# Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there are infinitely many solutions, express your answer in terms of the parameters t and/or s.)   `x + 2y + z = −2 ` , `−2x − 3y − z = 3` ,`3x + 6y + 3z = −6` .

The only solution for this system of linear equations is `x = 0, y = -31, z = 60.`

## Expert Answers

Hello!

The equations are `x + 2y + z = -2, ` `-2x - 3y - z = 33 , ` `x + 6y + 3z = -6 . ` It may be expressed by the augmented matrix

`( ( 1 , 2 , 1 , -2 ) , ( -2 , -3 , -1 , 33 ) , ( 1 , 6 , 3 , -6 ) )`

that consists of the coefficients and the free terms.

Forward elimination makes the matrix upper triangular. First, multiply the first row by 2 and add to the second and also multiply the first row by -1 and add to the third. The matrix becomes

`( ( 1 , 2 , 1 , -2 ) , ( 0 , 1 , 1 , 29 ) , ( 0 , 4 , 2 , -4 ) ).`

Then multiply the second row by -4 and add to the third row:

`( ( 1 , 2 , 1 , -2 ) , ( 0 , 1 , 1 , 29 ) , ( 0 , 0 , -2 , -120 ) ).`

Now the matrix is upper triangular (has all zeros under the main diagonal). Divide the third row by -2 to get

`( ( 1 , 2 , 1 , -2 ) , ( 0 , 1 , 1 , 29 ) , ( 0 , 0 , 1 , 60 ) ).`

Now we can easily perform the back elimination to make the matrix a diagonal one. Multiply the third row by -1 and add to the first and second rows:

`( ( 1 , 2 , 0 , -62 ) , ( 0 , 1 , 0 , -31 ) , ( 0 , 0 , 1 , 60 ) ).`

Finally, multiply the second row by -2 and add to the first:

`( ( 1 , 0 , 0 , 0 ) , ( 0 , 1 , 0 , -31 ) , ( 0 , 0 , 1 , 60 ) ).`

Now the left part is an identity 3x3 matrix and the last column contains the solution: x = 0, y = -31, z = 60.

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