For the system S of linear equations x+y+z=3, | vedclass

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(D) The given system of equations is: $x+y+z=3$ $2x+2y-z=3$ $x+y+\lambda z=1$ The coefficient matrix $A$ is given by: $A = \begin{bmatrix} 1 & 1 & 1 \

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$S$ is consistent for all $\lambda \in R$

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(D) The given system of equations is: $x+y+z=3$ $2x+2y-z=3$ $x+y+\lambda z=1$ The coefficient matrix $A$ is given by: $A = \begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & -1 \ 1 & 1 & \lambda \end{bmatrix}$ The determinant $|A|$ is: $|A| = 1(2\lambda + 1) - 1(2\lambda + 1) + 1(2-2) = 0$ Since $|A| = 0$, the system does not have a unique solution for any $\lambda \in R$. Now, let us check for consistency using the augmented matrix $[A|B]$: $[A|B] = \begin{bmatrix} 1 & 1 & 1 & | & 3 \ 2 & 2 & -1 & | & 3 \ 1 & 1 & \lambda & | & 1 \end{bmatrix}$ Applying $R_2 ightarrow R_2 - 2R_1$ and $R_3 ightarrow R_3 - R_1$: $\begin{bmatrix} 1 & 1 & 1 & | & 3 \ 0 & 0 & -3 & | & -3 \ 0 & 0 & \lambda-1 & | & -2 \end{bmatrix}$ From $R_2$, we get $-3z = -3 \implies z = 1$. Substituting $z=1$ into $R_3$: $(\lambda-1)(1) = -2 \implies \lambda = -1$. If $\lambda = -1$, the system is consistent (infinitely many solutions). If $\lambda 
eq -1$, the system is inconsistent (no solution). Thus, the statement '$S$ is consistent for all $\lambda \in R$' is incorrect.

For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is:

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