If the system of equations : gathered x+y+z=5 x+2y+3z=9 x+3y+ z= gathered has infinitely m…

Question

If the system of equations :

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

has infinitely many solutions, then the value of $\lambda +\mu $ is :

Accepted Answer

$\begin{gathered} D=\lambda -5 \ {D}_{1}=\lambda +\mu -18 \ {D}_{2}=4\lambda -2\mu +6 \ {D}_{3}=\mu -13 \ x=\frac{{D}_{1}}{D};y=\frac{{D}_{2}}{D};z=\frac{{D}_{3}}{D} \end{gathered}$

For the system to have infinitely many solutions, the determinant of the coefficient matrix and all the corresponding determinants must be zero.

$\begin{gathered} D={D}_{1}={D}_{2}={D}_{3}=0 \end{gathered}$

First, set $D=0$:

$\begin{gathered} \lambda -5=0\Rightarrow \lambda =5 \end{gathered}$

Next, set ${D}_{3}=0$:

$\begin{gathered} \mu -13=0\Rightarrow \mu =13 \end{gathered}$

Now check ${D}_{1}=0$ using $\lambda =5$ and $\mu =13$:

$\begin{gathered} \lambda +\mu -18=5+13-18=0 \end{gathered}$

Also check ${D}_{2}=0$:

$\begin{gathered} 4\lambda -2\mu +6=4(5)-2(13)+6=20-26+6=0 \end{gathered}$

So the values are consistent. Therefore,

$\begin{gathered} 	herefore \lambda +\mu =5+13=18 \end{gathered}$

Solutions · 1