Question

Use Gauss- Jordan elimination method to solve the following system:(final answer in an ordered triplet)

2x+4y+6z=18
4x+5y+6z=24
3x+y-2z=4

Answer 1

(4, -2, 3)

Explanation:

You want the final augmented coefficient matrix to look like ...

`\left[\begin{array}c1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]`

The left portion is an identity matrix, and the right column is the solution vector.

To get there, you do a series of row operations. The usual Gauss-Jordan elimination algorithm has you start by arranging the rows so the highest leading coefficient is in the first row. Dividing that row by that coefficient immediately generates a bunch of fractions, so gets messy quickly. Instead, we'll start by dividing the given first row by 2 to make its leading coefficient be 1:

x + 2y +3z = 9

Subtracting 4 times this from the second row makes the new second row be ...

0x -3y -6x = -12

And dividing that row by -3 makes it ...

0x +y +2z = 4

Continuing the process of zeroing out the first column, we can subtract the third row from 3 times the first to get ...

0x +5y +11z = 23

After these operations, our augmented matrix is ...

`\left[\begin{array}c1&2&3&9\\0&1&2&4\\0&5&11&23\end{array}\right]`

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Conveniently, the second row has a 1 on the diagonal, so we can use that directly to zero the second column of the other rows. Subtracting 2 times the second row from the first, the new first is ...

1, 2, 3 -2 4 = 1, 0, -1

Subtracting 5 times the second row from the 3rd, the new 3rd row is ...

23 -50, 1, 2 = 0, 0, 1

After these operations, our augmented matrix is ...

`\left[\begin{array}ccc1&0&-1&1\\0&1&2&4\\0&0&1&3\end{array}\right]`

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Conveniently, the third row has 1 on the diagonal, so we can use that directly to zero the third column of the other rows.

Adding the third row to the first, the new first row is ...

1, 0, -1 + 3 = 4

Subtracting twice the third row from the second gives the new second row ...

0, 1, 2 -20, 0, 1 = 0, 1, 0

So, our final augmented matrix is ...

`\left[\begin{array}ccc1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]`

This tells us the solution is (x, y, z) = (4, -2, 3).

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Comment on notation

It is a bit cumbersome to write the equations represented by each row of the matrix, so we switched to a bracket notation that just lists the coefficients in order. It is more convenient and less space-consuming, and illustrates the steps adequately. For your own work, you need to use a notation recognized by your grader, or explain any notation you may adopt as a short form.