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# ISI MStat PSB 2018 Problem 1 | System of Linear Equations

This is a beautiful sample problem from ISI MStat PSB 2018 Problem 1. This is based on finding the real solution of a system of homogeneous equations. We provide a detailed solution with prerequisites mentioned explicitly.

## Problem- ISI MStat PSB 2018 Problem 1

Find all real solutions ${(x_{1}, x_{2}, x_{3}, \lambda)}$ for the system of equations
$x_{2}-3 x_{3}-x_{1} \lambda =0$
$x_{1}-3 x_{3}-x_{2} \lambda =0$
$x_{1}+x_{2}+x_{3} \lambda =0$

## Solution

We are given the system of homogeneous equations as follows ,

$-x_{1} \lambda + x_{2} - 3 x_{3}=0$
$x_{1}-x_{2} \lambda-3 x_{3} =0$
$x_{1}+x_{2}+x_{3} \lambda =0$

Let , $A= \begin{pmatrix} - \lambda & 1 & -3 \\ 1 & - \lambda & -3 \\ 1 & 1 & \lambda \end{pmatrix}$

As A is a $3 \times 3$ matrix so ,if the Rank of A is <3 then it has infinitely many solution and

if Rank of A is 3 then it has only trivial solution i.e $x_{1}=x_{2}=x_{3}=0$

Let's try to find the rank of matrix A from it's row echelon form ,

$A= \begin{pmatrix} - \lambda & 1 & -3 \\ 1 & - \lambda & -3 \\ 1 & 1 & \lambda \end{pmatrix}$ $R_{1} \leftrightarrow R_{2}$ $\begin{pmatrix} 1 & - \lambda & -3 \\ - \lambda & 1 & -3 \\ 1 & 1 & \lambda \end{pmatrix}$ $R_{1} \lambda + R_{2} \to R_{2} ,R_{3}- R_{1} \to R_{3}$ $\begin{pmatrix} 1 & - \lambda & -3 \\ 0 & 1 - {\lambda}^2 & -3-3 \lambda \\ 0 & 1 + \lambda & 3+ \lambda \end{pmatrix}$

Now see when the determinant of A =0 as then the Rank(A) will be <3 and then it has infinitely many solutions

Det(A)= $|A|=0 \Rightarrow (1- {\lambda}^2)(3+ \lambda)+{(1+ \lambda)}^2 3=0 \Rightarrow (1+ \lambda)( \lambda +2)( \lambda -3 )=0$ $\Rightarrow \lambda =-1 ,-2 ,3$

So, if $\lambda=-1,-2,3$ then Rank(A) <3 hence it has infinitely many solutions

Now from here we can say that if $\lambda \ne -1,-2,3$ then Rank (A) =3 then the system of homogeneous equations has only trivial solution i.e $x_{1}=x_{2}=x_{3}=0$

For $\lambda =-1$ system of homogeneous equation is as follows ,

$x_{1} + x_{2} - 3 x_{3}=0$
$x_{1}+x_{2} -3 x_{3} =0$
$x_{1}+x_{2}-x_{3} =0$

Solving this we get $x_{2}=-x_{1}$ and $x_{3}=0$ . Hence solution space is {$x_{1}(1,-1,0)$} , $x_{1} \epsilon \mathbb{R}$ .

Similarly , for $\lambda =-2,3$ we have solution space { $x_1(1,1,0)$ } and { $x_1 (1,1, -2/3 )$ } respectively .

Therefore , real solutions (x1,x2,x3,λ) for the system of equations are $(0,0,0, \lambda )$ , $\lambda \epsilon \mathbb{R}$ and $(x_{1},-x_{1},0,-1)$, $(x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1)$$x_{1} \epsilon \mathbb{R}$ .