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

If $a,\ b,\ c$ are non zeros, then the system of equations $\left( \alpha +a \right) x+\alpha y+\alpha z=0,\ \alpha x+\left( \alpha +b \right) y+\alpha z=0,\ \alpha x+\alpha y+\left( \alpha +c \right) z=0$ has a non trivial solution if

  1. ${ \alpha }^{ -1 }=-\left( { a }^{ -1 }+{ b }^{ -1 }+{ c }^{ -1 } \right) $
  2. ${ \alpha }^{ -1 }=a+b+c$
  3. $\alpha +a+b+c=1$
  4. $\alpha =a+b+c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Homogeneous steamy has non- trivial solution it's mean determinant is zero

So, $\begin{vmatrix} (\alpha +a) & \alpha  & \alpha  \ \alpha  & (\alpha +b) & \alpha  \ \alpha  & \alpha  & \alpha +c \end{vmatrix}=0$
$\Rightarrow (\alpha+a)[(\alpha +b)(\alpha +c)(\alpha +c)-\alpha ^2]-[\alpha ^2+ \alpha C-\alpha^2]+\alpha [\alpha ^2-\alpha^2-2b]=0$
$\Rightarrow (\alpha +a)[\alpha ^2+\alpha c \alpha b+b^c-\alpha^2]-\alpha^2c-\alpha ^2b=0$
$\Rightarrow \alpha 2/c+\alpha2/b+\alpha ac+\alpha ab+abc-\alpha ^2c-\alpha ^2b=0$
$\Rightarrow \alpha(ab+bc+ac)=-abc $
$\Rightarrow \alpha =\dfrac{-abc}{ab+bc+ac}$
$\Rightarrow \dfrac{1}{\alpha}=-\dfrac{-(ab+bc+ac)}{abc}$
$\Rightarrow \dfrac{1}{\alpha}=-\left(\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\right)$
$\Rightarrow \boxed{\alpha ^{-1}=-(a^{-1}+b^{-1}+c^{-1})}$

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

The system of equation $5x+2y=4$,$7x+3y=5$ are inconsistent.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$5x + 2y - 4 = 0$, $7x + 3y - 5 = 0$
$\dfrac{{{a _1}}}{{{a _2}}} = \dfrac{5}{7}$
$\dfrac{{{b _1}}}{{{b _2}}} = \dfrac{2}{3}$
$\dfrac{{{c _1}}}{{{c _2}}} = \dfrac{{ - 4}}{{ - 5}} = \dfrac{4}{5}$
Since $\dfrac{{{a _1}}}{{{a _2}}} \ne \dfrac{{{b _1}}}{{{b _2}}}$
The system of equations has unique solution
Therefore it is consistent
Thus it is false
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The system of equations

$\displaystyle 
\begin{matrix}kx +y+z=1&  & \
 x+ky+z=k&  & \
 x+y+kz=k^{2}&  &
\end{matrix}$
have no solution,if k equals ?

  1. 0

  2. 1

  3. -1

  4. -2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta =\begin{vmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{vmatrix}$

$\Rightarrow \Delta =(k-1)^{2}(k+2)$

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

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

For k=-2,
$D _{1}=\begin{vmatrix} 1 & 1 & 1 \ -2 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix}$
$D _{1}=9 \ne 0$

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

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

The system of equations
$\displaystyle x + y + z = 2$
$\displaystyle 2x - y + 3z = 5$
$\displaystyle x - 2y - z + 1 = 0$
written in matrix form is

  1. $\displaystyle \begin{bmatrix}

    x \\

    y \\

    z

    \end{bmatrix} \begin{bmatrix}

    1 & 1 & 1 \\

    2 & -1 & 3 \\

    1 & -2 & -1

    \end{bmatrix} = \begin{bmatrix}

    2 \\

    5 \\

    -1

    \end{bmatrix}$
  2. $\displaystyle \begin{bmatrix}

    1 & 1 & 1 \\

    2 & -1 & 3 \\

    1 & -2 & -1

    \end{bmatrix} \: \begin{bmatrix}

    x \\

    y \\

    z

    \end{bmatrix} = \begin{bmatrix}

    -2 \\

    -5 \\

    1

    \end{bmatrix}$
  3. $\displaystyle \begin{bmatrix}

    1 & 1 & 1 \\

    2 & -1 & 3 \\

    1 & -2 & -1

    \end{bmatrix} \: \begin{bmatrix}

    x \\

    y \\

    z

    \end{bmatrix} = \begin{bmatrix}

    2 \\

    5 \\

    -1

    \end{bmatrix}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given system of equations can be written as
$AX=B$

