Let the system of equations x+2y+3z=5,2x+3y+z=9,4x+3y+ z= have infinite number of solution…

Question

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

Accepted Answer

$\begin{gathered} x+2y+3z=5 \ 2x+3y+z=9 \ 4x+3y+\lambda z=\mu \end{gathered}$

For infinite following $\Delta={\Delta}_{1}={\Delta}_{2}={\Delta}_{3}=0$

$\begin{gathered} \Delta=|\begin{array}{ccc}1 \quad  2 \quad  3 \ 2 \quad  3 \quad  1 \ 4 \quad  3 \quad  \lambda \end{array}|=0\Rightarrow \lambda =-13 \ {\Delta}_{1}=|\begin{array}{ccc}5 \quad  2 \quad  3 \ 9 \quad  3 \quad  1 \ \mu  \quad  3 \quad  -13\end{array}|=0\Rightarrow \mu =15 \ {\Delta}_{2}=|\begin{array}{ccc}1 \quad  5 \quad  3 \ 2 \quad  9 \quad  1 \ 4 \quad  15 \quad  -13\end{array}|=0 \end{gathered}$

${\Delta}_{3}=\begin{vmatrix} 1 \quad  2 \quad  5 \ 2 \quad  3 \quad  9 \ 4 \quad  3 \quad  15 \end{vmatrix}=0$

For $\lambda =-13,\mu =15$ system of equation has infinite solution hence $\lambda +2\mu =17$

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