Tag: non-homogeneous linear equations

Questions Related to non-homogeneous linear equations

For the system of linear equations 2x + 3y + 5z = 9, 7x + 3y - 2z = 8 and 2x + 3y +$\lambda$z $=\mu$.Under what condition does the above system of equations have infinitely many solutions.

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

  1. $\lambda = 5$ and $\mu \neq 9$

  2. $\lambda = 5$ and $\mu = 9$

  3. $\lambda = 9$ and $\mu \neq 5$

  4. $\lambda = 9$ and $ \mu = 5$

Show answer
Correct Option: B
Explanation:

$2x+3y+5z=9$,

$ 7x+3y-2z=8$,
$ 2x+3y+λz=μ$ 
Now, for infinitely many solutions. If $2$ equations out of $3$ are same then we are left with only $2$ equations with $3$ variables, this will give infinite solutions. Now if $\lambda =5$, $\mu =9$, then equation (i) and (iii) will be same and the system will have infinite solutions. 
Hence, B is correct.

The system $2x+3y+z=5, 3x+y+5z=7, x+4y-2z=3$ has:

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

  1. Unique Solution

  2. Finite number of solutions

  3. Infinite Solutions

  4. No solution

Show answer
Correct Option: D
Explanation:

${2}x+{3}y+z={5}$

${3}x+y+{5}z={7}$
$x+{4}y-{2}z={3}$
$A=\left[ \begin{matrix} 2 & 3 & 1 \ 3 & 1 & 5 \ 1 & 4 & -2 \end{matrix} \right] \quad \quad D=\left[ \begin{matrix} 5 \ 7 \ 3 \end{matrix} \right] \quad \quad AD=\left[ \begin{matrix}2&3&1&5\3&1&5&7\1&4&-2&3\end{matrix} \right] $

Rank of  $AD=\left[ \begin{matrix}2&3&1&5\3&1&5&7\1&4&-2&3\end{matrix} \right] $

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

$\left[ \begin{matrix} 2 & 3 & 1 & 5 \ 0 & 7 & -7 & 1 \ 0 & 5 & -5 & 1 \end{matrix} \right] $

$R _{3}\rightarrow{7}R _{3}-{5}R _{2}$

$\left[ \begin{matrix} 2 & 3 & 1 & 5 \ 0 & 7 & -7 & 1 \ 0 & 0 & 0 & 2 \end{matrix} \right] $

Rank of $A=\left[ \begin{matrix} 2 & 3 & 1 \ 3 & 1 & 5 \ 1 & 4 & -2 \end{matrix} \right] $

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


$A=\left[ \begin{matrix} 2 & 3 & 1 \ 0 & 7 & -7 \ 0 & 5 & -5 \end{matrix} \right] $

Clearly Rank of A $\neq$ Rank of AD.
Hence, no solution.


If AX = B where A is $3 \times 3$ and X and B are $3\times 1$ matrices then which of the following is correct?

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

  1. If | A | = 0 then AX = B has infinite solutions

  2. If AX = B has infinite solutions then | A | = 0

  3. If (adj (A)) B = 0 and | A | $\neq$ 0 then AX = B has unique solution

  4. If (adj (A)) B $\neq$ 0 & |A| = 0 then AX = B has no solution

Show answer
Correct Option: A

The system of equations , $ ax+y+z = a-1 $ , $x+ay+z = a-1 $, $x+y+az = a-1 $has no solution, if a 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. $not\ -2$

Show answer
Correct Option: A
Explanation:
Determinant of Coefficient Matrix , $\begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{bmatrix}=0$
$a(a^2-1)-1(a-1)+1(1-a)=0$
$a^3-a-a+1+1-a=0$
$a^3-3a+2=0\Rightarrow $ If $a=1$
$a=1|\begin{matrix} 1 & 0 & -3 & 2 \\ 0 & 1 & 1 & -2 \end{matrix}$
           |$\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ $
            $\begin{matrix} 1 & 1 & -2 & 0 \end{matrix}$
$a^2+a-2=0\Rightarrow a^2+2a-a-2=0$
$a(a+2)-1(a+2)=0$
$a=-2,1$
$a$ is either $-2$ (or) $1$ 

The three distinct straight lines $ax+by+c=0$;$bx+cy+a=0$ and $cx+ay+b=0$ are concurrent then

