If the system of linear equations,x+y+z = 6,x | vedclass

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(C) For a system of linear equations to have more than two solutions,it must have infinitely many solutions. This occurs when the determinant of the c

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$13$

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(C) For a system of linear equations to have more than two solutions,it must have infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero and the augmented matrix satisfies the consistency condition.The coefficient matrix is $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 3 & 2 & \lambda \end{bmatrix}$.Setting the determinant $|A| = 0$:$|A| = 1(2\lambda - 6) - 1(\lambda - 9) + 1(2 - 6) = 0$$2\lambda - 6 - \lambda + 9 - 4 = 0$$\lambda - 1 = 0 \Rightarrow \lambda = 1$.Now,for infinitely many solutions,the determinant of the matrix formed by replacing the constant column must also be zero $(D_z = 0)$:$D_z = \begin{vmatrix} 1 & 1 & 6 \ 1 & 2 & 10 \ 3 & 2 & \mu \end{vmatrix} = 0$$1(2\mu - 20) - 1(\mu - 30) + 6(2 - 6) = 0$$2\mu - 20 - \mu + 30 - 24 = 0$$\mu - 14 = 0 \Rightarrow \mu = 14$.Finally,calculating $\mu - \lambda^{2}$:$\mu - \lambda^{2} = 14 - (1)^{2} = 14 - 1 = 13$.

If the system of linear equations,$x+y+z = 6$,$x+2y+3z = 10$,and $3x+2y+\lambda z = \mu$ has more than two solutions,then $\mu-\lambda^{2}$ is equal to

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