Question

2X1+2X3-x3+x5=6,-X1-x2+2X3-3X4+X5=0,X3+X4+X5=0,X3+X5+X5=0 SOLVE IT BY USING GUASS JORDON METHOD

Answer 1

As far as we have 4 equations and 5 variables the system has an infinite number of solutions. But let's find one of them.

Let's create an coefficient's matrix:

$\begin{pmatrix} 2 &0&1&0&1&6 \\ -1 &-1&2&3&1&0 \\ 0&0&1&1&1&0 \\ 0&0&1&0&2&0 \end{pmatrix}$

after that according to an Gauss algorithm, we need to made an 1 in upper left corner, let's make it by adding second row to the first one (than I'll write it as I = I + II, witch means, that the first row is the sum of first and second etc)

We'll get next matrix:

$\begin{pmatrix} 1 &-1&3&3&2&6 \\ -1 &-1&2&3&1&0 \\ 0&0&1&1&1&0 \\\ 0&0&1&0&2&0 \end{pmatrix}$

Then we need to have 0's in first column. II = II + I

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

Then to have 1 in second column and second row let's make next transformations:

II = II + III

$\begin{pmatrix} 1 &-1&3&3&2&6 \\ 0&-2&6&7&4&6 \\ 0&0&1&1&1&0 \\\ 0&0&1&0&2&0 \end{pmatrix}$

II = II / (-2)

$\begin{pmatrix} 1 &-1&3&3&2&6 \\ 0&1&-3&-3.5&-2&-3 \\ 0&0&1&1&1&0 \\\ 0&0&1&0&2&0 \end{pmatrix}$

To make a 0 in a first row second column we'll make I = I + II

$\begin{pmatrix} 1 &0&0&-0.5&0&3 \\ 0&1&-3&-3.5&-2&-3 \\ 0&0&1&1&1&0 \\\ 0&0&1&0&2&0 \end{pmatrix}$

Replace 3-rd row with 4-th:

$\begin{pmatrix} 1 &0&0&-0.5&0&3 \\ 0&1&-3&-3.5&-2&-3 \\ 0&0&1&0&2&0 \\ 0&0&1&1&1&0 \end{pmatrix}$

II = II + 3*III

IV = IV - III

$\begin{pmatrix} 1 &0&0&-0.5&0&3 \\ 0&1&0&-3.5&4&-3 \\ 0&0&1&0&2&0 \\ 0&0&0&1&-1&0 \end{pmatrix}$

I = I + 0.5*IV

II = II + 3.5*IV

$\begin{pmatrix} 1 &0&0&0&-0.5&3 \\ 0&1&0&0&0.5&-3 \\ 0&0&1&0&2&0 \\ 0&0&0&1&-1&0 \end{pmatrix}$

Let x5 = 2, then:

x1 - 0.5*x5 = 3 ==> x1 = 4

x2 + 0.5*x5 = -3 ==> x2 = -4

x3 + 2*x5 = 0 ==> x3 = -4

x4 - x5 = 0 ==> x4 = 2

One of the solutions is {4, -4, -4, 2, 2}