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

  1. $a+b+c=0$

  2. $a^{3}+b^{3}+c^{3}=3 abc$

  3. $a=b=c$

  4. $a^{2}+b^{2}+c^{2}=ab+bc+ca$

Show answer
Correct Option: A,B
Explanation:

The lines are concurrent,
$=>\begin{bmatrix}a&b& c\ b &c &a\ c & a& b\end{bmatrix}=0$
Applying,$ C _{1}=C _{1}+C _{2}+C _{3}$,
$=\begin{bmatrix}a+b+c&b& c\ a+b+c &c &a\ a+b+c & a& b\end{bmatrix}$
Applying, $R _{1}=R _{1}-R _{2}$ and $R _{2}=R _{2}-R _{3}$,
$=(a+b+c)\begin{bmatrix}0&b-c& c-a\ 0 &c-a &a-b\ 1 & a& b\end{bmatrix}$
Expanding by $C _{1}$,
$=(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$=> (a+b+c)=0$
or,
$(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$abc-a^{3}-b^{3}+abc+abc-c^{3}=0$
$a^{3}+b^{3}+c^{3}=3abc$
So options are A and B.

If $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$, then $x = $ ............ 

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

  1. $ \dfrac{a+b+ c}{3}$

  2. $ \dfrac{a + \omega^2 b + \omega c}{3}$

  3. $\dfrac{a + \omega b + \omega^2 c}{3}$

  4. $\dfrac{a+b+ c \omega}{3}$

Show answer
Correct Option: A
Explanation:

Given $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$.

Adding all the given equations gives

$3x+y(1+\omega+\omega^2)+z(1+\omega+\omega^2)=a+b+c$

$\Rightarrow 3x=a+b+c$     $\because 1+\omega+\omega^2=0$

$\therefore x=\displaystyle\frac{a+b+c}{3}$

Hence, option A.

If $f(x) = ax^2 + bx + c, a, b, c \in  R$ and equation $f(x)- x = 0$ has non-real roots $\alpha, \beta$.  Let $\gamma, \delta$ be the roots of $f(f(x)) - x = 0$ ($\gamma, \delta$ are not equal to $\alpha, \beta$). Then $\begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ is

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

  1. 0

  2. purely real

  3. purely imaginary

  4. none of these

Show answer
Correct Option: B
Explanation:

$f(x) = ax^2 + bx + c; a, b, c \in  R$ and equation $f(x)- x = 0$ has imaginary roots $\alpha, \beta$ and $\gamma, \delta$ be the roots of $f(f(x)) - x = 0$
since $\alpha,\beta$ are roots of $f(x)- x = 0$
$f(\alpha)-\alpha =0$ $\Rightarrow f(\alpha)=\alpha$
$f(\beta)-\beta=0$ $\Rightarrow f(\beta)=\beta$
$f(f(\alpha))-\alpha = f(\alpha)-\alpha =0$
$f(f(\beta))-\beta = f(\beta)-\beta =0$
$\therefore  \alpha,\beta$ are also roots of $f(f(x)) - x = 0$ ------(*)
$f(x)- x= ax^2+(b-1)x+c=0$
roots are imaginary.
i.e $\alpha,\beta$ are conjugate to each other and $D<0$
$\Rightarrow (b-1)^2-4ac<0$ -------(1)
$f(f(x)) - x = a(ax^2+bx+c)^2+b(ax^2+bx+c)+c-x=0$
$\Rightarrow \left(ax^2+(b-1)x+c\right)\left(a^2x^2+(ab+a)x+ac+b+1\right)=0$
$D=(ab+a)^2-4a^2(ac+b+1) = a^2\left((b-1)^2-4ac\right)-4a^2<0$    ($\because$ from (1))
$\therefore \gamma,\delta$ are also imaginary roots and conjugate to each other.
$\begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ $= -3\alpha\beta+\alpha^2\gamma+\beta^2\delta$ ------ (2)
$\alpha\beta$ is real
$\alpha^2\gamma$ is conjugate to $\beta^2\delta$
$\Rightarrow \alpha^2\gamma+\beta^2\delta$ is real
from (2)
$\therefore \begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ is purely real.
Hence, option B.


If $\displaystyle \omega$ is cube root of unity and $\displaystyle x + y + z = a$, $\displaystyle x + \omega y + \omega^{2} z = b$, $\displaystyle x + \omega^{2} y + \omega z = b$ then which of the following is not correct?

