For the system of linear equations 2x-y+3z=5 3x+2y-z=7 4x+5y+ z= , which of the following …

Question

For the system of linear equations

$2x-y+3z=5$

$3x+2y-z=7$

$4x+5y+\alpha z=\beta $,

which of the following is NOT correct?

Accepted Answer

Given system of linear equation is

$$
\begin{gathered}
2 x-y+3 z=5 \\
3 x+2 y-z=7 \\
4 x+5 y+\alpha z=\beta \\
\text { Now, } \Delta=\left|\[\begin{array}{ccc} 2 & -1 & 3 \ 3 & 2 & -1 \ 4 & 5 & \alpha \end{array}\end{gathered}$$\right|=7(\alpha+5)

\]

So, this system of equation has unique solution, if $\alpha \neq-5$

and $\Delta_1=\left|\begin{array}{ccc} 5 & -1 & 3 \ 7 & 2 & -1 \ \beta & 5 & \alpha \end{array}\right|=17 \alpha-5 \beta+30$

and $\Delta_2=\left|\begin{array}{ccc} 2 & 5 & 3 \ 3 & 7 & -1 \ 4 & \beta & \alpha \end{array}\right|=-11 \beta+\alpha+104$

and $\Delta_3=\left|\begin{array}{ccc} 2 & -1 & 5 \ 3 & 2 & 7 \ 4 & 5 & \beta \end{array}\right|=7(\beta-9)$

For infinitely many solutions,

$$
\Delta=\Delta_1=\Delta_2=\Delta_3=0
$$

$$
\begin{aligned}
& \text { So, }  \Delta=0 \\
& \Rightarrow 7(\alpha+5)=0 \\
& \Rightarrow \alpha=-5
\end{aligned}
$$

$$
\begin{aligned}
& \Delta_3 =0 \\
& \Rightarrow 7(\beta-9) =0 \\
& \Rightarrow  \beta  =9
\end{aligned}
$$

If $\alpha=-5$ and $\beta=8$, then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$.

Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.

Solutions · 1