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Regarding the first general solution (λ1), there is no problem, but what were the steps to obtain the second general solution (λ2) in the following problem:

$$\begin{eqnarray} x_1 - 2x_2 + x_3 - x_4 + x_5 = 0 \\ x_3 -x_4 + 3x_5 = -2 \\ x_4 - 2x_5 = 1 \\ 0 = a + 1 \end{eqnarray}$$

Only for a = −1 this system can be solved. A particular solution is

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix}$$

The general solution, which captures the set of all possible solutions, is

$$\left\{ x \in \mathbb{R}^5: \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1\\ 0 \end{bmatrix} + \lambda_1 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \\0 \end{bmatrix} + \lambda_2 \begin{bmatrix} 2 \\ 0 \\ -1 \\ 2 \\ 0 \end{bmatrix} , \lambda_1, \lambda_2 \in \mathbb{R} \right\}$$

Reference: 'Mathematics for Machine Learning' Marc Peter Deisenroth, A. Aldo Faisal, Cheng Soon Ong

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There are 5 unknowns in your system of equations. But you have got only 3 useful equations to solve it. Therefore, you have to let 2 of the unknowns have some arbitrary values, such as $\lambda_1$ and $\lambda_2 \in \mathbb{R}$.

According to the solution you have posted, someone has assigned that $x_2=\lambda_1$ and $x_5=\lambda_2$. Now, your system of equations, which is already in echelon form, looks like, $$\begin{eqnarray} x_1 + x_3 - x_4 = 2\lambda_1 - \lambda_2 \\ x_3 -x_4 = -2 - 3\lambda_2 \\ x_4 = 1 + 2\lambda_2 \\ 0 = a + 1 \end{eqnarray}$$

The solution of this system is, $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 2 + 2\lambda_1 + 2\lambda_2 \\ \lambda_1 \\ -1 - \lambda_2 \\ 1 + 2\lambda_2 \\ \lambda_2 \end{bmatrix}.$$

Therefore, the general solution of the original system of equations should be, $$\left\{ x \in \mathbb{R}^5: \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1\\ 0 \end{bmatrix} + \lambda_1 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \\0 \end{bmatrix} + \lambda_2 \begin{bmatrix} 2 \\ 0 \\ -1 \\ 2 \\ \mathbf{\color{red}{1}} \end{bmatrix} , \lambda_1, \lambda_2 \in \mathbb{R} \right\}$$

Did you see the red colored number at the very end of my answer? It does not agree with that of your answer. So, please check.

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  • $\begingroup$ Yes, the red colored number in your answer is correct. That was well noted and the solution is clear. Thanks $\endgroup$ – Alex Javarotti Feb 10 at 21:24

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