Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

If $\begin{pmatrix}1 & -tan  \theta\ tan  \theta & 1\end{pmatrix} \begin{pmatrix} 1 & tan  \theta\ - tan  \theta & 1\end{pmatrix}^{-1} = \begin{bmatrix} a& -b\ b & a\end{bmatrix}$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $a = cos 2 \theta$

  2. $a = 1$

  3. $b = sin 2 \theta$

  4. $b = -1$

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Correct Option: A,C
Explanation:
we have 

$ \begin{pmatrix} 1 & tan  \theta\\ - tan  \theta & 1\end{pmatrix}^{-1} = \dfrac{1}{1+tan^2\theta} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} $

$\therefore \begin{bmatrix} a& -b\\ b & a\end{bmatrix}=cos^2\theta \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix}$

We get 

$\begin{pmatrix}cos2\theta & -sin2\theta\\ sin2\theta & cos2\theta \end{pmatrix}$

$\therefore a=cos2\theta, b=sin2\theta$

$A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}, I = \begin{bmatrix}1 & 0 & 0\ 0& 1 & 0\ 0 & 0 & 1\end{bmatrix}$ and $A^{-1} = \left [ \dfrac{1}{6} (A^2 + cA + dI) \right]$

The value of $(c,d)$ is

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $(-6, -11)$

  2. $(6, 11)$

  3. $(-6, 11)$

  4. $(6, -11)$

Show answer
Correct Option: C
Explanation:

Given $A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}$
The characteristic equation of $A$ is given by 
$|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda  & 0 & 0 \ 0 & 1-\lambda  & 1 \ 0 & -2 & 4-\lambda  \end{vmatrix}=0$

$\Rightarrow {\lambda}^{3}-6{\lambda}^{2}+11\lambda-6=0$
$\Rightarrow A^{3}-6A^{2}+11A-6=0$    ($\because$ Every square matrix satisfies its characteristic equation )
$\Rightarrow A^{2}-6A+11I-6A^{-1}=0$
$\Rightarrow A^{-1}=\displaystyle \frac{1}{6}(A^{2}-6A+11I)$

Comparing this with $A^{-1} = \left [ \frac{1}{6} (A^2 + cA + dI) \right]$, we get $c=-6, d=11$
$\therefore (c,d) = (-6,11)$

Hence, option C.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are two non-singular matrices, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $A$ is similar to $B$

  2. $AB$ is similar to $BA$

  3. $AB$ is similar to $A^{-1}B$

  4. none of these

Show answer
Correct Option: B
Explanation:

$AB = (B^{-1}B)AB = B^{-1}(BA)B $

$\therefore$ $AB$ is similar to $BA$.

Hence, option B.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar matrices such that $det :(A) =1$, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $det : (B) = 1$

  2. $det: (A)+det: (B)=0$

  3. $det : (B) = -1$

  4. none of these

Show answer
Correct Option: A
Explanation:

As $A$ and $B$ are similar matrices there exists a non-singular matrix $P$ such that

                                      $A=P^{-1}:BP$

$\Rightarrow det : (A) = det: (P^{-1}:BP)$

                $=det: (P^{-1}): det : (B) : det : (P)$

                 $\displaystyle =\frac{1}{det: (P)}det : (B) : (det P)$

                $= det : B$

Thus, $det : (A) = 0 \Leftrightarrow det : (B) = 0 : and : det : (A) =1 \Leftrightarrow  det : (B) = 1$

Hence, option A.

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar and $B$ and $C$ are similar, then

maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

  1. $AB$ and $BC$ are similar

  2. $A$ and $C$ are similar

  3. $A + C$ and $B$ are similar

  4. none of these

Show answer
Correct Option: B
Explanation:

$A = P^{-1}:BP, B = Q^{-1} : C : Q$,

$\Rightarrow      A = P^{-1}(Q^{-1} : C : Q)P = (QP)^{-1}: C : QP$

Thus, $A$ is similar to $C$

Hence, option B.


