Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equations
                 $\lambda x + (sin  \alpha) y + (cos  \alpha) z = 0$
                  $x + (cos  \alpha) y + (sin  \alpha) z = 0$
                  $ - x + (sin  \alpha) y + (cos  \alpha) z = 0$
has a non-trivial solution. For $\lambda = 1$, find all values of $\alpha$ which are possible

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

  1. $\pi/8$

  2. $7\pi/8$

  3. $15\pi/8$

  4. $9\pi/8$

Show answer
Correct Option: A,B,D
Explanation:

To   find  all  values   of    $\alpha$   which   are  possible
For  $\lambda = 1 $
$ \begin{bmatrix}
1  & sin\alpha  & cos\alpha \
 1 & cos\alpha  & sin\alpha  \
 -1& sin\alpha   &  cos\alpha 
\end{bmatrix} = \begin{bmatrix}
0\
0\
0\end{bmatrix}$

On   Solving   this   equation, we   get
$ 2cos2\alpha = 0$
$ cos2\alpha  = 0 $
$ 2\alpha  = \dfrac{\pi }{2} $
 Finally  , we  get

$ \alpha  = \dfrac{\pi }{8}   $ 
 
$ \alpha  = \dfrac{7\pi }{8}   $     and

$ \alpha  = \dfrac{ 9\pi }{8}   $  

The number of distinct real roots of $\displaystyle \left | \begin{matrix}\sin x &\cos x  &\cos x \\cos x  &\sin x  &\cos x \\cos x  &\cos x  &\sin x \end{matrix} \right |= 0$ in the interval $\displaystyle -\frac{\pi }{4}\leq x\le\frac{\pi }{4}$ is

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

  1. 0

  2. 2

  3. 1

  4. 3

Show answer
Correct Option: C
Explanation:
$\displaystyle \left | \begin{matrix}\sin x &\cos x  &\cos x \\\cos x  &\sin x  &\cos x \\\cos x  &\cos x  &\sin x \end{matrix} \right |= 0$

Applying ${ C } _{ 1 }\rightarrow { C } _{ 1 }+{ C } _{ 2 }+{ C } _{ 3 }$, we get

$\begin{vmatrix} \sin { x } +2\cos { x }  & \cos { x }  & \cos { x }  \\ \sin { x } +2\cos { x }  & \sin { x }  & \cos { x }  \\ \sin { x } +2\cos { x }  & \cos { x }  & \sin { x }  \end{vmatrix}=0$

Taking $\left( \sin { x } +2\cos { x }  \right) $ common from ${C} _{1}$, we get

$ \Rightarrow \left( \sin { x } +2\cos { x }  \right) \begin{vmatrix} 1 & \cos { x }  & \cos { x }  \\ 1 & \sin { x }  & \cos { x }  \\ 1 & \cos { x }  & \sin { x }  \end{vmatrix}=0$

Applying ${R} _{2}\rightarrow{R} _{2}-{R} _{1};{R} _{3}\rightarrow{R} _{3}-{R} _{1}$

$\Rightarrow \left( \sin { x } +2\cos { x }  \right) \begin{vmatrix} 1 & \cos { x }  & \cos { x }  \\ 0 & \sin { x-\cos { x }  }  & 0 \\ 0 & 0 & \sin { x-\cos { x }  }  \end{vmatrix}=0\\ \Rightarrow \left( \sin { x } +2\cos { x }  \right) { \left( \sin { x } -\cos { x }  \right)  }^{ 2 }=0\\ \Rightarrow \sin { x } +2\cos { x } =0,\quad \sin { x } -\cos { x } =0\\ \Rightarrow \tan { x } =-2,\quad \tan { x } =1$

If the system of equations $2x+3y=7,(2a-b)y=21$ has infinitely many solutions, then -

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

  1. $a=1,b=5$

  2. $a=5,b=1$

  3. $a=-1,b=5$

  4. $a=5,b=-1$

Show answer
Correct Option: A

The system of equation $\displaystyle \alpha x+y+z=\alpha-1,:x+\alpha y+z=\alpha-1,:x+y+\alpha z=\alpha-1$ has no solution if $\alpha$ is

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

  1. either $-2$ or $1$

  2. $-2$

  3. $1$

  4. $2$

Show answer
Correct Option: B
Explanation:

Here, $D =\begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$

$\Rightarrow D =(\alpha-1)^{2}(\alpha+2)$

Taking $D=0$
$\Rightarrow (\alpha-1)^{2}(\alpha+2)=0$
$\Rightarrow \alpha=1, \alpha=-2$
At these values of $\alpha$, system can have either no solution or infinitely many solution.

