Question

Let $\lambda$ be a real number for which the system of linear equations:
$\begin{aligned} & x+y+z=6 \\\ & 4x+\lambda y-\lambda z=\lambda -2 \\\ & 3x+2y-4z=-5 \\\ \end{aligned}$
have infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:
A.${{\lambda }^{2}}-3\lambda -4=0$
B. ${{\lambda }^{2}}-\lambda -6=0$
C. ${{\lambda }^{2}}+3\lambda -4=0$
D. ${{\lambda }^{2}}+\lambda -6=0$ .

Answer 1

We will first start with writing the coefficient matrix for the given system of equations and then we will apply the condition of infinite solutions that is the determinant of coefficient matrix will be 0, after that we will get the value of $\lambda$. We will then check from the options what equation does $\lambda$ satisfy and we will get the answer.

Complete step-by-step answer :
Let us consider a system of equations given by :
$\begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\\ \end{aligned}$
Now, we know that this system of equations will have infinitely many solutions if the determinant of the coefficients matrix is 0. We can write it as:
$\left| \Delta \right|=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|=0$
Now we are given the following system of the equations in the question:

& x+y+z=6\text{ }.....................\text{ (i)} \\\ & 4x+\lambda y-\lambda z=\lambda -2\text{ }.....................\left( \text{ii} \right) \\\ & 3x+2y-4z=-5\text{ }....................\left( \text{iii} \right) \\\ \end{aligned}$$ We will first form the coefficient matrix for our system of the question. For that we will be writing the coefficients of x in the first column, coefficients of y in the second column and coefficients of z in the third column. So, we will have: $\left( \begin{matrix} 1 & 1 & 1 \\\ 4 & \lambda & -\lambda \\\ 3 & 2 & -4 \\\ \end{matrix} \right)$ Now, for the system of equations to have infinite solutions determinant of this matrix must be 0 $$\begin{aligned} & \Rightarrow \left| \Delta \right|=\left| \begin{matrix} 1 & 1 & 1 \\\ 4 & \lambda & -\lambda \\\ 3 & 2 & 4 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow 1\left( -4\lambda +2\lambda \right)-1\left( -16+3\lambda \right)+1\left( 8-3\lambda \right)=0 \\\ & \Rightarrow -2\lambda +16-3\lambda +8-3\lambda =0\Rightarrow -8\lambda +24=0\Rightarrow \lambda =\dfrac{24}{8} \\\ & \Rightarrow \lambda =3 \\\ \end{aligned}$$ We will now check from the options that which equation will $$\lambda =3$$ satisfy, So we see that the equation: ${{\lambda }^{2}}-\lambda -6=0$ will satisfy for $$\lambda =3$$ , ${{\lambda }^{2}}-\lambda -6\Rightarrow {{\left( 3 \right)}^{2}}-3-6\Rightarrow 9-3-6\Rightarrow 6-6=0=R.H.S$ **Hence, option B is correct.** **Note** : There can either be 0 solutions, exactly 1 solution, or infinitely many solutions. A system has infinite solutions when the lines are coinciding or con-current. This same idea extends for n equations with n unknowns. Be conscious while finding out the determinant as sign changes alternatively students can make mistakes there.