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.
Find all real solutions \({(x_{1}, x_{2}, x_{3}, \lambda)}\) for the system of equations
\(x_{2}-3 x_{3}-x_{1} \lambda =0\)
\(x_{1}-3 x_{3}-x_{2} \lambda =0\)
\(x_{1}+x_{2}+x_{3} \lambda =0\)
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 \( (x_{1},-x_{1},0,-1) \), \( (x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1) \)\( x_{1} \epsilon \mathbb{R} \) .
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.
Find all real solutions \({(x_{1}, x_{2}, x_{3}, \lambda)}\) for the system of equations
\(x_{2}-3 x_{3}-x_{1} \lambda =0\)
\(x_{1}-3 x_{3}-x_{2} \lambda =0\)
\(x_{1}+x_{2}+x_{3} \lambda =0\)
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 \( (x_{1},-x_{1},0,-1) \), \( (x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1) \)\( x_{1} \epsilon \mathbb{R} \) .
