Question

How do you write the matrix `[(1, -3, 0, -7), (-3, 10, 1, 23), (4, -10, 2, -24)]` using the row echelon form?

Answer 1

The matrix in reduced row echelon form is `((1,0,3,-1),(0,1,1,2),(0,0,0,0))`

Explanation:

The matrix is

`((1,-3,0,-7),(-3,10,1,23),(4,-10,2,-24))`

Replace `L_2` by `(L_2+3L_1)`, we get

`((1,-3,0,-7),(0,1,1,2),(4,-10,2,-24))`

Replace `L_3` by `(L_3-4L_1)`, we get

`((1,-3,0,-7),(0,1,1,2),(0,2,2,4))`

Replace `L_3` by `(L_3-2L_2)`, we get

`((1,-3,0,-7),(0,1,1,2),(0,0,0,0))`

Replace `L_1` by `(L_1+3L_2)`, we get

`((1,0,3,-1),(0,1,1,2),(0,0,0,0))`

This is the matrix in reduced row echelon form.

Answer 2

`[(1,-3, 0, |,-7), (0,1,1,|,2), (0,0,0,|,0)]`

Explanation:

Row echelon form has the form

`[(1,a_(1,2), a_(1,3), |,b_1), (0,1,a_(2,3),|,b_2), (0,0,1,|,b_3)]`

The available operations include

• Swap rows (`R`)
• Multiply/divide a row by a constant
• Add/subtract rows

Multiply Row 1 by `3` and add to Row 2. In short hand:

`3xxR_1+R_2 -> R_2` gives:
`[(1,-3, 0, |,-7), (0,1,1,|,2), (4,-10,2,|,-24)]`

`-4xxR_1+R_3 -> R_3` gives:
`[(1,-3, 0, |,-7), (0,1,1,|,2), (0,2,2,|,4)]`

Notice in this last step that that `R_3=2xxR_2`

`1/2R_3 -> R_3` gives:
`[(1,-3, 0, |,-7), (0,1,1,|,2), (0,1,1,|,2)]`

`-1xxR_2+R_3 -> R_3` gives:
`[(1,-3, 0, |,-7), (0,1,1,|,2), (0,0,0,|,0)]`

Because no further manipulation will produce the final form,

`[(1,a_(1,2), a_(1,3), |,b_1), (0,1,a_(2,3),|,b_2), (0,0,1,|,b_3)]`

We are finished.