Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=

  1. $z$

  2. $-z$

  3. $0$

  4. $2z$


Correct Option: B
Explanation:
$A=\begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\quad B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Given $AB=k\ B$
$AB\Rightarrow \begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\quad k=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
$\Rightarrow \begin{pmatrix} -5x & -8y & -7z \\ 3x & +5y & +4z \\ x2 & +3y & +3z \end{pmatrix}=\quad k\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
On adding all the elements on left and right side 
$\Rightarrow \ (-5x-8y-7z)+(3x+5y+47)+(2x+3y+3z)$
$=k(x+y+z)$
$\Rightarrow \ D=k(x+y+z)$
$k\neq 0$
$\Rightarrow \ x+y+z=0$
$\Rightarrow \ x+y=-z$

If $A = \left[ {\begin{array}{*{20}{c}}1&2\3&4\end{array}} \right]$, then $8A^{-4}$ is equal to

  1. $145A^{-1}+27I$

  2. $145A^{-1}-27I$

  3. $27I - 145A^{-1}$

  4. $29A^{-1} +9I$


Correct Option: C

If $A$ and $B$ are square matrices such that $B=-A^{-1}BA$, then 

  1. $AB+BA=0$

  2. $(A+B)^{o}=A^{2}+B^{2}$

  3. $(A+B)^{2}=A^{2}+2AB+B^{2}$

  4. $(A+B)^{2}=A+B$


Correct Option: A

If $A$ is a $2\times 2$ matrix such that $A^{2}-4A+3I=0$, then the inverse of $A+3I$ is equal to

  1. $\dfrac{1}{24}S-\dfrac{7}{24}I$

  2. $\dfrac{1}{21} A-\dfrac{7}{21}I$

  3. $\dfrac{7}{24}I+\dfrac{1}{24}A$

  4. $A-3I$`


Correct Option: A

If $A^{-1} = \alpha I + \beta I$ where $\alpha, \beta \in R$, then $\alpha + \beta$ is equal to (where $A^{-1}$ denotes inverse of matrix $A$)-

  1. $1$

  2. $\dfrac{4}{3}$

  3. $\dfrac{5}{3}$

  4. $\dfrac{1}{3}$


Correct Option: A

If $A=\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, find the values of $\alpha$ for which $A^2=B$.

  1. $\pm 1$

  2. $4$

  3. $0$

  4. No value


Correct Option: D
Explanation:

We have,

$A^2=B$
$\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 +0 & 0+0 \ \alpha +1 & 0+1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 & 0 \ \alpha +1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\alpha^2=1$ and $\alpha +1=5$
$\alpha =\pm 1$ and $\alpha =4$, which is not possible.
Hence, there is no value of $\alpha$ for which $A^2=B$ is true.

If $A=\left[ \begin{matrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{matrix} \right] $ and $10B=\left[ \begin{matrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{matrix} \right] $ where $B=A^{-1}$ then $\alpha$ is equal to-

  1. $2$

  2. $-1$

  3. $-2$

  4. $5$


Correct Option: A

The inverse of the matrix  $\left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \ { 3 } & { 3 } & { 0 } \ { 5 } & { 2 } & { - 1 } \end{array} \right]$  is

  1. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \ { 3 } & { 1 } & { 0 } \ { 9 } & { 2 } & { - 3 } \end{array} \right]$

  2. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \ { 3 } & { - 1 } & { 0 } \ { - 9 } & { - 2 } & { 3 } \end{array} \right]$

  3. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { 3 } & { 0 } & { 0 } \ { 3 } & { - 1 } & { 0 } \ { - 9 } & { - 2 } & { 3 } \end{array} \right]$

  4. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \ { - 3 } & { - 1 } & { 0 } \ { - 9 } & { - 2 } & { 3 } \end{array} \right]$


Correct Option: A

If $A=\left[ \begin{matrix} 1 & 0 & -1 \ 3 & 4 & 5 \ 0 & 6 & 7 \end{matrix} \right]$ and $A^{-1}=[\alpha _{ij}] _{3\times 3}$ then $\alpha _{23}=$

  1. $-1/5$

  2. $1/5$

  3. $-2/5$

  4. $2/5$


Correct Option: A

Let $P=\begin{bmatrix} \cos { \dfrac { \pi  }{ 9 }  }  & \sin { \dfrac { \pi  }{ 9 }  }  \ -\sin { \dfrac { \pi  }{ 9 }  }  & \cos { \dfrac { \pi  }{ 9 }  }  \end{bmatrix}$ and $\alpha,\ \beta,\ \gamma$ be non-zero real numbers such that $\alpha P^{6}+\beta P^{3}+\gamma 1$ is the zero matrix. Then, $(\alpha^{2}+\beta^{2}+\gamma^{2})^{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$ is

  1. $\pi$

  2. $\dfrac {\pi}{2}$

  3. $0$

  4. $1$


Correct Option: C