Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

If $A,B,C$ are the angles of a triangle, the system of equations, $(\sin A)x+y+z=\cos Ax+(\sin B)y+z=\cos B$
$x+y+(\sin C)z=1-\cos C$ has 

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

  1. No solutions

  2. Unique solution

  3. Infinitely many solutions

  4. Finitely many solutions

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Correct Option: A

The number of solutions of the equation $3x+3y-z=5,\ x+y+z=3,\ 2x+2y-z=3$

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

  1. $1$

  2. $0$

  3. $infinite$

  4. $Two$

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Correct Option: B

If the system of equation $x-ky-z=0,kx-y-z=0,x+y-z$ has a non-zero solution, the possible values of $k$ are

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

  1. $-1,2$

  2. $-1,1$

  3. $1,2$

  4. $0,1$

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Correct Option: A

The straight lines
$\left.\begin{matrix}
2kx-2y+3=0\
x+ky+2=0\
2x+k=0
\end{matrix}\right}k\in R$  pass through the same point for

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

  1. no real value of $k$

  2. exactly one real value of $k$

  3. three real values of $k$

  4. all real values of $k$

Show answer
Correct Option: B
Explanation:

Given equation of lines passes through same point i.e. lines are concurrent
$kx-2y+3=0\ 
x+ky+2=0\ 
2x+k=0$
$\Rightarrow \left| \begin{matrix} 2k & -2 & 3 \ 1 & k & 2 \ 2 & 0 & k \end{matrix} \right| =0$
$\Rightarrow k^{3}-2k-4=0$
$\Rightarrow (k-2)(k^2+2k+2)=0$
The discriminant of the quadratic expression is negative.
Hence there is only one real value of $k$

The system of equations

$\displaystyle 
\begin{matrix}kx +y+z=1&  & \
 x+ky+z=k&  & \
 x+y+kz=k^{2}&  &
\end{matrix}$
have no solution,if k equals ?

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

  1. 0

  2. 1

  3. -1

  4. -2

Show answer
Correct Option: D
Explanation:

$\Delta =\begin{vmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{vmatrix}$

$\Rightarrow \Delta =(k-1)^{2}(k+2)$

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

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

For k=-2,
$D _{1}=\begin{vmatrix} 1 & 1 & 1 \ -2 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix}$
$D _{1}=9 \ne 0$

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

Find the real value of $r$ for which the following system of linear equation has a non-trivial solution $2rx-2y+3z=0$$x+ry+2z=0$$2x+rz=0$

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

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Correct Option: A

The number of solutions of the system of equations $2x+y-z=7   ,   x-3y-2z=1 ,  x+4y-3z=5,$ are 

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

  1. 0

  2. 1

  3. 2

  4. infinitely many

Show answer
Correct Option: A
Explanation:
From given, we have,

$\Delta =\begin{vmatrix} 2 &1  &-1 \\ 1 & 3 &-2 \\  1& 4 &-3 \end{vmatrix}$

$=2(9-8)-1(-3-2)-1(4+3)$

$=0$

$\Delta _1=\begin{vmatrix} 7 &1  &-1 \\ 1 & 3 &-2 \\  5& 4 &-3 \end{vmatrix}$

$=7(9-8)-1(-3-10)-1(4+15)$

$=1\neq 0$

Hence, the given system has no solution.

The system of equations
$\displaystyle x + y + z = 2$
$\displaystyle 2x - y + 3z = 5$
$\displaystyle x - 2y - z + 1 = 0$
written in matrix form is

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

  1. $\displaystyle \begin{bmatrix}

    x \

    y \

    z

    \end{bmatrix} \begin{bmatrix}

    1 & 1 & 1 \

    2 & -1 & 3 \

    1 & -2 & -1

    \end{bmatrix} = \begin{bmatrix}

    2 \

    5 \

    -1

    \end{bmatrix}$

  2. $\displaystyle \begin{bmatrix}

    1 & 1 & 1 \

    2 & -1 & 3 \

    1 & -2 & -1

    \end{bmatrix} : \begin{bmatrix}

    x \

    y \

    z

    \end{bmatrix} = \begin{bmatrix}

    -2 \

    -5 \

    1

    \end{bmatrix}$

  3. $\displaystyle \begin{bmatrix}

    1 & 1 & 1 \

    2 & -1 & 3 \

    1 & -2 & -1

    \end{bmatrix} : \begin{bmatrix}

    x \

    y \

    z

    \end{bmatrix} = \begin{bmatrix}

    2 \

    5 \

    -1

    \end{bmatrix}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Given system of equations can be written as
$AX=B$

