Question

The system linear equations
$x-y+z=1$
$x+y-z=3$
$x-4y+4z=\alpha$ has

• A. A unique solution when $\alpha =2$
• B. A unique solution when $\alpha \ne 2$
• C. Infinite number of solution when $\alpha =2$
• D. Infinite number of solution when $\alpha =-2$

Answer 1

The correct option is D Infinite number of solution when $\alpha =-2$

$D=0D_1=0(\text{for}\alpha \in R)D_2=D_3=0(\text{for}\alpha =-2)$
$\text{at}\alpha =-2$ system has infinite number of solutions