Let the system of equations x + 2y + 3z = 5, 2x + 3y + z = 9, 4x + 3y + λz = μ

Question

Let the system of equations \(x+2 y+3 z=5,2 x+3 y+z\) \(=9,4 x+3 y+\lambda z=\mu\) have infinite number of solutions. Then \(\lambda+2 \mu\) is equal to :

(1) 28

(2) 17

(3) 22

(4) 15

Answer 1

Correct option is (2) 17

\(\mathrm x+2\mathrm y+3\mathrm z=5\)

\(2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}=9\)

\(4 \mathrm{x}+3 \mathrm{y}+\lambda \mathrm{z}=\mu\)

for infinite following \(\Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0\)

\(\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13\)

\(\Delta_{1}=\left|\begin{array}{ccc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15\)

\(\Delta_{2}=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\)

\(\Delta_{3}=\left|\begin{array}{ccc}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15\end{array}\right|=0\)

for \(\lambda=-13, \mu=15\) system of equation has infinite solution hence \(\lambda+2 \mu=17\).