Question

The system of linear equations $3x+y-z=2,x-z=1$ and $2x+2y+az=5$ has unique solution when

• A. $a\ne 3$
• B. $a\ne 4$
• C. $a\ne 5$
• D. $a\ne 2$
• E. $a\ne 1$

Answer 1

Answer: (D)

$a\ne 2$

$3x+y-z=2$

$x-z=1$

$2x+2y+az=5$

The system has a unigue solution if $D\ne 0$

$\left|\begin{array}{ccc}3& 1& -1\\ 1& 0& -1\\ 2& 2& a\end{array}\right|\ne 03(0+2)-1(a+2)-1(2-0)\ne 06-a-2-2\ne 0\Rightarrow a\ne 2$

So, option D is correct.