5 linear equations in 5 unknowns

Question

I need an example of 5 linearly independent equations with 5 variables. How can I write such a equation set. As an example:

                      0   0    0    0   1   |   -4
                      0   0    1   -1   0   |    3
                      1   0    0    0   1   |    2
                      0   7   -8    0   0   |  -14
                      2   0    0    0   0   |    2

But this is wrong. I need to build a correct one.

Answer 1

You can choose any $\hat{x}_{1} , \ldots , \hat{x}_{5}$, and use the system $$ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix}. $$ If you want something that is less obvious, you can start multiplying the rows by nonzero constants: $$ \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} 2 \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix} $$ (in this example, we multiply the first row by $2$), or adding multiples of one row to another row: $$ \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} 2 \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} - 3 \hat{x}_{5} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix} $$ (in this example, we add $-3$ times the fifth row to the third row).

Answer 2

Write in matrix form this system: $x_1=1,...,x_5=5$

Answer 3

Begin with a 5x5 random matrix, with positive diagonal. For example, $$ A=\begin{pmatrix} 1&0&-3&2&4\\ 5&6&7&8&-9\\ 1&1&1&1&1\\ 0&0&0&1&0\\ 2&-3&2&-3&4 \end{pmatrix} $$ Then compute the absolute row sums (i.e. sum up the absolute values of the entries in each row): \begin{align} 1+0+3+2+4=10\\ 5+6+7+8+9=35\\ 1+1+1+1+1=5\\ 0+0+0+1+0=1\\ 2+3+2+3+4=14. \end{align} Add these row sums to the diagonal of $A$: $$ A=\begin{pmatrix} 1+10&0&-3&2&4\\ 5&6+35&7&8&-9\\ 1&1&1+5&1&1\\ 0&0&0&1+1&0\\ 2&-3&2&-3&4+14 \end{pmatrix}. $$ The rows of $A$ are then linearly independent. If you want to make $A$ looks more random, further scramble the rows or columns of $A$. Then form a random but nice looking solution vector $x$ (so that you may control the look of the solution). Compute $b=Ax$ and let the coefficients of $b$ be the numbers on the RHS of the equations. The resulting system will always have a unique solution, which is your nice $x$.