Question

The set of values of $\lambda$ for which the system of linear equations.
$\begin{aligned} & x-2y-2z=\lambda x \\\ & x+2y+z=\lambda y \\\ & -x-y=\lambda z \\\ \end{aligned}$
has a non-trivial solution.
(a) contains more than two elements
(b) is a singleton
(c) is an empty set
(d) contains exactly two elements

Answer 1

To solve this question we will first rearrange the given system of linear equations to find the coefficients of x, y and z from each equation. After that we know that for the non-trivial solution, the determinant of the coefficient matrix should be zero so we will find the determinant then after solving the determinant we will get the values of $\lambda$.

Complete step by step answer:
We are given the system of linear equations as,
$\begin{aligned} & x-2y-2z=\lambda x \\\ & x+2y+z=\lambda y \\\ & -x-y=\lambda z \\\ \end{aligned}$
Now we will rearrange the above equations to find the coefficient of x, y, and z from the each equation, so we can also represent above equations as,

& \left( 1-\lambda \right)x-2y-2z=0 \\\ & x+\left( 2-\lambda \right)y+z=0 \\\ & x+y+\lambda z=0 \\\ \end{aligned}$$ Now we know that for the non-trivial solution determinant of the coefficient matrix will be zero, i.e. that determinant formed with the respective coefficients of x, y, and z from each equation will be zero. Hence we get, $\left| \begin{matrix} 1-\lambda & -2 & -2 \\\ 1 & 2-\lambda & 1 \\\ 1 & 1 & \lambda \\\ \end{matrix} \right|=0$ Now we will solve this above determinant to get the values of $$\lambda $$. As, this matrix is very simple so, we will directly expand this matric long row 1, hence we get $$\begin{aligned} & ((1-\lambda )(\lambda (2-\lambda )-1)-2(1-\lambda )-2(1-(2-\lambda ))=0 \\\ & \left( 1-\lambda \right)\left( -{{\lambda }^{2}}+2\lambda -1 \right)-2+2\lambda +2-2\lambda =0 \\\ \end{aligned}$$ Cancelling out similar terms we get, $$\left( 1-\lambda \right)\left( -{{\lambda }^{2}}+2\lambda -1 \right)=0$$ $$\left( 1-\lambda \right)\left( {{\lambda }^{2}}-2\lambda +1 \right)=0$$ Now we know $$\left( {{\lambda }^{2}}-2\lambda +1 \right)$$ can also be written as$${{\left( 1-\lambda \right)}^{2}}$$, hence we get $${{\left( 1-\lambda \right)}^{3}}=0$$ $\Rightarrow \lambda =1$ Hence $\lambda $ has only one value so it is a singleton set, **Hence option (b) is the correct answer.** **Note:** While forming the determinant you need to be careful because many students directly from the determinant without any rearrangement of the equations first which is wrong, so always make sure to rearrange the equations before forming the determinant. And remember that for the non-trivial solution determinant of the coefficient matrix will be zero.