Mathematics · Quantitative Aptitude

Linear Equations

196 Questions

Linear equations involve solving for unknown variables in single or multi variable systems. These questions test algebraic manipulation and logical consistency skills. They are a core component of quantitative aptitude and advanced mathematics tests.

Solving simultaneous equationsSingle variable equationsSystem consistency checksIndeterminate equationsMatrix form solutions

Linear Equations Questions

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

The system of equation $\displaystyle \alpha x+y+z=\alpha-1,:x+\alpha y+z=\alpha-1,:x+y+\alpha z=\alpha-1$ has no solution if $\alpha$ is

  1. either $-2$ or $1$
  2. $-2$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $D =\begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$

$\Rightarrow D =(\alpha-1)^{2}(\alpha+2)$

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

For $\alpha=1$, equations takes the form $x+y+z=0$
Hence, infinitely many solution for $\alpha=1$.

For $\alpha=-2$,
$D _{1}=\begin{vmatrix} -3 & 1 & 1 \ -3 & -2 & 1 \ -3 & 1 & -2 \end{vmatrix}$
$D _{1}=27 \ne 0$

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

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

The solution set of the equation $\left| \begin{matrix} 2 & 3 & x \ 2 & 1 & { x }^{ 2 } \ 6 & 7 & 3 \end{matrix} \right| =0$ is 

  1. $\left\{ 1,-3 \right\} $
  2. $\left\{ 1,-1 \right\} $
  3. $\left\{ 1,3 \right\} $
  4. $\theta $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expanding the determinant: 2(1*3 - 7*x^2) - 3(2*3 - 6*x^2) + x(2*7 - 6*1) = 0. 2(3 - 7x^2) - 3(6 - 6x^2) + x(8) = 0. 6 - 14x^2 - 18 + 18x^2 + 8x = 0. 4x^2 + 8x - 12 = 0. x^2 + 2x - 3 = 0. (x+3)(x-1) = 0. Roots are 1 and -3.

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

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

  1. $c-b:a-c:b-a$
  2. $b+c:c+a:a+b$
  3. $a:b:c$
  4. $\displaystyle \frac{a}{b}:\frac{b}{c}:\frac{c}{a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $x:y:z=(c-b):(a-c):(b-a)$

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

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

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

  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 $k \in R$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
$y=1-2k ; x=2+k$

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

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

  1. $\lambda=4, \mu=11$
  2. $\lambda=3, \mu=10$
  3. $\lambda=2, \mu=8$
  4. $\lambda=1, \mu=11$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\Delta =0,{ \Delta  } _{ 1 }\Rightarrow \lambda =3\Rightarrow \mu =10$

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

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

  1. a unique solution

  2. no solution

  3. two solutions

  4. infinite solutions
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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.

  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$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. Unique Solution

  2. Finite number of solutions

  3. Infinite Solutions

  4. No solution
Reveal answer Fill a bubble to check yourself
D Correct answer
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.


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

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 

  1. either $-2\ or\ 1$
  2. $-2$
  3. $1$
  4. $not\ -2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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) $1$ 
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

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 $k\neq 3$ 

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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. $k \neq 3$

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 elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

The system $\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$ has no solution, if

  1. $a=-5,b\ne 5$
  2. $a=-5, b=5$
  3. $a\ne -5, b=5$
  4. $a\ne -5,b\ne 5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 
$\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$
The augment matrix of given system is
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 3 & 5 & -3 & | & b \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 2 }\rightarrow { R } _{ 2 }-3{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 8 & a-4 & | & -4 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 2 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 0 & a+5 & | & 5-b \end{bmatrix}$
We know that
For no solution
Rank A < Rank $\left[ A|B \right] $
$\therefore$ $a+5=0$ and $5-b\ne 0$
$a=-5, b\ne 5$