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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%264%5C%5C0%261%260%26-2%5C%5C0%260%261%263%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%263%269%5C%5C0%261%262%264%5C%5C0%265%2611%2623%5Cend%7Barray%7D%5Cright%5D)
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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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-1%261%5C%5C0%261%262%264%5C%5C0%260%261%263%5Cend%7Barray%7D%5Cright%5D)
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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 ...
This tells us the solution is (x, y, z) = (4, -2, 3).
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<em>Comment on notation</em>
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.
