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 :
Solution
$$
\begin{aligned}
& x+2 y+3 z=5 \\\\
& 2 x+3 y+z=9 \\\\
& 4 x+3 y+\lambda z=\mu
\end{aligned}
$$
<br/><br/>For infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$
<br/><br/>$\begin{aligned} & \Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13 \\\\ & \Delta_1=\left|\begin{array}{llc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15 \\\\ & \Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\end{aligned}$
<br/><br/>$$
\Delta_3=\left|\begin{array}{ccc}
1 & 2 & 5 \\
2 & 3 & 9 \\
4 & 3 & 15
\end{array}\right|=0
$$
<br/><br/>For $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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