Question

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

Answer 1

Answer:

  (4, -2, 3)

Step-by-step explanation:

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

$\left[\begin{array}{ccc|c}1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]$

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 ...

$\left[\begin{array}{ccc|c}1&2&3&9\\0&1&2&4\\0&5&11&23\end{array}\right]$

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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 | 9} -2{0, 1, 2 | 4} = {1, 0, -1 | 1}

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

  {0, 5, 11 | 23} -5{0, 1, 2 | 4} = {0, 0, 1 | 3}

After these operations, our augmented matrix is ...

$\left[\begin{array}{ccc|c}1&0&-1&1\\0&1&2&4\\0&0&1&3\end{array}\right]$

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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 | 1} + {0, 0, 1 | 3} = {1, 0, 0 | 4}

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

  {0, 1, 2 | 4} -2{0, 0, 1 | 3} = {0, 1, 0 | -2}

So, our final augmented matrix is ...

$\left[\begin{array}{ccc|c}1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]$

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.