If the system of equations x+2y+3z=3 4x+3y-4z=4 [8x+4y- z=9+] has infinitely many solution…

Question

If the system of equations

$x+2y+3z=3$

$4x+3y-4z=4$

$$8x+4y-\lambda z=9+\mu$$

has infinitely many solutions, then the ordered pair ($$\lambda,\mu$$) is equal to :

Accepted Answer

For infinite many solution, $\Delta=0$ and $\Delta_x=0$

$$
\begin{aligned}
& \Delta=\left|\[\begin{array}{ccc} 1 & 2 & 3 \ 4 & 3 & -4 \ 8 & 4 & -\lambda \end{array}\end{aligned}$$\right|=0 \\
\& \Rightarrow  1(-3 \lambda+16)-2(-4 \lambda+32)+3(16-24)=0 \\
\& \Rightarrow  16-3 \lambda+8 \lambda-64-24=0 \Rightarrow 5 \lambda=72 \\
\& \therefore  \lambda=\frac{72}{5} \\
\& \Delta\_x=\left|$$\begin{array}{ccc} 3 & 2 & 3 \ 4 & 3 & -4 \ 9+\mu & 4 & -\lambda \end{array}$$\right|=0

\]

$$
\begin{aligned}
& \Rightarrow \quad 3(-3 \lambda+16)-2(-4 \lambda+36+4 \mu)+3(16-27-3 \mu)=0 \\
& \Rightarrow-9 \lambda+48+8 \lambda-72-8 \mu-33-9 \mu=0 \\
& \Rightarrow-\lambda-17 \mu=57 \\
& \Rightarrow-17 \mu=57+\lambda \\
& \therefore -\mu=\frac{57+\frac{72}{5}}{17}\\
& \Rightarrow \mu=\frac{-357}{85}=\frac{-21}{5}
\end{aligned}
$$

$$
\text { Thus, }(\lambda, \mu) \equiv\left(\frac{72}{5}, \frac{-21}{5}\right)
$$

Solutions · 1