The system of equations $2 x+6 y=-11$, $6 x+20 y-6 z=-3$ and $6 y-18 z=-1$ are
The system of equations $2 x+6 y=-11$, $6 x+20 y-6 z=-3$ and $6 y-18 z=-1$ are
- inconsistent
- consistent with unique solution
- consistent with countable infinite mary solutions
- consistent with infinitely many solutions
Solution
$
\begin{gathered}
\text { Given, } 2 x+6 y=-11 \\
6 x+20 y-6 z=-3 \\
6 y-18 z=-1
\end{gathered}
$
Using Cramer's rule,
$
\begin{aligned}
& D=\left|\begin{array}{ccc}
2 & 6 & 0 \\
6 & 20 & -6 \\
0 & 6 & -18
\end{array}\right| \\
& =2(20 \times(-18)-6 \times(-6)-6(6 \times(-18)-(-6) \times 0) \\
& =2(-360+36)-6(-108) \\
& =2(-360+36)-6(-108) \\
& =2(-324)+648 \\
& =-648+648=0 \\
& D_1=\left|\begin{array}{ccc}
-11 & 6 & 0 \\
-3 & 20 & -6 \\
-1 & 6 & -18
\end{array}\right|=-11(-324)-6(54-6) \\
& =-3564-6(48) \\
& =-3564-288=-3852 \neq 0 \\
& D_2=\left|\begin{array}{ccc}
2 & -11 & 0 \\
6 & -3 & -6 \\
0 & -1 & -18
\end{array}\right|=2(54-6)+11(-108) \\
& =2(48)+(-1188)=96-1188=-1092 \neq 0 \\
& D_3=\left|\begin{array}{ccc}
2 & 6 & -11 \\
6 & 20 & -3 \\
0 & 6 & -1
\end{array}\right|=2(-20+18)-6(-6)-11(36) \\
& =-4+36-396=-364 \neq 0 \\
& \because D=0, D_1 \neq D_2 \neq D_3 \neq 0 \\
&
\end{aligned}
$
$\Rightarrow$ Inconsistent solution
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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