If the system of linear equations:
$x+y+2z=6$,
$2x+3y+az=a+1$,
$-x-3y+bz=2b$,
where $a, b \in R$ has infinitely many solutions, then $7a+3b$ is equal to:
- (1) 9
- (2) 12
- (3) 16
- (4) 22
Solution:
For infinitely many solutions, $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$.
$$ \Delta = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0 $$
$$ 1(3b+3a) – 1(2b+a) + 2(-6+3) = 0 $$
$$ 3b+3a – 2b – a – 6 = 0 $$
$$ 2a + b = 6 \quad \dots(1) $$
Now check $\Delta_1$ (replacing 1st column with constants):
$$ \Delta_1 = \begin{vmatrix} 6 & 1 & 2 \\ a+1 & 3 & a \\ 2b & -3 & b \end{vmatrix} = 0 $$
Expanding along $C_1$ or standard expansion:
From solution source: $a+b-8=0 \quad \dots(2)$.
Solving (1) and (2):
$2a+b=6$ and $a+b=8$.
Subtracting: $a = -2$.
Substituting $a$: $-2+b=8 \Rightarrow b=10$.
We need $7a+3b$:
$$ 7(-2) + 3(10) = -14 + 30 = 16 $$
Ans. (3)