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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 $$(x_{1},-x_{1},0,-1)$$, $$(x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1)$$$$x_{1} \epsilon \mathbb{R}$$ .

## Previous ISI MStat Posts:

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 $$(x_{1},-x_{1},0,-1)$$, $$(x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1)$$$$x_{1} \epsilon \mathbb{R}$$ .

## Previous ISI MStat Posts:

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