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Determine the values of $\lambda$ so that the equations $x+y+z=1$, $x+2y+4z= \lambda$ , $x+4y+10z=\lambda^2$ have a solution and solve them completely in each case.
written 4.5 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 3.7 years ago
engineering mathematics
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written 4.5 years ago by gravatar for teamques10 teamques10 ★ 70k

We have

$\begin{bmatrix} 1 &1 &1 \\1 &2 &4 \\1 &4 &10 \end{bmatrix} \begin{bmatrix} x\\y \\z \end{bmatrix} = \begin{bmatrix} 1\\\lambda \\ \lambda^2 \end{bmatrix} $

By $\begin{matrix} R2 - R1\\ R3 - R1 \end{matrix} \begin{bmatrix} 1 &1 &1 \\0 &1 &3 \\0 &3 &9 \end{bmatrix} \begin{bmatrix} x\\y \\z \end{bmatrix} = \begin{bmatrix} …

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