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Solve the system of linear equations using the Gauss-Jordan elimination method. 5x + 3y = 16 −2x + y = −13 (x, y) =

User Patrick W
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5.3k points

2 Answers

6 votes

Answer:

This is it:

Explanation:

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User Simlev
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5.5k points
6 votes

Answer:

The solution for this system is
x = 5, y = 3.

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:


5x + 3y = 16


-2x + y = -13

This system has the following augmented matrix.


\left[\begin{array}{ccc}5&3&16\\-2&1&-13\end{array}\right]

The first step is dividing the first line by 5. So:


L_(1) = (L_(1))/(5)

We now have


\left[\begin{array}{ccc}1&(3)/(5)&(16)/(5)\\-2&1&-13\end{array}\right]

Now i want to reduce the first row, so I do:


L_(2) = L_(2) + 2L_(1)

So we have


\left[\begin{array}{ccc}1&amp;(3)/(5)&amp;(16)/(5)\\0&amp;(11)/(5)&amp;-(33)/(5)\end{array}\right][\tex].</p><p>Now, the first step to reduce the second row is:</p><p>[tex]L_(2) = (5L_(2))/(11)

So we have:


\left[\begin{array}{ccc}1&amp;(3)/(5)&amp;(16)/(5)\\0&amp;1&amp;-3\end{array}\right].

Now, to reduce the second row, we do:


L_(1) = L_(1) - (3L_(2))/(5)

And the augmented matrix is:


\left[\begin{array}{ccc}1&amp;0&amp;5\\0&amp;1&amp;-3\end{array}\right]

The solution for this system is
x = 5, y = 3.

User Jayh
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4.6k points