Question

The set of all values of $\lambda $ for which the system of linear equations:
$2{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}}$ ; $2{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}}$;$-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}$has a non – trivial solution
( a ) is an empty set
( b ) is a singleton set
( c ) contain two elements
( d ) contains more than two elements

Answer 1

Hint: What we will do to solve this question is, we will first write the system of linear equation $2{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}}$ ; $2{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}}$;$-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}$ in matrix form that is AX = B and then we will put $\left| A \right|=0$ as it is given that system of equation has non trivial solution and then we will evaluate for how many numbers of $\lambda $ does $\left| A \right|=0$.

Complete step-by-step answer:
In question it is given that $2{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}}$ ; $2{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}}$;$-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}$has non – trivial solution for some values of $\lambda $, then we have to find how many elements are there in set of values of $\lambda $.
Before we solve this let us ee what is trivial and non-trivial solution.
If the system of equation in which the determinant of the coefficient is zero, then the solution is called non – trivial solution and id=f the system of equation in which the determinant of the coefficient matrix is not zero but the solution are x = y = z then solution is trivial.

$2{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}}$ ;
$2{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}}$;
$-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}$
Re-arranging the equations, we get
$2{{x}_{1}}-\lambda {{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=0$
$2{{x}_{1}}-3{{x}_{2}}-\lambda {{x}_{2}}+2{{x}_{3}}=0$
$-{{x}_{1}}+2{{x}_{2}}-\lambda {{x}_{3}}=0$
Taking common factors out and collecting coefficients, we get
$(2-\lambda ){{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=0$
$2{{x}_{1}}-(3+\lambda ){{x}_{2}}+2{{x}_{3}}=0$
$-{{x}_{1}}+2{{x}_{2}}-\lambda {{x}_{3}}=0$
Now writing linear equations in matrix form which is AX = B, where A is matrix of coefficient of linear equation, x is matrix of variables in linear equation and B is matrix of output of linear equations.
$\left( \begin{matrix}
   2-\lambda & -2 & 1 \\
   2 & -(3+\lambda ) & 2 \\
   -1 & 2 & -\lambda \\
\end{matrix} \right)$ $\left( \begin{align}
  & {{x}_{1}} \\
 & {{x}_{2}} \\
 & {{x}_{3}} \\
\end{align} \right)$ =\[\left( \begin{align}
  & 0 \\
 & 0 \\
 & 0 \\
\end{align} \right)\]
Now, for non – trival solution, determinant of matrix A i.e $\left| \begin{matrix}
   2-\lambda & -2 & 1 \\
   2 & -(3+\lambda ) & 2 \\
   -1 & 2 & -\lambda \\
\end{matrix} \right|$ must be zero
Now , determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
So, $\left| \begin{matrix}
   2-\lambda & -2 & 1 \\
   2 & -(3+\lambda ) & 2 \\
   -1 & 2 & -\lambda \\
\end{matrix} \right|=(2-\lambda )[-(3+\lambda )(-\lambda )-2\cdot 2]-2[(-2)\cdot (-\lambda )-2\cdot 1]+(-1)[2(-2)+(3+\lambda )\cdot 1]$
Now, as system of equation has non – trivial solution so, $\left| A \right|=0$
On simplifying, we get
\[(2-\lambda )[{{\lambda }^{2}}+3\lambda -4]-2[2\lambda -2]+(-1)[-4+3+\lambda ]=0\]
\[(2-\lambda )[{{\lambda }^{2}}+3\lambda -4]+2[-2\lambda +2]+1[1-\lambda ]=0\]
\[(\lambda -1)({{\lambda }^{2}}+2\lambda -3)=0\]
\[{{(\lambda -1)}^{2}}(\lambda +3)=0\]
\[\lambda =1,-3\]
So, $\left| A \right|=0$ for \[\lambda =1,-3\]

So, the correct answer is “Option c”.

Note: while solving the question based on system of linear equation, always find the values of matrix A, matrix X and matrix B which is AX = B, where A is matrix of coefficient of linear equation, x is matrix of variables in linear equation and B is matrix of output of linear equations. Always remember that, determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$ and system of equation has non – trivial solution so, $\left| A \right|=0$. As finding determinant is very calculative so try to avoid calculation error.