-1} \ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}$ , $n \in N$

  1. True

  2. False


Correct Option: A

If $A = \begin{bmatrix}1\ 2\ 3
\end{bmatrix}$ then $AA^{1}$.

  1. $40$

  2. $\begin{bmatrix} 1\ 4\ 3 \end{bmatrix}$

  3. $\begin{bmatrix} 1 & 2 & 3\ 2 & 4 & 6\ 3 & 6 & 9\end{bmatrix}$

  4. None of these


Correct Option: C

If f (x)= $\left[ {\begin{array}{{20}{c}}  {\cos \,x}&{ - \sin \,x}&0 \   {\sin \,x}&{\cos \,x\,}&0 \   0&0&1 \end{array}} \right]$ and $\left[ {\begin{array}{{20}{c}}  {\cos \,x}&0&{\sin \,x} \   0&1&0 \   { - \sin \,x}&0&{\cos \,x} \end{array}} \right]$
then ${\left[ {f\left( x \right)g\left( y \right)} \right]^{ - 1}}$ is equal to 

  1. $f\left( { - x} \right)g\left( { - y} \right)$

  2. $f\left( {{x^{ - 1}}} \right)g\left( {{y^{ - 1}}} \right)$

  3. $g\left( { - y} \right)f\left( { - x} \right)$

  4. $g\left( {{y^{ - 1}}} \right)f\left( {{x^{ - 1}}} \right)$


Correct Option: B

If for the matrix $A.A^3=1$, then $A^{-1}=$

  1. $A^2$

  2. $A^3$

  3. $A$

  4. none of these


Correct Option: A

Let $A$ be a square matrix such that $A^2 = A$ and $|A| \neq 0$, then choose the correct option.

(A' represents transpose of matrix A)

  1. $A = A'$

  2. $A = -A'$

  3. $A' =-I$

  4. $A = -I$


Correct Option: A
Explanation:

$A$ is a square matrix such that $A^2 = A$ and $|A| \neq 0$
$\Rightarrow A^{-1}A^2=A^{-1}A=I$
$\Rightarrow A=I$
$\therefore A=A'=I$
Hence, option A.

For two matrices $A$ and $B$, if $AB=0$, then

  1. $A=0$ and $B=0$

  2. $A=0$ or $B=0$

  3. it is not necessary that $A=0$ or $B=0$

  4. all above are false


Correct Option: C
Explanation:

It is not necessary that, if $AB=0$ then $A=0$ or $B=0$


Take example, $A=\begin{bmatrix}0&0\0&1 \end{bmatrix}, B=\begin{bmatrix}0&1\0&0 \end{bmatrix}$

Clearly $AB=\begin{bmatrix}0&0\0&0 \end{bmatrix}$ but $A,B\neq 0$

For any non-singular matrix A, $ \displaystyle A^{-1} $ =

  1. $|A|adj A$

  2. $\dfrac{1}{|A| adj A}$

  3. $\dfrac{adj A}{|A|}$

  4. None of the above


Correct Option: C
Explanation:

Singular matrix is square matrix whose determinant is equal to Zero.

So, non-singular matrix $A$ is a matrix whose determinant is non-zero.
$\implies$ inverse of $A$ i.e. $A^{-1}$ exist.
$\therefore A^{-1}=\dfrac{adj A}{|A|}$ .... $[|A|\neq 0]$

If $A=\begin{bmatrix} \cos { \alpha  }  & -\sin { \alpha  }  \ \sin { \alpha  }  & \cos { \alpha  }  \end{bmatrix}$, $B=\begin{bmatrix} \cos { 2\beta  }  & \sin { 2\beta  }  \ \sin { 2\beta  }  & -\cos { 2\beta  }  \end{bmatrix}$, where 0 < $\beta$ < ${ \pi  }/{ 2 }$, then prove that $BAB=$ ${ A }^{ -1 }$.

  1. True

  2. False


Correct Option: A

Let $A$ be a $3\times 2$ matrix with real entries. Let $H = A(A^{T}A)^{-1}A^{T}$ where $A^{T}$ is the transpose of $A$ and let $I$ be the identity matrix of order $3\times 3$. Then

  1. $H^{2} = I$

  2. $H^{2} = -I$

  3. $H^{2} = H$

  4. $H^{2} = -H$


Correct Option: C
Explanation:

We know that from the propeties of inverse matrices,

${ \left( A.B \right)  }^{ -1 }={ B }^{ -1 }.A^{ -1 }$
$\Longrightarrow H=A{ A }^{ -1 }{ \left( { A }^{ T } \right)  }^{ -1 }{ A }^{ T }=I\ \Longrightarrow H.H={ H }^{ 2 }=I.H=H$

If $A^3 = 0$ then $1 + A + A^2$ is equal to

  1. I + A

  2. $(I + A)^{-1}$

  3. I - A

  4. $(I - A)^{-1}$


Correct Option: D
Explanation:
since the eigenvalue of matrix $A$ is zero

$(I-A) $ is invertible we have

$(I - A) (I + A + A^2) = I - A^3 = I - 0 = I$

$(I - A) (I + A + A^2)  = I$

$\therefore I + A + A^2 = (I - A)^{-1}$