Question

Consider the system of equations
ax + y + z = 1
x + ay + z = 1
x + y + az = 1, then which of the following statement(s) is/are correct?

• A. if a = 2, then the system has unique solution
• B. if a = 1, then the system has infinite solutions
• C. if a = –2, then the system has no solution
• D. if a = 2, then the system has infinite solutions

Answer 1

Answer: (A), (B), (C)

The correct options are
A

if a = 2, then the system has unique solution

B

if a = 1, then the system has infinite solutions

C

if a = –2, then the system has no solution

$\Delta =\left|\begin{array}{ccc}a& 1& 1\\ 1& a& 1\\ 1& 1& a\end{array}\right|=(a-1{)}^2(a+2)$

${\Delta }_x=\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& 1& a\end{array}\right|=(a-1{)}^2$

${\Delta }_y=\left|\begin{array}{ccc}a& 1& 1\\ 1& 1& 1\\ 1& 1& a\end{array}\right|=(a-1{)}^2$

${\Delta }_z=\left|\begin{array}{ccc}a& 1& 1\\ 1& a& 1\\ 1& 1& 1\end{array}\right|=(a-1{)}^2$

Clearly, if a = 2, then $\Delta \ne 0\Rightarrow$ system is consistent with unique solution.
If a = 1, then all three equations are identical, so system has infinite solutions
If a = -2, then $\Delta =0$ but ${\Delta }_x\ne 0,{\Delta }_y\ne 0,{\Delta }_z\ne 0.$
Thus, system is inconsistent.