Solve the system of linear equations using the Gauss-Jordan elimination method.2x + 3y − 2z = 122x − 3y + 2z = −44x − y + 3z = −4

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We are asked to solve the following system using the Gauss-Jordan elimination method:


We put the coefficients into an augmented matrix:


The idea is to get the leading digit in each row to 1 with only zeros below it. Let R1 be row 1, R2 row 2 and R3 be row 3.

We are allowed to multiply any row by a scalar, swap any rows, and replace any row with the sum of multiples of any two rows.

Take 1/2 * R1, then 2R1-R2 to R2, then 4R1-R3 to R3 to get:

`([1,3/2,-1,6],[0,6,-4,16],[0,7,-7,28]) ==> ([1,3/2,-1,6],[0,3,-2,8],[0,1,-1,4])` (From 1/2 R@ and 1/7R3)

Now take 1/3R2; then R2-R3 to R3:

`([1,3/2,-1,6],[0,1,-2/3,8/3],[0,0,1/3,-4/3]) ==> ([1,3/2,-1,6],[0,1,-2/3,8/3],[0,0,1,-4])` (R3 * 3)

Now 2/3R3+R2 to R2; R3+R1 to R1:

`([1,3/2,0,2],[0,1,0,0],[0,0,1,-4]) ==> ([1,0,0,2],[0,1,0,0],[0,0,1,-4])` (-3/2R2+R1 to R1)

Thus the solution is x=2, y=0, and z=-4.

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