Question

If the system of linear equations :

x + y + 2z = 6,

2x + 3y + az = a + 1,

-x - 3y + bz = 2b,

where \(\mathrm{a}, \mathrm{b} \in \mathbf{R},\) has infinitely many solutions, then \(7 a+3 b\) is equal to :

(1) 9

(2) 12

(3) 16

(4) 22

Answer 1

Correct option is (3) 16

For \(\infty\) many solution

\( \Delta=\Delta_1=\Delta_2=\Delta_3=0 \)

\(\Delta=0 \)

\(\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & 2 \\ 2 & 3 & \mathrm{a} \\ -1 & -3 & \mathrm{~b} \end{array}\right|=0 \end{aligned}\)   

\((3 b+3 a)-1(2 b+a)+2(-6+3)=0 \)

\(2 a+b=6 \)      ....(1)

\(\Delta_3=0\)

\(\left|\begin{array}{ccc} 1 & 1 & 6 \\ 2 & 3 & a+1 \\ -1 & -3 & 2 b \end{array}\right|=0\)  

(6b + 3a + 3) - 1(4b + a + 1) + 6(-6 + 3) = 0  

\(2 a+2 b=16 \ldots(2)\)

Solve(1) and (2)

a + b = 8

2a + b = 6

\(\begin{aligned} & -\quad-\quad- \\ & \hline-a=2 \\ \end{aligned}\) 

a = -2, b = 10 

7a + 3b 

-14 + 30 = 16