Appendix-A

Cramer’s Rule to Solve Linear Equations:

The solution of simultaneous equations, in circuit theory, may be easily solved using Cramer’s rule. Cramer’s rule requires the use of the concept of determinant. A determinant is a scalar defined on a square array of numbers, or matrix, such as

Two simultaneous equations in unknowns x and y, \(a_1x+b_1y=c_1~~~and~~~a_2x+b_2y=c_2\)

have the solutions

\(x=\frac{\begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}}=\frac{c_1b_2-c_2b_2}{a_1b_2-a_2b_1}\)

and

\(y=\frac{\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}}=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}\)

Three simultaneous equations in unknowns x, y, and z \(a_1x+b_1y+c_1z=d_1~~~,~~~a_2x+b_2y+c_2z=d_2~~~and~~~a_3x+b_3y+c_3z=d_3\)

A third-order, or 3 × 3, determinant such as

\(det(A) = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}\)

\[det(A) = a_1(b_2c_3 − c_2b_3) − b_1(a_2c_3 − c_2a_3) + c_1(a_2b_3 − b_2a_3)\]

\(x=\frac{\begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_2 & b_3 & c_3 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}}}\)

,

\(y=\frac{\begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}}}\)

and

\(z=\frac{\begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}}}\)

Example A.1 Use Cramer’s rule to solve the system

\(i_1 + 2i_2 + i_3 = 6\) \(i_1 + i_2 − 2i_3 = 1\) \(i_1 − i_2 + i_3 = 0\)

Solution:

det(A) = \[\begin{vmatrix} 1 & 2 & 1 \\ 1 & 1 & -2 \\ 1 & -1 & 1 \end{vmatrix}\]

\[det(A) = 1((1)(1) − (-2)(-1)) − 2((1)(1) − (-2)(1)) + 1((1)(-1) − (1)(1))=-9\]

\(i_1=\frac{\begin{vmatrix} 6 & 2 & 1 \\ 1 & 1 & -2 \\ 0 & -1 & 1 \end{vmatrix}}{det(A)}=\frac{-9}{-9}=1\)

,

\(i_2=\frac{\begin{vmatrix} 1 & 6 & 1 \\ 1 & 1 & -2 \\ 1 & 0 & 1 \end{vmatrix}}{det(A)}=\frac{-18}{-9}=2\)

and

\(i_3=\frac{\begin{vmatrix} 1 & 2 & 6 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{vmatrix}}{det(A)}=\frac{9}{-9}=-1\)