For $\alpha=1$, equations takes the form $x+y+z=0$
Hence, infinitely many solution for $\alpha=1$.

For $\alpha=-2$,
$D _{1}=\begin{vmatrix} -3 & 1 & 1 \ -3 & -2 & 1 \ -3 & 1 & -2 \end{vmatrix}$
$D _{1}=27 \ne 0$

So, $D=0$, at least one $D _{1}\ne 0$
Hence, system has no solution for $\alpha=-2$

The solution set of the equation $\left| \begin{matrix} 2 & 3 & x \ 2 & 1 & { x }^{ 2 } \ 6 & 7 & 3 \end{matrix} \right| =0$ is 

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

  1. $\left{ 1,-3 \right} $

  2. $\left{ 1,-1 \right} $

  3. $\left{ 1,3 \right} $

  4. $\theta $

Show answer
Correct Option: A

If a,b,c$\in $ R. Than the system of the equation is :$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1.\ \ \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } +\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1.\ \ \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } -\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1\ \ has\quad $.

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

  1. No solution

  2. a unique solution

  3. infinirty many solution

  4. finietil many solution

Show answer
Correct Option: A

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

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

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

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

  3. $x = -\dfrac {1}{3}, 4 = -\dfrac {2}{5}$

  4. Not possible to find

Show answer
Correct Option: D
Explanation:

If the given matrices are equal then we have the following equations,

$3x+7=0$
Or, $ x=-\dfrac{7}{3}.$......(1).

And 
$5=y-2$
Or, $ y=7$.......(2)

And
$y+1=8$
Or, $y=7$.......(3)

And
$ 2-3x=4$
Or, $x=-\dfrac{2}{3}.$....(4)

From (1) and (4) we get contradicting result i.e. why $x$ and $y$ can't be found simultaneously.
So not possible to find $x$ and $y$.

Suppose $a _1, :a _2,: ... $ are real numbers, with $a _1\neq 0$. If $a _1, :a _2,:a _3,:...$ are in A.P.  Then,

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

  1. $A=\begin{bmatrix}a _1&a _2 &a _3 \a _4 &a _5 &a _6 \a _5 &a _6 &a _7 \end{bmatrix}$ is singular

  2. the system of equations $a _1x+a _2y+a _3z=0, : a _4x+a _5y+a _6z=0,:a _7x+a _8y+a _9z=0$ has infinite number of solutions

