Writing given equations in matrix form AX=B

$ \therefore \left[ \begin{array}{ccc} 1& 1& 1\\ 1& 2& 4\\ 1& 4& 10 \end{array}\right] \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ \lambda\\ \lambda^2 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 - R_1 , R_3 \rightarrow R_3-R_2 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& 1\\ 0& 1& 3\\ 0& 3& 9 \end{array}\right] \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ \lambda-1\\ \lambda^2-1 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3- 3R_2 \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 1& 1\\ 0& 1& 3\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ \lambda-1\\ \lambda^2-1-3\lambda+3 \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ \lambda-1\\ \lambda^2-3\lambda+2 \end{array}\right] \\ \; \\ $

In order that given equations have solutions, 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 to rank of augmented matrix [A : B]

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

Rank of A=2 Since rank of [A:B] should also be 2, it is necessary that $ \lambda^2-3\lambda+2 \;=\; 0$

$ \\ \therefore \lambda^2-2\lambda - \lambda+2 \;=\; 0 \\ \; \\ \therefore \lambda \;= \; 1,2 \\ \; \\ \; \\ $

For $ \lambda =1$

$ \\ \therefore \left[ \begin{array}{ccc} 1& 1& 1\\ 0& 1& 3\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ 0\\ 0 \end{array}\right] \\ \; \\ \; \\ \therefore x+y+z \;=\; 1 \\ y+3z\;=\;0 \\ \; \\ $

No. of unknowns (3) > Rank of A (2)

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

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Let z=k

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$ \therefore y+3k=0 \; \; \; \therefore y\;=\; -3k \\ \; \\ x+yz\;=\;1 \; \; \; x-3k+k=1 \; \; \; \;\therefore x\;=\; 2k+1 \\ \; \\ \; \\ \; \\ \; \\ For \lambda =2 \therefore \left[ \begin{array}{ccc} 1& 1& 1\\ 0& 1& 3\\ 0& 0& 0 \end{array}\right] \left[ \begin{array}{c} x\\ y\\ z \end{array}\right] \;=\; \left[ \begin{array}{c} 1\\ 1\\ 0 \end{array}\right] \\ \; \\ \; \\ \therefore x+y+z \;=\; 1 \\ y+3z\;=\;1 \\ \; \\ $

Here also we have to assign 3-2=1 parameter say k. Let
$ z=k $

$ \therefore y=1-3k \\ \; \\ \therefore x+y+z=1 \\ \; \\ \therefore x+1-3k+k=1 \\ \; \\ \therefore x=2k $