Consider the circuit shown in Figure 7.1.2. Let I1, I2, and I3 be the currents through the capacitor, resistor, and inductor, respectively. Likewise, let V1, V2, and V3 be the corresponding voltage drops. The arrows denote the arbitrarily chosen directions in which currents and voltage drops will be taken to be positive. a. Applying Kirchhoffs second law to the upper loop in the circuit, show that V1 V2 = 0. (15) In a similar way, show that V2 V3 = 0. (16) b. Applying Kirchhoffs first law to either node in the circuit, show that I1 + I2 + I3 = 0. (17) c. Use the current-voltage relation through each element in the circuit to obtain the equations CV_ 1 = I1, V2 = RI2, L I_ 3 = V3. (18) d. Eliminate V2, V3, I1, and I2 among equations (15) through (18) to obtain CV_ 1 = I3 V1 R , L I_ 3 = V1. (19) Observe that if we omit the subscripts in equations (19), then we have the system (2) of this section. 1

Section 1.5 Reduced Echelon Form 10 r If we get then the solution is 01 s x1= r and x =2s. 0 r 0 s If we get 001 t then the solution is x1= r, x2= s and x =3t. With many systems, particularly rectangular systems, this isn't possible. We need a form to show us when the augmented matrix is "as reduced as possible"....