Equations : $ x_1+2x_2+x_3=3 \; \; ; \; \; x_1+x_2+x_3= \lambda \; \; ; \; \; 3x_1+x_2+3x_3= \lambda^2 $

Writing given equations in matrix form AX=B,

$ \therefore \left[ \begin{array}{ccc} 1& 2& 1\\ 1& 1& 1\\ 3& 1& 3 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ \lambda\\ \lambda^2 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2-R_1 \; , \; R_3 \rightarrow R_3-3R_1 \\ \; \\ \; \left[ \begin{array}{ccc} 1& 2& 1\\ 0& -1& 0\\ 0& -5& 0 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ \lambda-3\\ \lambda^2-9 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3-5R_2 \\ \; \\ \; \left[ \begin{array}{ccc} 1& 2& 1\\ 0& -1& 0\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ \lambda-3\\ \lambda^2-9-5\lambda+15 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3-5R_2 \\ \; \\ \; \left[ \begin{array}{ccc} 1& 2& 1\\ 0& -1& 0\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ \lambda-3\\ \lambda^2-5\lambda+6 \end{array}\right] \\ \; \\ \; \\ $

In order that given equations have a solution, it is necessary to have given set of equations consistent.

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Given equations are consistent if rank of matrix A is equal of augmented matrix [A:B]

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Augmented matrix = [A:B] = $ \left[ \begin{array}{ccc|c} 1& 2& 1 & 3 \\ 0& -1& 0 & \lambda-3 \\ 0& 0& 0 & \lambda^2-5\lambda+6 \end{array}\right] \\ \; \\ \; \\ $

Its rank should be 2. For that, $ \lambda^2-5\lambda+6 \;=\; 0 \\ \; \\ \lambda \;=\; 2,3 $

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For $\lambda \;=\; 2 $

$ \\ \left[ \begin{array}{ccc} 1& 2& 1\\ 0& -1& 0\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ -1\\ 0 \end{array}\right] \\ \; \\ \; \\ \therefore x_1 + 2 x_2 + x_3 \;=\; 3 \\ \; \\ -x_2 + 0x_3 \;=\; -1 \; \; \; \therefore x_2=1 \\ \; \\ x_1 + 2(1) + x_3 =3 \; \; \; \therefore x_1+x_3=1 \\ \; \\ $

No. of unknowns=3 $ \; \; \; \;$ Rank=2

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$ \therefore$ Equations have infinite solutions and we have to assign 3-2=1 parameter, say k.

Let, $ x_3 =k \\ \; \\ \therefore x_1 \;=\; 1-k \\ \; \\ \; \\ \; \\ $

For $\lambda \;=\; 3 $

$ \\ \left[ \begin{array}{ccc} 1& 2& 1\\ 0& -1& 0\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{ccc} x_1\\ x_2\\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 3\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ \therefore x_1+2x_2+x_3 \;=\;3 \\ \; \\ \; \\ x_2=0 \\ \; \\ \; \\ x_1 + x_3 \;= \; 0 \\ \; \\ $

Here also, Equations have infinite solutions and we have to assign one parameter k .

Let, $ x_3 \;=\; k \\ \; \\ x_1 \;=\; -k $