Question

question

http://farm8.staticflickr.com/7037/7006635430_f3729567e3.jpg


Here is the answer:

http://farm6.staticflickr.com/5453/7006635542_fa0551d73d.jpg

Answer 1

My question is why doesn't the following method work:

Switch Z over into the first column, move row 1 to row 3, then exchange rows 1 and 2 so that the new matrix is


1 -5 4 7
-3 -1 2 5
0 -3 2 3

Multiply the first row by 3 and add that to equation 2 to get:

1 -5 4 7
0 5 8 35
0 -3 2 3

Multiply row 2 by 3/5 and add that to equation 3 to get

1 -5 4 7
0 5 8 35
0 0 (34/5) 24

By this time I realized I was on the wrong track but I don't why what I did was illegal.

Answer 2

switching columns is fine as long as you keep track of what went where; since addition is commutative, column switching is fine.

Multiply the first row by 3 and add that to equation 2 to get:


1 -5 4 7 ; *3
-3 -1 2 5
0 -3 2 3



3 -15 12 21
-3 -1 2 5
0 -3 2 3



1 -5 4 7
0 -16 14 26
0 -3 2 3

Answer 3

simplify the second row by factoring out a 2

1 -5 4 7
0 -8 7 13
0 -3 2 3

divide second row by -8 and last row by 3


1 -5 4 7
0 1 -7/8 -13/8
0 -1 2/3 1

and work form there