The system of linear equations lambda x + 2y | vedclass

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Question

(D) The determinant of the coefficient matrix $D$ is given by:$D = \begin{vmatrix} \lambda & 2 & 2 \ 2\lambda & 3 & 5 \ 4 & \lambda & 6 \end{vmatrix

Vedclass · Accepted Answer

no solution when $\lambda = 2$

Vedclass · Answer

(D) The determinant of the coefficient matrix $D$ is given by:$D = \begin{vmatrix} \lambda & 2 & 2 \ 2\lambda & 3 & 5 \ 4 & \lambda & 6 \end{vmatrix}$Expanding along the first row:$D = \lambda(18 - 5\lambda) - 2(12\lambda - 20) + 2(2\lambda^2 - 12)$$D = 18\lambda - 5\lambda^2 - 24\lambda + 40 + 4\lambda^2 - 24$$D = -\lambda^2 - 6\lambda + 16 = -(\lambda^2 + 6\lambda - 16) = -(\lambda + 8)(\lambda - 2) = (\lambda + 8)(2 - \lambda)$For $\lambda = 2$,$D = 0$. We check the consistency using Cramer's rule or by substituting $\lambda = 2$ into the equations:$2x + 2y + 2z = 5$$4x + 3y + 5z = 8$$4x + 2y + 6z = 10$Subtracting the first equation multiplied by $2$ from the second: $(4x + 3y + 5z) - 2(2x + 2y + 2z) = 8 - 10 \implies -y + z = -2 \implies y - z = 2$.Subtracting the first equation multiplied by $2$ from the third: $(4x + 2y + 6z) - 2(2x + 2y + 2z) = 10 - 10 \implies -2y + 2z = 0 \implies y - z = 0$.Since $y - z = 2$ and $y - z = 0$ are contradictory,the system has no solution for $\lambda = 2$.

The system of linear equations $\lambda x + 2y + 2z = 5$,$2\lambda x + 3y + 5z = 8$,and $4x + \lambda y + 6z = 10$ has:

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