Question

Consider the system of equations
$x+y+z=4$
$2x+y+3z=6$
$x+2y+pz=q$
Let $L$ denotes the value of $p$ if the system of equations has no solution.
and $M$ denotes the value of $q$ if the system of equations has infinite solutions.
Find the value of $(L^2+M$.

Answer 1

Answer: 0036

$x+y+z=4$ ... (1)
$2x+y+3z=6$ ... (2)
$x+2y+pz=q$ ... (3)
Solving (1) and (2) $\Rightarrow x=2-2z$ and $y=2+z$
Put in equation (3), we get
$pz=q-6$
Hence, for unique solution $p\ne 0,q\in R$
for no solution we must have $p=0,q\ne 6$
for infinite solution $p=0$ and $q=6$
$\therefore L=0,M=6\Rightarrow L^2+M^2=0+36=36$