Answer to Question #350832 in Linear Algebra for Busi

Question

The following were obtained by applying Kirchoff’s laws to an electric circuit

2IA +IB −IC = −8

−IA +IB +IC = 3

−2IA +4IC = 18

Answer 1

Augmented matrix

$$\begin{pmatrix} 2 & 1 & -1 & & -8 \\ -1 & 1 & 1 & & 3 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}$$

$R_1=R_1/2$

$$\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ -1 & 1 & 1 & & 3 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}$$

$R_2=R_2+R_1$

$$\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 3/2 & 1/2 & & -1 \\ -2 & 0 & 4 & & 18 \\ \end{pmatrix}$$

$R_3=R_3+2R_1$

$$\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 3/2 & 1/2 & & -1 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}$$

$R_2=2R_2/3$

$$\begin{pmatrix} 1 & 1/2 & -1/2 & & -4 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}$$

$R_1=R_1-R_2/2$

$$\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 1 & 3 & & 10 \\ \end{pmatrix}$$

$R_3=R_3-R_2$

$$\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 8/3 & & 32/3 \\ \end{pmatrix}$$

$R_3=3R_3/8$

$$\begin{pmatrix} 1 & 0 & -2/3 & & -11/3 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}$$

$R_1=R_1+2R_3/3$

$$\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 1/3 & & -2/3 \\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}$$

$R_2=R_2-R_3/3$

$$\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 0 & & -2\\ 0 & 0 & 1 & & 4 \\ \end{pmatrix}$$

$$I_A=-1, I_B=-2, I_C=4$$