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If the system of equations $x+k y+3 z=-2,4 x+3 y+k z$ $=14,2 x+y+2 z=3$ can be solved by matrix inversion method then
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Verified Answer
The correct answer is:
$k \neq 0$ and $\frac{9}{2}$
$\begin{aligned}
& \text { (a) } x+k y+3 z=-2 \\
& 4 x+3 y+k z=14 \\
& 2 x+y+2 z=3 \\
& |A|=\left|\begin{array}{lll}
1 & k & 3 \\
4 & 3 & k \\
2 & 1 & 2
\end{array}\right| \\
& =(6-k)+k(2 k-8)+3(4-6)=2 k^2-9 k
\end{aligned}$
To get inverse by matrix inversion method, $A^{-1}$ should exist.
$\Rightarrow A$ must be non-singular.
$\begin{aligned}
& |A| \neq 0 \\
\Rightarrow & 2 k^2-9 k \neq 0 \Rightarrow k(2 k-9) \neq 0 \Rightarrow k \neq 0, \frac{9}{2}
\end{aligned}$
& \text { (a) } x+k y+3 z=-2 \\
& 4 x+3 y+k z=14 \\
& 2 x+y+2 z=3 \\
& |A|=\left|\begin{array}{lll}
1 & k & 3 \\
4 & 3 & k \\
2 & 1 & 2
\end{array}\right| \\
& =(6-k)+k(2 k-8)+3(4-6)=2 k^2-9 k
\end{aligned}$
To get inverse by matrix inversion method, $A^{-1}$ should exist.
$\Rightarrow A$ must be non-singular.
$\begin{aligned}
& |A| \neq 0 \\
\Rightarrow & 2 k^2-9 k \neq 0 \Rightarrow k(2 k-9) \neq 0 \Rightarrow k \neq 0, \frac{9}{2}
\end{aligned}$
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