Tag: non-homogeneous linear equations

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Questions Related to non-homogeneous linear equations

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

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)

  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
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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 $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
$\displaystyle =a^{2}+b^{2}+c^{2}-\frac{1}{2}\left [ \left ( a+b+c \right )^{2}- a^{2}-b^{2}-c^{2} \right ]$
$\displaystyle =\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )\left [ \because a+b+c=0 \right ]$
$\displaystyle \Rightarrow x= \pm \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

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

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

  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 $\alpha, \beta, \gamma$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

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

  1. infinite number of solutions

  2. no solution

  3. unique solution

  4. finitely many solutions

  5. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 $\begin{bmatrix}b _{1}\ b _{2}\ b _{3}
\end{bmatrix}\epsilon \  S$?

  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 $x + 4y - 5z = b _{3}$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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) $D\neq 0$.

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

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:

  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) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let u=e^x and v=e^y. The system is au+bv=c and pu+qv=d. By Cramer's Rule, u = Delta2/Delta1 and v = Delta3/Delta1. Since u=e^x and v=e^y, x=log(Delta2/Delta1) and y=log(Delta3/Delta1).

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

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?


  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
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.