# How to solve this system without WolframAlpha

How can I solve this system without using WolframAlpha or any other program? $$\begin{equation} \begin{cases} 2x_1+\lambda x_2+3x_3+4x_4=1\\x_1-x_2+9x_3+7x_4=3\\3x_1+5x_2+\lambda x_3+5x_4=1\\x_1+2x_2+x_4=0 \end{cases} \end{equation}$$

• What methods have you tried? – Zuriel Feb 7 '15 at 16:13
• row reducing does not work? – abel Feb 7 '15 at 16:13
• @abel I tried reducing but for me didn't work – Ali Mostafa Feb 7 '15 at 16:15
• Try to do it with Maple – Leox Feb 7 '15 at 16:20
• Why are people suggesting programs to solve it when the question specifically states that it should be solved without programs? The system is a system of linear equations, so if you know linear algebra it should not be too difficult. Set it up as matrix equation $\mathbf{A}\mathbf{x} = \mathbf{b}$. – Eff Feb 7 '15 at 16:23

## 2 Answers

Use the last equation to replace $x_1$ with $-2x_2-x_4$.
Use the second equation to replace $x_2$ with $3x_3+2x_4-1$.
Now you have two equations in two unknowns.
Write it as a $2\times2$ matrix equation, and solve it to get $x_3$ and $x_4$ as functions of $\lambda$.
Then substitute back to find $x_2$ then $x_1$.

i will use $k$ for $\lambda.$ look at the augmented matrix

\begin{align}\pmatrix{2&k&3&4&1&|a\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\1&2&0&1&1&|d} &\to \pmatrix{1&2&0&1&1&|d\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\2&k&3&4&1&|a} \\ &\to \pmatrix{1&2&0&1&1&|d\\0&-3&9&6&-2&|b-d\\0&-1&k&3&-2&|c-3d\\0&k-4&3&2&-1&|a-2d} \\ &\to \pmatrix{1&2&0&1&1&|d\\0&1&-k&-3&2&|-c+3d\\0&0&9-3k&0&4&|b-d+3(-c+3d)\\ 0&0&3-k(4-k)&2-3(4-k)&-1+2(4-k)&|a-2d+(4-k)(-c+3d)} \end{align}

now you have to look at different cases. for example if $k = 3.$