Question

How do you solve `x+2y+3w+4z = 10`, `2x+y+w-z = 1`, `3x+y+4w+3z=22`, and `-2x+6y+4w+20z = 18` using matrices?

Answer 1

See explanation...

Explanation:

Write these equations as a `4 xx 5` matrix, then perform a sequence of row operations until the left hand `4 xx 4` matrix is the identity matrix. Then the right hand column will be the solution.

Start with:

`((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (3, 1, 4, 3, 22), (-2, 6, 4, 20, 18))`

Add row `2` to row `4` to get:

`((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (3, 1, 4, 3, 22), (0, 7, 5, 19, 19))`

Subtract rows `1` and `2` from row `3` to get:

`((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (0, -2, 0, -2, 11), (0, 7, 5, 19, 19))`

Subtract twice row `1` from row `2` to get:

`((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -2, 0, -2, 11), (0, 7, 5, 19, 19))`

Add row `2` plus twice row `3` to row `4` to get:

`((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -2, 0, -2, 11), (0, 0, 0, 6, 22))`

Multiply row `3` by `3` to get:

`((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -6, 0, -6, 33), (0, 0, 0, 6, 22))`

Subtract twice row `2` from row `3` to get:

`((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))`

Multiply row `2` by `-1/3` to get:

`((1, 2, 3, 4, 10), (0, 1, 5/3, 3, 19/3), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))`

Subtract twice row `2` from row `1` to get:

`((1, 0, -1/3, -2, -8/3), (0, 1, 5/3, 3, 19/3), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))`

Divide row `3` by `10` to get:

`((1, 0, -1/3, -2, -8/3), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 6, 22))`

Add `1/3` row `3` to row `1` to get:

`((1, 0, 0, -8/5, -3/10), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 6, 22))`

Divide row `4` by `6` to get:

`((1, 0, 0, -8/5, -3/10), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))`

Add `8/5` row `4` to row `1` to get:

`((1, 0, 0, 0, 167/30), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))`

Subtract `5/3` row `3` from row `2` to get:

`((1, 0, 0, 0, 167/30), (0, 1, 0, 1, -11/2), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))`

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

`((1, 0, 0, 0, 167/30), (0, 1, 0, 0, -55/6), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))`

Subtract `6/5` row `4` from row `3` to get:

`((1, 0, 0, 0, 167/30), (0, 1, 0, 0, -55/6), (0, 0, 1, 0, 27/10), (0, 0, 0, 1, 11/3))`

Provided I have made no arithmetic errors:

`{ (x = 167/30), (y = -55/6), (w = 27/10), (z = 11/3) :}`