Which of the given values of $x$ and $y$ make the following pair of matrices equal.
$\displaystyle \begin{bmatrix} 3x+7 & 5 \ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4 \end{bmatrix}$

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $\displaystyle x=\frac { -1 }{ 3 } ,y=7$

  2. Not possible to find

  3. $\displaystyle y=7,x=\frac { -2 }{ 3 } $

  4. $\displaystyle x=\frac { -1 }{ 3 } ,y=\frac { -2 }{ 3 } $

Show answer
Correct Option: B
Explanation:

$\displaystyle \begin{bmatrix} 3x+7 & 5

\ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4

\end{bmatrix}$
Equating the corresponding elements, we get,
$\displaystyle 3x+7=0\Rightarrow x=-\frac { 7 }{ 3 } $
$\displaystyle 5=y-2\Rightarrow y=7$
$\displaystyle y+1=8\Rightarrow y=7$
$\displaystyle 2-3x=4\Rightarrow x=-\frac { 2 }{ 3 } $
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.


Let $A$ be the set of all $3 \times  3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.

The number of matrices $\mathrm{A}$ in $d$ for which the system of linear equations $\mathrm{A}\begin{bmatrix}\mathrm{X}\\mathrm{Y}\\mathrm{Z}\end{bmatrix}=\begin{bmatrix}1\0\0\end{bmatrix}$ has a unique solution, is

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. less than 4

  2. at least 4 but less than 7

  3. atleast 7 but less than 10

  4. at least 10

Show answer
Correct Option: B
Explanation:

$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 0 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
either $\mathrm{b}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0$
$ \Rightarrow  2$ matrices
$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 0\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & \mathrm{o} & \mathrm{c}\\mathrm{b} & \mathrm{c} & \mathrm{o}\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{b}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices.
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
If $\mathrm{a}=\mathrm{b}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{a}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{b}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
$\Rightarrow $ there will be only 6 matrices.

Solve the following system of equations by consistency- in consistency method $x+y+z=6,\ x-y+z=2,\ 2x-y+3z=9$

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $1,3,2$

  2. $2,3,4$

  3. $5,2,6$

  4. $2,5,7$

Show answer
Correct Option: A

Number of real values of $'a'$ for which the system of equations $2ax-2y+3z=0, x+ay+2z=0$ and $2x+az=0$ has a non-trivial solution, is equal to

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: A

Let $X=\begin{bmatrix} { x } _{ 1 } \ { x } _{ 2 } \ { x } _{ 3 } \end{bmatrix};A=\begin{bmatrix} 1 & -1 & 2 \ 2 & 0 & 1 \ 3 & 2 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix}$. If $AX=B$, then $X$ is equal to

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

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

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

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

  4. $\begin{bmatrix} 0 \ 2 \ 1 \end{bmatrix}$

Show answer
Correct Option: D
Explanation:
$AX=B$

$X=A^{-1}B$

$A^{-1}=\dfrac{1}{|A|}adj(A)$

Given,

$X=\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}$

$A=\begin{bmatrix}1 &-1  &2 \\  2& 0 &1 \\  3& 2 &1 \end{bmatrix}$

$B=\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}$

$|A|=\begin{vmatrix}1 &-1  &2 \\ 2 & 0 &1 \\  3&2  &1 \end{vmatrix}$

$=1(0-2)+1(2-3)+2(4-0)=-2-1+8=5$

$C _A=\begin{bmatrix}-2 &1  &4 \\  5& -5 &-5 \\  -1& 3 &2 \end{bmatrix}$

$adj(A)=C _A^T=\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$A^{-1}=\dfrac{1}{5}\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$X=A^{-1}B$

$=\dfrac{1}{5}\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}$

$=\dfrac{1}{5}\begin{bmatrix}-6+5-4\\ 3-5+12\\ 12-5+8\end{bmatrix}$

$=\dfrac{1}{5}\begin{bmatrix}-5\\ 10\\ 15\end{bmatrix}$

$=\begin{bmatrix} \left(\dfrac{-5}{5}\right)\\ \left(\dfrac{10}{5}\right)\\ \left(\dfrac{15}{5}\right)\end{bmatrix}$

$\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}=\begin{bmatrix}-1\\ 2\\ 3\end{bmatrix}$