Question

The existence of the unique solution of the system of equations
$x+y+z=\beta$
$5x-y+\alpha z=10$
$2x+3y-z=6$
depends on

• A. $\alpha$ only
• B. $\beta$ only
• C. $\alpha$ and $\beta$ both
• D. neither $\beta$ nor $\alpha$

Answer 1

Answer: (A)

Given, system of equation is $x+y+z=\beta$
$5x-y+\alpha z=10$ and $2x+3y-z=6$
For unique solution $\left|\begin{array}{ccc}1& 1& 1\\ 5& -1& \alpha \\ 2& 3& -1\end{array}\right|\ne 0$
$\Rightarrow 1\left(1-3\alpha \right)-1\left(-5-2\alpha \right)+1\left(15+2\right)\ne 0$
$\Rightarrow 1-3\alpha +5+2\alpha +17\ne 0$
$\Rightarrow -\alpha +23\ne 0$
$\Rightarrow \alpha \ne 23$
Hence, for the existence of the unique solutionthe system of equations depend on $\alpha$ only