Question 457 of 472HardMCQ
If the system of equations \[x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1\] has infinitely many solutions, then \[(2 \mu+3 \lambda)\] is equal to :
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Official solution
\[\begin{aligned}
& x+4 y-z=\lambda \\
& 7 x+9 y+\mu z=-3 \\
& 5 x+y+2 z=-1 \\
& {\left[\[\begin{array}{|ccccccc|}1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2\end{array}\end{aligned}\]\right]\left[\[\begin{gathered} x \\ y \\ z \end{gathered}\]\right]=\left[\[\begin{gathered} \lambda \\ -3 \\ -1 \end{gathered}\]\right]} \\
\& A=\left[\[175491−1μ2\]\right], B=\left[\[\begin{gathered} \lambda \\ -3 \\ -1 \end{gathered}\]\right] \\
\& A X=B \\
\& X=A\^{-1} B \\
\& \quad=\frac{\operatorname{adj} A}{|A|} B
\]
If ∣A∣=0 and \[(\operatorname{adj} A) \cdot B=0\], system has infinitely many solutions.
\[\begin{aligned}
& |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\
& \Rightarrow 18-\mu-56+20 \mu+38=0 \\
& \Rightarrow 19 \mu=0 \\
& \Rightarrow \mu=0 \\
& \text { Also adjA }=\left[\[\begin{array}{|ccccccc|}18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19\end{array}\end{aligned}\]\right] \\
\& (\text { adj } A) \cdot B=0 \\
\& {\left[\[18−14−38−97199−7−19\]\right]\left[\[\begin{gathered} \lambda \\ -3 \\ -1 \end{gathered}\]\right]=\left[\[\begin{gathered} 0 \\ 0 \\ 0 \end{gathered}\]\right]} \\
\& 18 \lambda+27-9=0 \\
\& \Rightarrow 18 \lambda=-18 \\
\& \Rightarrow \lambda=-1 \\
\& \Rightarrow 2 \mu+3 \lambda=3(-1)=-3
\] Answer the question to unlock