Tag: non-homogeneous linear equations

Questions Related to non-homogeneous linear equations

If $ \displaystyle a+b+c=0$ then value of $ \displaystyle (s) $ of $x$ which makes $\displaystyle \begin{vmatrix}
a-x &c  &b \
 c&b-x  &a \
b & a &c-x
\end{vmatrix}$ zero is (are)

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

  1. $\displaystyle x=0 $

  2. $\displaystyle x=\sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

  3. $\displaystyle x=- \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

  4. None of these

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

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ and taking $\displaystyle a+b+c-x $ common from the first row, we obtain
$\displaystyle \Delta =\left ( a+b+c-x \right )\begin{vmatrix}
1 &1  &1 \ 
c &b-x  &a \ 
b &a  &c-x 
\end{vmatrix}$
Applying  $\displaystyle C _{2}\rightarrow C _{2}-C _{1}$ and $\displaystyle C _{3}\rightarrow C _{3}-C _{1}$ we obtain
$\displaystyle \Delta =\begin{vmatrix}
1 &0  &0 \ 
c &b-c-x  &a-c \ 
b &a-b  &c-b-x 
\end{vmatrix}\left [ \because a+b+c=0 \right ]$
Expanding along $\displaystyle R _{1}$ we get
$\displaystyle \Delta =x\left [ \left ( b-c-x \right )\left (c-b-x  \right )-\left ( a-b \right )\left ( a-c \right ) \right ] $
$\displaystyle \Delta =x\left [ \left ( a-b \right )\left (a-c  \right )-\left ( x+b-c\right )\left ( x-b+c \right ) \right ] $
$\displaystyle  =x\left [ a^{2}-ab-ac+bc-x^{2}+b^{2}+c^{2}-2bc \right ]$
$\displaystyle \Delta =x\left [ a^{2}+b^{2}+c^{2}-bc-ab-ac-x^{2} \right ]$
$\displaystyle \Delta =0$ implies $\displaystyle x =0$ or $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
Now $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
$\displaystyle =a^{2}+b^{2}+c^{2}-\frac{1}{2}\left [ \left ( a+b+c \right )^{2}- a^{2}-b^{2}-c^{2} \right ]$
$\displaystyle =\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )\left [ \because a+b+c=0 \right ]$
$\displaystyle \Rightarrow x= \pm \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

Consider the system of equations:
$x+y+z=0$
$\alpha x+\beta y+\gamma z=0$
$\alpha^2 x+\beta^2 y+\gamma^2 z=0$
Then the system of equations has

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

  1. A unique solution for all values $\alpha, \beta, \gamma$

  2. Infinite number of solutions if any two of $\alpha,\beta, \gamma$ are equal

  3. A unique solution if $\alpha, \beta, \gamma$ are distinct

  4. More than one, but finite number of solutions depending on values of $\alpha, \beta, \gamma$

Show answer
Correct Option: B,C
Explanation:
$x+y+z=0$
$\alpha x+\beta y+\gamma z=0$
${ \alpha  }^{ 2 }x+{ \beta  }^{ 2 }y+{ \gamma  }^{ 2 }z=0$
$\triangle =\begin{vmatrix} 1 & 1 & 1 \\ \alpha  & \beta  & \gamma  \\ { \alpha  }^{ 2 } & { \beta  }^{ 2 } & { \gamma  }^{ 2 } \end{vmatrix}$
If any of the two values $\left( \alpha ,\beta  \right) $ or $\left( \alpha ,\gamma  \right) $ or $\left( \beta ,\gamma  \right) $ are equal then $\triangle =0$
Infinite solution
Option B
For all different values of $\alpha ,\beta ,\gamma $
$\triangle \neq 0$
Unique solution
Option C

The following system of equations
$x+y+z=1$
$2x+2y+2z=3$
$3x+3y+3z=4$ has

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

  1. infinite number of solutions

  2. no solution

  3. unique solution

  4. finitely many solutions

  5. none of these

Show answer
Correct Option: B
Explanation:

$D=\begin{vmatrix}1 &1  &1 \ 2 & 2 & 2\ 3 & 3 & 3\end{vmatrix}=0$
$D _1=0,$     $D _2=0$,      $D _3=0$
Let $z=t$
$x+y=1-t$
$2x+2y=3-2t$
Since both the lines are parallel hence no value of x and y
Hence there is no solution of the given equations.