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

  1. $\displaystyle x = \frac{a + b + c}{3}$

  2. $\displaystyle y = \frac{a + b \omega^{2} + \omega c}{3}$

  3. $\displaystyle x = \frac{a + b \omega + \omega^{2} c}{3}$

  4. None of these

Show answer
Correct Option: D
Explanation:

Given $x+y+z=a $...(i)

$\displaystyle x+\omega y+\omega ^{2}z= b$ ...(ii)

$\displaystyle x+\omega ^{2}y+\omega z= c$ ...(iii)

By adding (i), (ii) and (iii), we get

$\displaystyle x= \frac{a+b+c}{3}$

Hence option A is correct.

Again $\displaystyle \left ( i \right )+\left ( ii \right )\times \omega ^{2}+\left ( iii \right )\times \omega $, we get

$\displaystyle 3y= a\omega ^{3}+b\omega ^{2}+c\omega$

$\displaystyle y= \frac{a+b\omega ^{2}+c\omega }{3}$

Hence, option B is correct.

Similarly, $\displaystyle \left ( i \right )+\left ( ii \right )\times \omega +\left ( iii \right )\times \omega ^{2}$ we get

$\displaystyle z= \frac{a+b\omega +c\omega ^{2}}{3}$

Hence, option C is correct

Consider the system of equations $x-2y+3z=-1,
-x+y-2z=k , x-3y+4z=1$ 

STATEMENT - 1 : The system of equations has no solutions for $k\neq 3$ and 
STATEMENT - 2 : The determinant $\begin{vmatrix}
1 & 3 & -1\
-1 & -2& k\
1& 4& 1
\end{vmatrix}$ $\neq 0$ for $k\neq 3$ 

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

  1. Statement-1 is true, statement - 2 is true,

    statement - 2 is a correct explanation for

    statement -

  2. Statement -1 is true, statement - 2 is true,

    statement -2 is a not a correct explanation for

    statement - 1

  3. Statement -1 is true, statement -2 is false

  4. Statement -1 is false, statement - 2 is true

Show answer
Correct Option: A
Explanation:

For the solution of the system of equations
$x - 2y + 3z = -1$
$-x + y - 2z = 2$
$kx - 3y + 4z = 1$, 
$\begin{bmatrix}
1 &-2  &3 \ 
-1 &1  &-2 \ 
k &-3  &4 
\end{bmatrix} 
$
If determinant of the matrix is zero than the given lines are coplanar and if the determinant is non zero the they are non co-planar.
The solution for the system of equations exist when the lines are coplanar.
Hence, $1(-2 - 4k) - 3(-1 - k) -1(-4 + 2) \neq 0$
$\Rightarrow -2 - 4k + 3 + 3k + 2 \neq 0$
i.e. $k \neq 3$

The values of $\theta $ lying between $\theta =0$ and $\theta =\dfrac {\pi}{2}$ and satisfying the equation
$\begin{vmatrix}
1+\sin ^{2}\theta  & \cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & 1+\cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & \cos ^{2}\theta  & 1+4\sin 6\theta
\end{vmatrix}$
are given by

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

  1. $\dfrac {\pi }{36}, \dfrac{5\pi}{ 36}$

  2. $\dfrac{7\pi}{36}, \dfrac{11\pi}{3}$

  3. $\dfrac{5\pi }{36}, \dfrac{7\pi }{36}$

  4. $\dfrac{11\pi}{36}, \dfrac{\pi }{36}$

Show answer
Correct Option: B
Explanation:

$\begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & 1+\cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 1+4\sin  6\theta  \end{vmatrix}=0$

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

$\Rightarrow \begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0$

Applying ${ C } _{ 1 }\rightarrow { C } _{ 1 }+{ C } _{ 2 }$

$\Rightarrow \begin{vmatrix} 2 & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ 0 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0\ \Rightarrow 2+4\sin  6\theta =0\Rightarrow \sin  6\theta =-\cfrac { 1 }{ 2 } \ \Rightarrow 6\theta =n\pi +{ \left( -1 \right)  }^{ n }\left( -\cfrac { \pi  }{ 6 }  \right) \ \Rightarrow \theta =\cfrac { n\pi  }{ 6 } +{ \left( -1 \right)  }^{ n+1 }\left( \cfrac { \pi  }{ 36 }  \right) \ \Rightarrow \theta =\cfrac { 7\pi  }{ 36 } ,\cfrac { 11\pi  }{ 36 } $