$x_1+2x_2+x_3=2 \ x_1+x_2+x_3= \lambda \ 3x_1+x_2+3x_3 = \lambda^3$ These given set of equations can be written in Matrix form as AX=B $\begin{bmatrix} 1 &2 &1 \1 &1 &1 \3 &1 &3 \end{bmatrix} \begin{bmatrix} x_1\x_2 \x_3 \end{bmatrix} = \begin{bmatrix} 3\\lambda \\lambda^2 \end{bmatrix}$ The augmented Matrix A:B is $\begin{bmatrix} 1 &2 &1 &3 \1 &1 &1 &\lambda \3 &1 &3 &\lambda^2 \end{bmatrix} $ $R_2-R_1, \ R_3-3R_1$ $\begin{bmatrix} 1 &2 &1 &3 \0 &-1 &0 &\lambda-3 \0 &-5 &0 &\lambda^2 -9 \end{bmatrix} $ $R_3-5R_2$ $\begin{bmatrix} 1 &2 &1 &3 \0 &-1 &0 &\lambda-3 \0 &0 &0 &\lambda^2-5\lambda+6 \end{bmatrix}$ Rank of A=r=2, Number of variables = n=3 ∴, for the system to have a solution, the rank of the augmented matrix A:B must be equal to rank of A and \lt3. For rank of A:B to be =2, $\lambda^2-5\lambda+6=0 \ (\lambda -2 )(\lambda-3)=0 \ \lambda-2=0, \ \lambda-3=0 \ \lambda =2,3$ Hence for λ=2 or λ=3, the given system has a solution, $for \ \lambda =3,$ $\begin{bmatrix} 1 &2 &1 &3 \0 &-1 &0 &0 \0 &0 &0 &0 \end{bmatrix} $ $x_2=0, \ x_1+2x_2+x_3=0 \x_1+x3=3 \ Let \ x_1 = t \ x_3 = 3-t$ Hence, the system has infinite solution of $(x_1, x_2, x_3)= (t,0,3-t)$ $for \ t \ \in \ R \ and \ \lambda =3.$ $for \lambda =2,$ $\begin{bmatrix} 1 &2 &1 &3 \0 &-1 &0 &-1 \0 &0 &0 &0 \end{bmatrix}$ $-x_2=-1 \ x_2=1 \ x_1+2x_2+x_3=3 \ Let \ x_1=t \ t+2+x_3=3 \ x_3=1-t $ Hence, the system has infinite solution of $(x_1, x_2,x _3) = (t, 1 , 1-t)$ $for \ t \ \in \ R \ and \ \lambda =2$