Let $S$ be the set of all column matrices $\begin{bmatrix}b _{1}\b _{2} \ b _{3}
\end{bmatrix}$ such that $b _{1}, b _{2}, b _{3} \  \epsilon \  \mathbb {R}$ and the system of equation (in real variables) 
$-x + 2y + 5z = b _{1}$
$2x - 4y + 3z = b _{2}$
$x - 2y + 2z = b _{3}$
has at least one solution. Then, which of the following system(s) (in real variables) has/have at least one solution of each $\begin{bmatrix}b _{1}\ b _{2}\ b _{3}
\end{bmatrix}\epsilon \  S$?

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

  1. $x + 2y + 3z = b _{1}, 4y + 5z = b _{2}$ and $x + 2y + 6z = b _{3}$

  2. $x + y + 3z = b _{1}, 5x + 2y + 6z = b _{2}$ and $-2x - y - 3z = b _{3}$

  3. $-x + 2y - 5z = b _{1}, 2x - 4y + 10z = b _{2}$ and $x - 2y + 5z = b _{3}$

  4. $x + 2y + 5z = b _{1}, 2x + 3z = b _{2}$ and $x + 4y - 5z = b _{3}$

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

We find $D = 0$, where $D$ is the determinant formed by the coefficients of $x,\ y,\ z$ in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions.
Let $\alpha P _{1} + \lambda P _{2} = P _{3}$
$\Rightarrow P _{1} + 7P _{2} = 13P _{3}$
$\Rightarrow b _{1} + 7b _{2} = 13b _{3}$
(A) $D\neq 0\Rightarrow$ unique solution for any $b _{1}, b _{2}, b _{3}$
(B) $D = 0$ but $P _{1} + 7P _{2} \neq 13P _{3}$
(C) $D = 0$ Also $b _{2} = -2b _{1}, b _{3} = -b _{1}$
Satisfied $b _{1} + 7b _{2} = 13b _{3}$ (Actually all three planes are co-incident)
(D) $D\neq 0$.

If $a{ e }^{ x }+b{ e }^{ y }=c;\quad p{ e }^{ x }+q{ e }^{ y }=d$ and $\quad { \Delta  } _{ 1 }=\begin{vmatrix} a & b \ p & q \end{vmatrix};{ \Delta  } _{ 2 }=\begin{vmatrix} c & b \ d & q \end{vmatrix};{ \Delta  } _{ 3 }=\begin{vmatrix} a & c \ p & d \end{vmatrix}$ then the value of $(x,y)$ is:

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

  1. $\left( \cfrac { { \Delta } _{ 2 } }{ { \Delta } _{ 1 } } ,\cfrac { { \Delta } _{ 3 } }{ { \Delta } _{ 1 } } \right) $

  2. $\left( \log { \cfrac { { \Delta } _{ 2 } }{ { \Delta } _{ 1 } } } ,\log { \cfrac { { \Delta } _{ 3 } }{ { \Delta } _{ 1 } } } \right) $

  3. $\left( \log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 3 } } } ,\log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 2 } } } \right) $

  4. $\left( \log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 2 } } } ,\log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 3 } } } \right) $

Show answer
Correct Option: B

System of equations
$x + 2y + z = 0, 2x + 3y- z = 0 $ and $(tan\theta) x + y -3z = 0$ has non-trivial solution then number of value(s) of $\theta \epsilon (-\pi,\pi)$ 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: C
Explanation:

For non-trivial solutions
$\begin{vmatrix}
1 &2  &1 \ 
2 &3  &-1 \ 
tan\theta &1  &-3 
\end{vmatrix}=0$
$\Rightarrow 6-5\tan\theta =0$
$\Rightarrow\displaystyle \tan \theta=\frac{6}{5}$
Hence, the number of solutions in $(-\pi,\pi)$ is 2
Hence, option 'C' is correct.

The number of values of $\theta \in (0,\pi )$ for which the system of linear equations
x+3y+7z=0
x+4y+7z=0
$(\sin { 3\theta  } )x+(\cos { 2\theta  } )y+2z=0$
has a non trivial solution is :

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

  1. one

  2. three

  3. four

  4. two

Show answer
Correct Option: A