Okay, here we have this:
Considering the provided system of equation, we are going to solve it using Gauss-Jordan elimination, so we obtain the following:
So first let's reduce the matrix to echelon form, we have:
[tex]\begin{gathered} \begin{bmatrix}{5} & {6} & -{4} \\ {2} & -{4} & {3} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix} \\ \end{gathered}[/tex]
We perform the operation R2=R2-(2/5)R1:
[tex]\begin{bmatrix}{5} & {6} & -{4} \\ {0} & {-\frac{32}{5}} & {\frac{23}{5}} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix}[/tex]
Now we apply the operation: R2=-(5/32)R2
[tex]\begin{bmatrix}{5} & {6} & {-4} \\ {0} & {1} & {-\frac{23}{32}} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix}[/tex]
Now we apply the operation: R1=R1-6R2
[tex]\begin{bmatrix}{5} & {0} & {\frac{5}{6}} \\ {0} & {1} & {-\frac{23}{32}} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix}[/tex]
And finally we apply the operation: R1=1/5R1
[tex]\begin{bmatrix}{1} & {0} & {\frac{1}{16}} \\ {0} & {1} & {-\frac{23}{32}} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix}[/tex]
Finally we obtain that there is only one solution, the solution is: (1/16, -23/32).
