Tag: system of simultaneous equations

Questions Related to system of simultaneous equations

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

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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

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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

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

Use matrix to solve the following system of equations
$x+ y +z = 3$

$x +2y+ 3z= 4$
$2x+3y +4z= 7$

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

  1. $x = 2 + k, :y = -1 - 2k, :z = -k $ where $k \in R$

  2. $x = 2 + k, :y = 1 - 2k, :z = k $ where $k \in R$

  3. $x = -2 - k, :y = 1 - 2k, :z = -k $ where $k \in R$

  4. $x = -2 + k, :y = -1 + 2k, :z = -k $ where $k \in R$

Show answer
Correct Option: B
Explanation:

Given system of equations can be written as
$AX=B$
where $A=\begin{bmatrix} { 1 } & { 1 } & { 1 } \ { 1 } & { 2 } & 3\ { 2 } & 3 & 4 \end{bmatrix}$ 
$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$ ;$B=\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

Here, $|A|=0$
Now, we will find $(adj A)B$

$adj A=C^{T}={\begin{bmatrix} { -1 } & { 2 } & { -1 } \ { -1 } & { 2 } & -1 \ { 1 } & -2 & 1 \end{bmatrix}}^T$

$\Rightarrow adj A=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}$

Now, $(adj A)B=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

$\Rightarrow (adj A)B=\begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
$\Rightarrow (adj A)B=O$

Hence,the system of equations has infinitely many solutions.
Let $z=k$ where $k\in R$
Then 
$x+y=3-k$
$x+2y=4-3k$
Solving these eqns, we get 
$y=1-2k ; x=2+k$

Investigate for what values of $\lambda, \mu$ the simultaneous equation $x+y+z=6; x+2y+3z=10$ & $x+2y+\lambda z=\mu$ have an infinite number of solutions

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

  1. $\lambda=4, \mu=11$

  2. $\lambda=3, \mu=10$

  3. $\lambda=2, \mu=8$

  4. $\lambda=1, \mu=11$

Show answer
Correct Option: B
Explanation:

$\Delta =\begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 2 & \lambda  \end{vmatrix}=1\left( 2\lambda -6 \right) -1\left( \lambda -3 \right) +1\left( 2-2 \right) =2\lambda -6-\lambda +3+0=\lambda -3\ \Delta _{ 1 }=\begin{vmatrix} 6 & 1 & 1 \ 10 & 2 & 3 \ \mu  & 2 & \lambda  \end{vmatrix}=6\left( 2\lambda -6 \right) -1\left( 10\lambda -3\mu  \right) +1\left( 20-2\mu  \right) \ =12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16$

For infinite solution $\Delta =0,{ \Delta  } _{ 1 }\Rightarrow \lambda =3\Rightarrow \mu =10$

The equations $x+4y-2z=3$, $3x+y+5z=7$ and $2x+3y+z=5$ have

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

  1. a unique solution

  2. no solution

  3. two solutions

  4. infinite solutions

Show answer
Correct Option: B
Explanation:

$\Delta=\begin{vmatrix} 1 &4  &-2  \3  &1  &5  \  2&3  &1  \end{vmatrix}=1(1-15)-4(3-10)-2(9-2)=0$
and $\Delta _1$ is not zero
Therefore, it has no solution

Let $a,\ b,\ c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x=cy+bz,\ y=az+cx$ and $z=bx+ay$. Then $a^{2}+b^{2}+c^{2}+2abc$ 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. $-1$

Show answer
Correct Option: B
Explanation:

For $x,y,z$ not to be simultaneously zero

determinant of the coefficients should be zero.
$\left| \begin{matrix}-1 & c & b\ c& -1 & a \ b & a & -1 \end{matrix}\right|=0$
$\Rightarrow a^2+b^2+c^2+2abc = 1$

One of the roots of $\begin{vmatrix} x+a & b & c\  a & x+b & c\  a & b & x+c \end{vmatrix}=0$ is :

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

  1. $abc$

  2. $a+b+c$

  3. $-(a+b+c)$

  4. $-abc$

Show answer
Correct Option: C
Explanation:
For the determinant
$ column\ 1=column\ 1+column\ 2+column \ 3\\ \left| \begin{matrix} x+a+b+c & b & c \\ x+a+b+c & \quad x+b & c \\ x+a+b+c & b & x+c \end{matrix} \right| =0\\ \\ row2=row2-row1\\ row3=row3-row1\\ \begin{vmatrix} x+a+b+c & \quad b & \quad c \\ 0 & \quad x & \quad 0 \\ 0 & \quad 0 & \quad x \end{vmatrix}=0$
Expanding the determinant 
$\Rightarrow  (x+a+b+c)(x.x)-b.0+c.0=0$ 
$\Rightarrow { x }^{ 2 }(x+a+b+c)=0$ 
Then, the roots are $x=0$ or $x=-(a+b+c)$