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

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.

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.