Question

The system of equations,
$\begin{align}
  & ax+y+z=a-1, \\
 & x+ay+z=a-1, \\
 & x+y+az=a-1 \\
\end{align}$
Has no solution if a is
A. Either -2 or 1.
B. -2
C. 1
D. Not -2

Answer 1

Hint: In this question, we are given three equations in terms of 'a' and we need to find the value for which the system of equations has no solution. For this, we will first form a coefficient matrix A. Then we will find the determinant of the matrix, for no solution, the determinant should be equal to zero. So, solving equations formed from determinants will give us the value of 'a'. Equation formed will be cubic so we will find three values of 'a'.

Complete step-by-step answer:
Here we are given the equation as:
$\begin{align}
  & ax+y+z=a-1, \\
 & x+ay+z=a-1, \\
 & x+y+az=a-1 \\
\end{align}$
Let us make coefficient matrix of these equations we get:
$A=\left| \begin{matrix}
   a & 1 & 1 \\
   1 & a & 1 \\
   1 & 1 & a \\
\end{matrix} \right|$.
We have to find the value of 'a' such that these equations have no solution. We know that, if the determinant of the coefficient matrix is equal to zero, then the equation has no solution.
So let us put the value of the determinant of matrix A equal to A.
Let us first simplify the determinant using row and column operation.
Determinant A is $A=\left| \begin{matrix}
   a & 1 & 1 \\
   1 & a & 1 \\
   1 & 1 & a \\
\end{matrix} \right|$.
Adding columns first, second and third to the first column i.e. Applying the operation ${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}$ we get:
$A=\left| \begin{matrix}
   a+2 & 1 & 1 \\
   a+2 & a & 1 \\
   a+2 & 1 & a \\
\end{matrix} \right|=0$.
Taking (a+2) common from the first column of determinant we get (property of determinant).
$A=\left( a+2 \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & a & 1 \\
   1 & 1 & a \\
\end{matrix} \right|=0$.
Now, applying row operation on second row and third row given by ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\text{ and }{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ we get:
$A=\left( a+2 \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & a-1 & 0 \\
   0 & 0 & a-1 \\
\end{matrix} \right|=0$.
Now let us find the determinant using first column we get:
$\begin{align}
  & \left| A \right|=\left( a+2 \right)\left[ 1\left( a-1 \right)\left( a-1 \right)-0+0 \right] \\
 & \Rightarrow \left| A \right|=\left( a+2 \right)\left( a-1 \right)\left( a-1 \right) \\
\end{align}$.
Since determinant is supposed to be equal to 0, so $\left| A \right|=0$.
i.e. $\left( a+2 \right)\left( a-1 \right)\left( a-1 \right)=0$.
Hence the value of 'a' is given by $a+2=0\text{ and }a-1=0$.
Therefore, a = -2, 1
Now putting in the value of a = 1 in the given system of equations, we find that the three equations become the same. Therefore, it will have an infinite solution. So only a = -2 is our required value of 'a' for which given system of equations has no solution.

So, the correct answer is “Option B”.

Note: Students should know properties of determinants for solving this question. Students make a mistake of selecting option A for two found values of 'a' but they should note that a = 1 makes equation same which gives infinite solution.