Question

The number of solutions of equation
$x_{2}-x_{3}=1,-x_{1}+2 x_{3}=2, x_{1}-2 x_{2}=3$ is

• A. zero
• B. one
• C. two
• D. infinite

Answer 1

Answer: (A)

The system is $0 x_{1}+x_{2}-x_{3}=1$
$-x_{1}+0 x_{2}+2 x_{3}=2$
$x_{1}-2 x_{2}+0 x_{3}=3 $
$\Rightarrow \begin{bmatrix}0 & 1 & -1 \\ -1 & 0 & 2 \\ 1 & -2 & 0\end{bmatrix}\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3}\end{bmatrix}=\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$ or $A X=B$
Clearly $|A|=0$
Now Adj $A=\begin{bmatrix}4 & 2 & 2 \\ 2 & 1 & 1 \\ 2 & 1 & 1\end{bmatrix}$
$\therefore (\text{Adj} A) B \neq 0 $
$\Rightarrow $ system is inconsistent