College Algebra, 2013a

608 Systems of Equations and Matrices
⎡
⎢
⎣
1 0 0.7 0 0 0 1.7
0 1 −3.5 0 0 −1.5 −4
0 0 0 1 0 0.6 1.6
0 0 0 0 1 0 1
Decoding this system from the matrix, we get
⎧
R 1 +0.7R 3 = 1.7
⎪⎨
R 2 − 3.5R 3 − 1.5R 6 = −4
R ⎪⎩
4 +0.6R 6 = 1.6
R 5 = 1
We can solve for R 1 , R 2 , R 4 and R 5 leaving R 3 and R 6 as free variables. Labeling R 3 = s
and R 6 = t, we have R 1 = −0.7s +1.7, R 2 = 3.5s +1.5t − 4, R 4 = −0.6t +1.6 and
R 5 = 1. Since resistance values are always positive, we need to restrict our values of s
and t. We know R 3 = s > 0 and when we combine that with R 1 = −0.7s +1.7 > 0,
we get 0 < s < 16
7 . Similarly, R 6 = t > 0andwithR 4 = −0.6t +1.6 > 0, we find
0 < t < 8 3 . In order visualize the inequality R 2 = 3.5s +1.5t − 4 > 0, we graph the
line 3.5s +1.5t − 4=0onthest-plane and shade accordingly. 8 Imposing the additional
conditions 0