http://www.ck12.org Chapter 8. Systems and Matrices
Solution:With four equations and three unknowns there must be at least one dependent equation. The simplest
method of seeing linearly dependence is to notice that one equation is just a multiple of the other. In this case the
fourth equation is just six times the second equation and so it is dependent.
Most people will not notice that the third equation is also dependent. It is common to start doing a problem and
notice somewhere along the way that all the variables in a row disappear. This means that the original equations
were dependent. In this case, the third equation is the first equation plus two times the second equation. This means
they are dependent.
Example C
Reduce the following system to a system of two equations and two unknowns.
3 x+ 2 y+z= 7
4 x+ 0 y+z= 6
6 x−y+ 0 z= 5Solution: Strategically swapping rows so that the zero coefficients do not live on the diagonal is a clever starting
move.
Step 1: Swap rows 2 and 3.
3 x+ 2 y+z= 7
6 x−y+ 0 z= 5
4 x+ 0 y+z= 6Step 2: Scale row 3 by a factor of 3. Subtract 2 times row 1 from for 2.
3 x+ 2 y+z= 7
0 x− 5 y− 2 z=− 9
12 x+ 0 y+ 3 z= 18Step 3: Subtract 4 times row 1 from row 3.
3 x+ 2 y+z= 7
0 x− 5 y− 2 z=− 9
0 x− 8 y−z=− 10Step 4: Scale the second row by 8 and the third row by 5.
3 x+ 2 y+z= 7
0 x− 40 y− 16 z=− 72
0 x− 40 y− 5 z=− 50