Question

How do you solve `((5, 1, 4), (2, -3, -4), (7, 2, -6))X=((5), (2), (5))`?

Answer 1

`X = ((99/107),(-28/107),(17/107))`

Explanation:

Starting with the augmented matrix:

`((5,1,4,|,5),(2,-3,-4,|,2),(7,2,-6,|,5))`

Perform a sequence of row operations until the left hand side `3xx3` matrix is the identity matrix. Then the right hand column will be `X`.

Subtract row 1 and 2 from row 3 to get:

`((5,1,4,|,5),(2,-3,-4,|,2),(0,4,-6,|,-2))`

Divide row 1 by `5` to get:

`((1,1/5,4/5,|,1),(2,-3,-4,|,2),(0,4,-6,|,-2))`

Subtract twice row 1 from row 2 to get:

`((1,1/5,4/5,|,1),(0,-17/5,-28/5,|,0),(0,4,-6,|,-2))`

Multiply row `2` by `-5/17` to get:

`((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,4,-6,|,-2))`

Subtract `4` times row 2 from row 3 to get:

`((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,0,-214/17,|,-2))`

Multiply row `3` by `-17/214` to get:

`((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,0,1,|,17/107))`

Subtract `1/5` row 2 from row 1 to get:

`((1,0,8/17,|,1),(0,1,28/17,|,0),(0,0,1,|,17/107))`

Subtract `8/17` row 3 from row 1 to get:

`((1,0,0,|,99/107),(0,1,28/17,|,0),(0,0,1,|,17/107))`

Subtract `28/17` row 3 from row 2 to get:

`((1,0,0,|,99/107),(0,1,0,|,-28/107),(0,0,1,|,17/107))`

So:

`X = ((99/107),(-28/107),(17/107))`