$\begin{bmatrix} 1 & 1 & 1 \ 2 & -1 & 3 \ 1 & -2 & -1 \end{bmatrix}: \begin{bmatrix} x \ y \ z \end{bmatrix}=\begin{bmatrix} 2 \ 5 \ -1 \end{bmatrix}$

If the following system of equations possess a non-trivial solution over the set of rationals
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$,
then x,y,z are in the ratio 

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

  1. $ \displaystyle \frac{15}{2} : 1 : 3$

  2. $ \displaystyle \frac{15}{2} : -1 : - 3$

  3. $ \displaystyle \frac{15}{2} : 1 : - 3$

  4. $ \displaystyle -\frac{15}{2} : 1 : - 3$

Show answer
Correct Option: D
Explanation:

For non trivial solution

$\Delta = 0$

$\therefore \begin{vmatrix}1 & k & 3\ 3 & k & -2\ 2 & 3 & -4\end{vmatrix} = 0$

applying $R _2 \rightarrow R _2 - 3R _1$ and $R _3 \rightarrow R _3 - 2 R _1$

$\therefore \begin{vmatrix}1 & k & 3\ 0 & -2k & -11\ 0 & 3-2k & -10\end{vmatrix} = 0$

$\Rightarrow \begin{vmatrix}-2k & -11 \3-2k  & -10\end{vmatrix} = 0$

$\Rightarrow 20k + 33- 22k = 0$

$\therefore k = \dfrac{33}{2}$

Putting the value of $k$ in the given equations. Then equations become

$\displaystyle x + \dfrac{33}{2} y + 3z = 0$               ...(i)

$\displaystyle 3x + \dfrac{33}{2} y - 2z = 0$              ...(ii)

$2x + 3y - 4z = 0$                           .....(iii)

Multiply (i) by 3 and subtract from (ii) then we get

$-33y - 11z = 0$

or   $z = - 3y$             ...(iv)

again multiply (i) by 2 and subtract from (iii) then we get

$-30y - 10z = 0$

$\therefore z = - 3y$                 ....(v)

Now let $y = \lambda,$

$ \therefore z = - 3 \lambda$

from (iii), $2x + 3 \lambda + 12 \lambda = 0$

$\therefore x = \displaystyle - \frac{15 \lambda}{2}$

$\therefore x : y : z = - \displaystyle \frac{15}{2} : 1 : - 3$

If the system of equations $ax+by+c=0$ $bx+cy+a=0$  ,$cx+ay+b=0$ has a solution then the system of equations $(b+c)x+(c+a)y+(a+b)z=0$   ,$(c+a)x+(a+b)y+(b+c)z=0$  , $(a+b)x+(b+c)y+(c+a)z=0$ has 

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

  1. only one solution

  2. no solution

  3. infinite number of solutions

  4. none of these

Show answer
Correct Option: C
Explanation:

$ax+by+c=0,bx+cy+a=0,cx+ay+b=0$
Gives
$\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$
Now $(b+c)x+(c+a)y+(a+b)z=0,(c+a)x+(a+b)y+(b+c)z=0,(a+b)x+(b+c)y+(c+a)z=0$
gives
$\begin{vmatrix} b+c & c+a & a+b \ c+a & a+b & b+c \ a+b & b+c & c+a \end{vmatrix}=\begin{vmatrix} b & c & a \ c & a & b \ a & b & c \end{vmatrix}+\begin{vmatrix} c & a & b \ a & b & c \ b & c & a \end{vmatrix}=\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}+\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=2\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$