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Solve the following system of equations in field $\Bbb R$ depending on the real parameter $\lambda$:

$$\lambda x_1+2x_2+x_3+x_4=1$$ $$2x_1-x_2-x_4=1$$ $$4x_1+3x_2+2x_3+\lambda x_4=3$$ $$5x_2+2x_3+3x_4=1$$ My attempt: $$\begin{pmatrix} \lambda &2 &1 &1 | &1\\ 2 &-1 &0 &-1 | &1\\ 4 &3 &2 &\lambda | &3\\ 0 &5 &2 &3 | &1\\ \end{pmatrix}\sim \begin{pmatrix} 2 &-1 &0 &-1 |&1\\ \lambda &2 &1 &1 |&1\\ 4 &3 &2 &\lambda |&3\\ 0 &5 &2 &3| &1\\ \end{pmatrix}\sim \begin{pmatrix} 2 &-1 &0 &-1 | &1\\ 0 &\frac{\lambda+4}{2} &1 &\frac{\lambda+2}{2}| &\frac{2-\lambda}{2}\\ 0 &5 &2 &2+\lambda | &1\\ 0 &5 &2 &3|&1\\ \end{pmatrix}\sim\begin{pmatrix} 2 &-1 &0 &-1| &1\\ 0 &\frac{\lambda+4}{2} &1 &\frac{\lambda+2}{2} |&\frac{2-\lambda}{2}\\ 0 &5 &2 &2+\lambda | &1\\ 0 &0 &0 &1-\lambda| &0\\ \end{pmatrix}$$

For $\lambda\neq 1$ the system has no solution. For $\lambda=1$ we have:

$$\begin{pmatrix} 2 &-1 &0 &-1 |&1\\ 0 &\frac{5}{2} &1 &\frac{3}{2} | &\frac{1}{2}\\ 0 &5 &2 &3 |&1\\ 0 &0 &0&0| &0\\ \end{pmatrix}\sim\begin{pmatrix} \\2 &0 &\frac{2}{5} &-\frac{2}{5}|&\frac{6}{5}\\ 0 &\frac{5}{2} &1 &\frac{3}{2} |& \frac{1}{2}\\ 0 &0 &0&0| &0\\ 0 & 0 &0 &0| &0 \end{pmatrix}\sim\begin{pmatrix} 5 &0 &1 &-1| &3\\ 0 &5 &2 &3 |&1\\ 0 &0 &0 &0|&0\\ 0 &0 &0 &0|&0 \end{pmatrix}$$

$$ \Rightarrow 5x_1+x_3-x_4=3$$ $$5x_2+2x_3+3x_4=1$$

$$\Rightarrow x_1=\frac{3}{5}-\frac{x_3}{5}+\frac{x_4}{5}$$ $$x_2=\frac{1}{5}-\frac{2x_3}{5}-\frac{3x_4}{5}$$

$$X=\begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}=\begin{pmatrix} \frac{3}{5}\\ \frac{1}{5}\\ 0\\ 0 \end{pmatrix}+t\begin{pmatrix} -\frac{1}{5}\\ -\frac{2}{5}\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} \frac{1}{5}\\ -\frac{3}{5}\\ 0\\ 1 \end{pmatrix}t,s\in\Bbb R$$

Is this correct?

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    $\begingroup$ Why do you think that for $\lambda\ne 1$ has no solution? $\endgroup$ Jan 31, 2017 at 6:55
  • $\begingroup$ @mathlover oh no, that's wrong...thanks! $\endgroup$
    – lmc
    Jan 31, 2017 at 7:15
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    $\begingroup$ Determinant of your matrix is non-zero unless $\lambda=1$. Thus for $\lambda\neq1$, you will have a unique solution. $\endgroup$
    – Jan
    Jan 31, 2017 at 7:18
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    $\begingroup$ Looks like $$(x_1,x_2,x_3,x_4)=(0,-1,3,0)$$ $\endgroup$
    – polfosol
    Jan 31, 2017 at 7:20
  • $\begingroup$ @polfosol Ok, I managed to get the same solution in case where $\lambda\neq 1$. Is what I did above alright for $\lambda=1$? $\endgroup$
    – lmc
    Jan 31, 2017 at 10:46

1 Answer 1

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I haven't solved any similar problems before. But my guess is, it looks like $X$ should be a linear combination of the eigenvectors of the matrix of coefficients (correct me if I am wrong). So with $$A=\begin{pmatrix} \lambda &2 &1 &1\\ 2 &-1 &0 &-1\\ 4 &3 &2 &\lambda\\ 0 &5 &2 &3\\ \end{pmatrix}$$ the eigenvectors (for $\lambda=1$) are: $$\pmatrix{1\\0\\2\\2},\pmatrix{1\\-3\\0\\5},\pmatrix{-1\\-2\\5\\0},\pmatrix{0\\0\\0\\0}$$

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