Answer:
The solution for this system is [tex]x = 5, y = 3[/tex].
Step-by-step explanation:
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]5x + 3y = 16[/tex]
[tex]-2x + y = -13[/tex]
This system has the following augmented matrix.
[tex]\left[\begin{array}{ccc}5&3&16\\-2&1&-13\end{array}\right][/tex]
The first step is dividing the first line by 5. So:
[tex]L_{1} = \frac{L_{1}}{5}[/tex]
We now have
[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\-2&1&-13\end{array}\right][/tex]
Now i want to reduce the first row, so I do:
[tex]L_{2} = L_{2} + 2L_{1}[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&\frac{11}{5}&-\frac{33}{5}\end{array}\right][\tex].
Now, the first step to reduce the second row is:
[tex]L_{2} = \frac{5L_{2}}{11}[/tex]
So we have:
[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&1&-3\end{array}\right][/tex].
Now, to reduce the second row, we do:
[tex]L_{1} = L_{1} - \frac{3L_{2}}{5}[/tex]
And the augmented matrix is:
[tex]\left[\begin{array}{ccc}1&0&5\\0&1&-3\end{array}\right][/tex]
The solution for this system is [tex]x = 5, y = 3[/tex].
Answer:
This is it:
Step-by-step explanation:
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