$\begin{bmatrix} 1 & 1 & 1 \ 2 & -1 & 3 \ 1 & -2 & -1 \end{bmatrix}: \begin{bmatrix} x \ y \ z \end{bmatrix}=\begin{bmatrix} 2 \ 5 \ -1 \end{bmatrix}$

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

If the following system of equations possess a non-trivial solution over the set of rationals
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$,
then x,y,z are in the ratio 

  1. $ \displaystyle \frac{15}{2} : 1 : 3$
  2. $ \displaystyle \frac{15}{2} : -1 : - 3$
  3. $ \displaystyle \frac{15}{2} : 1 : - 3$
  4. $ \displaystyle -\frac{15}{2} : 1 : - 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For non trivial solution

$\Delta = 0$

$\therefore \begin{vmatrix}1 & k & 3\ 3 & k & -2\ 2 & 3 & -4\end{vmatrix} = 0$

applying $R _2 \rightarrow R _2 - 3R _1$ and $R _3 \rightarrow R _3 - 2 R _1$

$\therefore \begin{vmatrix}1 & k & 3\ 0 & -2k & -11\ 0 & 3-2k & -10\end{vmatrix} = 0$

$\Rightarrow \begin{vmatrix}-2k & -11 \3-2k  & -10\end{vmatrix} = 0$

$\Rightarrow 20k + 33- 22k = 0$

$\therefore k = \dfrac{33}{2}$

Putting the value of $k$ in the given equations. Then equations become

$\displaystyle x + \dfrac{33}{2} y + 3z = 0$               ...(i)

$\displaystyle 3x + \dfrac{33}{2} y - 2z = 0$              ...(ii)

$2x + 3y - 4z = 0$                           .....(iii)

Multiply (i) by 3 and subtract from (ii) then we get

$-33y - 11z = 0$

or   $z = - 3y$             ...(iv)

again multiply (i) by 2 and subtract from (iii) then we get

$-30y - 10z = 0$

$\therefore z = - 3y$                 ....(v)

Now let $y = \lambda,$

$ \therefore z = - 3 \lambda$

from (iii), $2x + 3 \lambda + 12 \lambda = 0$

$\therefore x = \displaystyle - \frac{15 \lambda}{2}$

$\therefore x : y : z = - \displaystyle \frac{15}{2} : 1 : - 3$

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

If the system of equations $ax+by+c=0$ $bx+cy+a=0$  ,$cx+ay+b=0$ has a solution then the system of equations $(b+c)x+(c+a)y+(a+b)z=0$   ,$(c+a)x+(a+b)y+(b+c)z=0$  , $(a+b)x+(b+c)y+(c+a)z=0$ has 

  1. only one solution

  2. no solution

  3. infinite number of solutions

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ax+by+c=0,bx+cy+a=0,cx+ay+b=0$
Gives
$\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$
Now $(b+c)x+(c+a)y+(a+b)z=0,(c+a)x+(a+b)y+(b+c)z=0,(a+b)x+(b+c)y+(c+a)z=0$
gives
$\begin{vmatrix} b+c & c+a & a+b \ c+a & a+b & b+c \ a+b & b+c & c+a \end{vmatrix}=\begin{vmatrix} b & c & a \ c & a & b \ a & b & c \end{vmatrix}+\begin{vmatrix} c & a & b \ a & b & c \ b & c & a \end{vmatrix}=\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}+\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=2\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$

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

Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equations
                 $\lambda x + (sin  \alpha) y + (cos  \alpha) z = 0$
                  $x + (cos  \alpha) y + (sin  \alpha) z = 0$
                  $ - x + (sin  \alpha) y + (cos  \alpha) z = 0$
has a non-trivial solution. For $\lambda = 1$, find all values of $\alpha$ which are possible

  1. $\pi/8$
  2. $7\pi/8$
  3. $15\pi/8$
  4. $9\pi/8$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

To   find  all  values   of    $\alpha$   which   are  possible
For  $\lambda = 1 $
$ \begin{bmatrix}
1  & sin\alpha  & cos\alpha \
 1 & cos\alpha  & sin\alpha  \
 -1& sin\alpha   &  cos\alpha 
\end{bmatrix} = \begin{bmatrix}
0\
0\
0\end{bmatrix}$

On   Solving   this   equation, we   get
$ 2cos2\alpha = 0$
$ cos2\alpha  = 0 $
$ 2\alpha  = \dfrac{\pi }{2} $
 Finally  , we  get

$ \alpha  = \dfrac{\pi }{8}   $ 
 
$ \alpha  = \dfrac{7\pi }{8}   $     and

$ \alpha  = \dfrac{ 9\pi }{8}   $