Question
The system of equations $x-k y-z=0, k x-y-z=0$ and $x+y-z=0$ has non trivial solutions. Then the possible values of $k$ are(A) $\pm 1$(B) $\pm 2$(C) 0(D) $\pm 4$
Step 1
So, we write the system of equations in matrix form and find the determinant: \[ \begin{bmatrix} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \\ \end{bmatrix} \] Show more…
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