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 $(L^2 + M%2)$.

Answer 1

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