If the system of linear equations 2x + 3y - z | vedclass

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(B) For the system of linear equations to have no solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.$D = \left|\begin{array

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$\lambda = -7$

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(B) For the system of linear equations to have no solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.$D = \left|\begin{array}{ccc} 2 & 3 & -1 \ 1 & 1 & 1 \ 1 & -1 & |\lambda| \end{array}ight| = 0$Expanding along the first row:$2(1 \cdot |\lambda| - 1(-1)) - 3(1 \cdot |\lambda| - 1(1)) - 1(1(-1) - 1(1)) = 0$$2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-2) = 0$$2|\lambda| + 2 - 3|\lambda| + 3 + 2 = 0$$-|\lambda| + 7 = 0 \Rightarrow |\lambda| = 7 \Rightarrow \lambda = \pm 7$.Now,we check the consistency for $\lambda = 7$ and $\lambda = -7$.For $\lambda = 7$,the third equation is $x - y + 7z = 24$. Adding $(2)$ and $(3)$ gives $3x + 4y = 2$. Solving the system shows it is consistent.For $\lambda = -7$,the third equation is $x - y + 7z = -32$. Substituting $\lambda = -7$ into the system leads to a contradiction,meaning the system is inconsistent (no solution).Thus,$\lambda = -7$ is the correct condition.

If the system of linear equations $2x + 3y - z = -2$; $x + y + z = 4$; $x - y + |\lambda|z = 4\lambda - 4$ (where $\lambda \in R$) has no solution,then:

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