If the system of linear equations : x + y + 2z = 6, 2x + 3y + az = a + 1, –x – 3y + bz = 2b, where a, b ∈ R, has infinitely many solutions, then 7a + 3b is equal to :

Δ=11223a−1−3b=0⇒2a+b−6=0Δ1=11623a+1−1−32b=0⇒a+b−8=0Solving (1) + (2) a=−2,b=10⇒7a+3b=16

If the system of linear equations : x + y + 2z = 6, 2x + 3y + az = a + 1, –x – 3y + bz = 2b, where a, b ∈ R, has infinitely many solutions, then 7a + 3b is equal to :

Δ=11223a−1−3b=0⇒2a+b−6=0Δ1=11623a+1−1−32b=0⇒a+b−8=0Solving (1) + (2) a=−2,b=10⇒7a+3b=16

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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 :

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Detailed Solution

$\begin{array}{l} Δ = |\begin{array}{ccc} 1 & 1 & 2 \\ 2 & 3 & a \\ - 1 & - 3 & b \end{array}| = 0 \\ \Rightarrow 2 a + b - 6 = 0 \\ Δ_{1} = |\begin{array}{ccc} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ - 1 & - 3 & 2 b \end{array}| = 0 \\ \Rightarrow a + b - 8 = 0 \end{array}$
Solving (1) + (2)
$\begin{aligned} a = - 2, b = 10 \\ \Rightarrow 7 a + 3 b = 16 \end{aligned}$