The system of equations kx + 2y - z = 1,(k - | vedclass

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Question

(B) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coeffic

Vedclass · Accepted Answer

$k = -1$

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(B) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coefficient matrix is:$D = \begin{vmatrix} k & 2 & -1 \ 0 & k-1 & -2 \ 0 & 0 & k+2 \end{vmatrix}$Since this is an upper triangular matrix,the determinant is the product of the diagonal elements:$D = k(k-1)(k+2)$For a unique solution,$D 
eq 0$:$k(k-1)(k+2) 
eq 0$This implies $k 
eq 0, k 
eq 1, k 
eq -2$.Comparing this with the given options,if $k = -1$,the condition $D 
eq 0$ is satisfied because $(-1)(-1-1)(-1+2) = (-1)(-2)(1) = 2 
eq 0$.Therefore,the system has a unique solution when $k = -1$.

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

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