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The system of linear equations $\mathrm{kx}+\mathrm{y}+\mathrm{z}=1, \mathrm{x}+\mathrm{ky}+\mathrm{z}=1$ and
$\mathrm{x}+\mathrm{y}+\mathrm{kz}=1$ has a unique solution under which one of the following conditions?
Options:
$\mathrm{x}+\mathrm{y}+\mathrm{kz}=1$ has a unique solution under which one of the following conditions?
Solution:
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Verified Answer
The correct answer is:
$k \neq 1$ and $k \neq-2$
Linear equations $K x+y+z=1$
$x+K y+z=1$
$x+y+K z=1$
$\left[\begin{array}{ccc}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$A X=B$
$X=A^{-1} B$
Linear equantion will have unique solution when $A^{-1}$ exist:
$|A| \neq 0$
$\left[\begin{array}{lll}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right] \neq 0$
$K\left(K^{2}-1\right)-1(K-1)+1(1-K) \neq 0$
$\Rightarrow K^{3}-K-K+1+1-K \neq 0$
$\Rightarrow K^{3}-3 K+2 \neq 0$
$(K-1)\left(K^{2}+K-2\right) \neq 0$
$(K-1)(K-1)(K+2) \neq 0$
$K \neq 1, K \neq 1$ and $K \neq-2$
$\Rightarrow K \neq 1$ and $K \neq-2$
$x+K y+z=1$
$x+y+K z=1$
$\left[\begin{array}{ccc}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$A X=B$
$X=A^{-1} B$
Linear equantion will have unique solution when $A^{-1}$ exist:
$|A| \neq 0$
$\left[\begin{array}{lll}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right] \neq 0$
$K\left(K^{2}-1\right)-1(K-1)+1(1-K) \neq 0$
$\Rightarrow K^{3}-K-K+1+1-K \neq 0$
$\Rightarrow K^{3}-3 K+2 \neq 0$
$(K-1)\left(K^{2}+K-2\right) \neq 0$
$(K-1)(K-1)(K+2) \neq 0$
$K \neq 1, K \neq 1$ and $K \neq-2$
$\Rightarrow K \neq 1$ and $K \neq-2$
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