  3. $B=\begin{bmatrix}a _1&ia _2 \ ia _2 & a _1\end{bmatrix}$ is non singular

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:
Given ${ a } _{ 1 },{ a } _{ 2 }...$are in AP
${ a } _{ 1 }\neq 0$
Option A$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 2 } & { a } _{ 3 } \\ { a } _{ 4 } & { a } _{ 5 } & { a } _{ 6 } \\ { a } _{ 7 } & { a } _{ 8 } & { a } _{ 9 } \end{bmatrix}$
$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 1 }+d & { a } _{ 1 }+2d \\ { a } _{ 1 }+3d & { a } _{ 1 }+4d & { a } _{ 1 }+5d \\ { a } _{ 1 }+4d & { a } _{ 1 }+5d & { a } _{ 1 }+6d \end{bmatrix}\quad { C } _{ 1 }\rightarrow { C } _{ 2 }-{ C } _{ 2 }$
$=\begin{bmatrix} d & { a } _{ 1 }+d & { a } _{ 1 }+2d \\ d & { a } _{ 1 }+4d & { a } _{ 1 }+5d \\ d & { a } _{ 1 }+5d & { a } _{ 1 }+6d \end{bmatrix}{ C } _{ 2 }\rightarrow { C } _{ 3 }-{ C } _{ 2 }$
$=\begin{bmatrix} d & d & { a } _{ 1 }+2d \\ d & d & { a } _{ 1 }+5d \\ d & d & { a } _{ 1 }+6d \end{bmatrix}$
${ C } _{ 1 }\& { C } _{ 2 }$ are same $\Rightarrow \triangle =0$
The matrix is singular-correct
Option B - co efficients matrix$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 2 } & { a } _{ 3 } \\ { a } _{ 4 } & { a } _{ 5 } & { a } _{ 6 } \\ { a } _{ 7 } & { a } _{ 8 } & { a } _{ 9 } \end{bmatrix}$
They are in AP
$\Rightarrow \triangle =0$
$\therefore $They have infinite solutions
Option B is also correct
Option C $B=\begin{bmatrix} { a } _{ 1 } & i{ a } _{ 2 } \\ i{ a } _{ 2 } & { a } _{ 1 } \end{bmatrix}$
$\Rightarrow \triangle ={ a } _{ 1 }^{ 2 }-\left( { i }^{ 2 }{ a } _{ 2 }^{ 2 } \right) $
$={ a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 }$
$={ a } _{ 1 }^{ 2 }+{ \left( { a } _{ 1 }+d \right)  }^{ 2 }$
$=2{ a } _{ 1 }^{ 2 }+{ d }^{ 2 }+2{ a } _{ 2 }d>0$
Non singular
Option A,B, C are true

if $x= -5 $ is a root of $\displaystyle \Delta =\begin{vmatrix}
2x+1 & 4  & 8 \
2 & 2x  & 2 \
7 & 6  & 2x
\end{vmatrix}=0$ then the other two roots are

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

  1. $3 , 3.5$

  2. $1 , 3.5$

  3. $3 , 6$

  4. $2 , 6$

Show answer
Correct Option: B
Explanation:

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ we get 

$\displaystyle \Delta =\begin{vmatrix}
2x+10 &2x+10  &2x+10 \ 
 2 & 2x  & 2 \ 
 7&6  & 2x
\end{vmatrix}$

Taking $2x+10$ common from $\displaystyle R _{1}$ and applying 

$\displaystyle C _{2}\rightarrow C _{2}-C _{1},C _{3}\rightarrow C _{3}-C _{1}$ we get

$\displaystyle \Delta =2\left ( x+5 \right )\begin{vmatrix}
1 &0  &0 \ 
 2 & 2x-2  & 0 \ 
 7&-1  & 2x-7
\end{vmatrix}$

$\displaystyle =2\left ( x+5 \right )\left ( 2x-2 \right )\left ( 2x-7 \right )$

Thus $\displaystyle \Delta =0 \ \Rightarrow x=-5,1,3.5$

$\displaystyle \therefore $ the other two roots are $1$ and $3.5$

Given the system of equations
$(b+c)(y+z)-ax=b-c$
$(c+a)(z+x)-by=c-a$
$(a+b)(x+y)-cz=a-b$
(where $a+b+c\neq 0$); then $x:y:z$ is given by

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

  1. $c-b:a-c:b-a$

  2. $b+c:c+a:a+b$

  3. $a:b:c$

  4. $\displaystyle \frac{a}{b}:\frac{b}{c}:\frac{c}{a}$

Show answer
Correct Option: A
Explanation:

$(b+c)(y+z)-ax=b-c\$       ..............(1)


$ (c+a)(z+x)-by=c-a\$       ..............(2)

$ (a+b)(x+y)-cz=a-b$             .............(3)

Adding all three equation

$\left( x+y+z \right) \left( a+b+c \right) =0\\$

$ \Rightarrow x+y+z=0\\$

$ \Rightarrow y+z=-x$

Substituting this in the first (1) equation

$(b+c)(-x)-ax=b-c$

$x=\cfrac { c-b }{ a+b+c } $

Similarly, we get

$y=\cfrac {a-c }{ a+b+c } ,z=\cfrac { b-a }{ a+b+c } $

Hence $x:y:z=(c-b):(a-c):(b-a)$