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Question: The set of all values of \(\lambda \) for which the system of linear equations: \(2{{x}_{1}}-2{{x}...

The set of all values of λ\lambda for which the system of linear equations:
2x1−2x2+x3=λx12{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}} ; 2x1−3x2+2x3=λx22{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}};−x1+2x2=λx3-{{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

Explanation

Solution

What we will do to solve this question is, we will first write the system of linear equation 2x1−2x2+x3=λx12{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}} ; 2x1−3x2+2x3=λx22{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}};−x1+2x2=λx3-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}} in matrix form that is AX = B and then we will put ∣A∣=0\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 λ\lambda does ∣A∣=0\left| A \right|=0.

Complete step-by-step answer:
In question it is given that 2x1−2x2+x3=λx12{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}} ; 2x1−3x2+2x3=λx22{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}};−x1+2x2=λx3-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}has non – trivial solution for some values of λ\lambda , then we have to find how many elements are there in set of values of λ\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.

2x1−2x2+x3=λx12{{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=\lambda {{x}_{1}} ;
2x1−3x2+2x3=λx22{{x}_{1}}-3{{x}_{2}}+2{{x}_{3}}=\lambda {{x}_{2}};
−x1+2x2=λx3-{{x}_{1}}+2{{x}_{2}}=\lambda {{x}_{3}}
Re-arranging the equations, we get
2x1−λx1−2x2+x3=02{{x}_{1}}-\lambda {{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=0
2x1−3x2−λx2+2x3=02{{x}_{1}}-3{{x}_{2}}-\lambda {{x}_{2}}+2{{x}_{3}}=0
−x1+2x2−λx3=0-{{x}_{1}}+2{{x}_{2}}-\lambda {{x}_{3}}=0
Taking common factors out and collecting coefficients, we get
(2−λ)x1−2x2+x3=0(2-\lambda ){{x}_{1}}-2{{x}_{2}}+{{x}_{3}}=0
2x1−(3+λ)x2+2x3=02{{x}_{1}}-(3+\lambda ){{x}_{2}}+2{{x}_{3}}=0
−x1+2x2−λx3=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.
(2−λ−21 2−(3+λ)2 −12−λ )\left( \begin{matrix} 2-\lambda & -2 & 1 \\\ 2 & -(3+\lambda ) & 2 \\\ -1 & 2 & -\lambda \\\ \end{matrix} \right) (x1 x2 x3 )\left( \begin{aligned} & {{x}_{1}} \\\ & {{x}_{2}} \\\ & {{x}_{3}} \\\ \end{aligned} \right) =(0 0 0 )\left( \begin{aligned} & 0 \\\ & 0 \\\ & 0 \\\ \end{aligned} \right)
Now, for non – trival solution, determinant of matrix A i.e ∣2−λ−21 2−(3+λ)2 −12−λ ∣\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 3×33\times 3 is evaluated as,
∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a32a23)−a21(a12a33−a32a13)+a31(a23a12−a22a13)\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, ∣2−λ−21 2−(3+λ)2 −12−λ ∣=(2−λ)[−(3+λ)(−λ)−2⋅2]−2[(−2)⋅(−λ)−2⋅1]+(−1)[2(−2)+(3+λ)⋅1]\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, ∣A∣=0\left| A \right|=0
On simplifying, we get
(2−λ)[λ2+3λ−4]−2[2λ−2]+(−1)[−4+3+λ]=0(2-\lambda )[{{\lambda }^{2}}+3\lambda -4]-2[2\lambda -2]+(-1)[-4+3+\lambda ]=0
(2−λ)[λ2+3λ−4]+2[−2λ+2]+1[1−λ]=0(2-\lambda )[{{\lambda }^{2}}+3\lambda -4]+2[-2\lambda +2]+1[1-\lambda ]=0
(λ−1)(λ2+2λ−3)=0(\lambda -1)({{\lambda }^{2}}+2\lambda -3)=0
(λ−1)2(λ+3)=0{{(\lambda -1)}^{2}}(\lambda +3)=0
λ=1,−3\lambda =1,-3
So, ∣A∣=0\left| A \right|=0 for λ=1,−3\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 3×33\times 3 is evaluated as,
∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a32a23)−a21(a12a33−a32a13)+a31(a23a12−a22a13)\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, ∣A∣=0\left| A \right|=0. As finding determinant is very calculative so try to avoid calculation error.