Question

If $S$ is the set of distinct values of ' $b$' for which the following system of linear equations
$$\begin{gathered} x+y+z=1 \\ x+ay+z=1 \\ ax+by+z=0 \end{gathered}$$
has no solution, then $\mathrm{S}$ is

• A. an infinite set
• B. a finite set containing two or more elements
• C. a singleton set
• D. an empty set

Answer 1

The correct option is C

a singleton set

Explanation for the correct option.

Step 1: Find the value of $a$

For the system of equations with no solution, the determinant of the coefficient matrix should be $0$

$\begin{array}{c}\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& b& 1\end{array}\end{array}\end{array}=0$

On solving the determinant we get:
$$\begin{gathered} \Rightarrow 1(a-b)-1(1-a)+1\left(b-a^2\right)=0 \\ \Rightarrow a-b-1+a+b-a^2=0 \\ \Rightarrow 2a-1-a^2=0 \\ \Rightarrow -(a-1{)}^2=0 \\ \Rightarrow a=1 \end{gathered}$$
Step 2: Find the value of $b$

For $\mathrm{a}=1$, the equations become:
$$\begin{gathered} x+y+z=1 \\ x+y+z=1 \\ x+by+z=0 \end{gathered}$$
From the above three equations, we can see that if $b=1$, the system will be inconsistent and hence will produce no solution.

For $b\equiv 1$, the system will produce infinite solutions.
So, S has to be a singleton set $\left\{1\right\}$ for no solution.

Hence, option C is correct.