Question

The following system of equations $x_1+x_2+2x_3=1,x_1+2x_2+3x_3=2,x_1+4x_2+\alpha x_3=4$ has a unique solution. The only possible value for $\alpha$ is

• A. 0
• B. either 0 (or) 1
• C. one of 0,1 (or) -1
• D. any real number other than 5

Answer 1

The correct option is D any real number other than 5

Given AX = B

$\Rightarrow \left[\begin{array}{ccc}1& 1& 2\\ 1& 2& 3\\ 1& 4& \alpha \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 4\end{array}\right]$
$[A:B]=\left[\begin{array}{ccccc}1& 1& 2& ⋮& 1\\ 1& 2& 3& ⋮& 2\\ 1& 4& \alpha & ⋮& 4\end{array}\right]$
$R_2\to R_2-R_1;R_3\to R_3-R_1$
$\sim \left[\begin{array}{ccccc}1& 1& 2& ⋮& 1\\ 0& 1& 1& ⋮& 1\\ 0& 3& \alpha -2& ⋮& 3\end{array}\right]$
$R_3\to R_3-3R_2$
$\sim \left[\begin{array}{ccccc}1& 1& 2& ⋮& 1\\ 0& 1& 1& ⋮& 1\\ 0& 0& \alpha -5& ⋮& 0\end{array}\right]$

To have unique solution, $\rho \left(A\right)=\rho (A:B)$ = Number of unknowns = 3
$\Rightarrow \alpha \ne 5$
$\therefore$ For all values of $\alpha \ne 5$, the system will have unique solution.