Exam 5: Power Flows

Correct Answer:

Verified

$\left[\begin{array}{ccc}5 & -2 & -3 \\-5 & 7 & -2 \\-3 & -3 & 8\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{c}4 \\-10 \\6\end{array}\right]$
There are $N-1=2$ Gauss elimination steps.
During step 1, subtract $A_{21} / A_{11}=-5 / 5=-1$ times Eq. 1 from Eq. 2, and subtract $A_{31} / A_{11}=-3 / 5$ times Eq. 1 from Eq. 3.
$\left[\begin{array}{ccc}5 & -2 & -3 \\0 & 5 & -5 \\0 & -\frac{21}{5} & \frac{31}{5}\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{c}4 \\-6 \\\frac{42}{5}\end{array}\right]$
which is $A^{(1)} X=y^{(1)}$
During step 2, subtract $A_{32}^{(1)} / A_{22}^{(1)}=\frac{-21 / 5}{5}=-\frac{21}{25}$ times Eq. 2 from Eq. 3.
$\left[\begin{array}{ccc}5 & -2 & -3 \\0 & 5 & -5 \\0 & 0 & 2\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{c}4 \\-6 \\\frac{84}{25}\end{array}\right]$
which is triangularized.
Via back substitution, $x_{3}=1.68 ; \quad x_{2}=0.48 ; \quad x_{1}=2$
By Cramer's rule,
$$\begin{aligned}& \Delta=\left|\begin{array}{ccc}5 & -2 & -3 \\-5 & 7 & -2 \\-3 & -3 & 8\end{array}\right|=50 ; \quad \Delta_{1}=\left|\begin{array}{ccc}4 & -2 & -3 \\-10 & 7 & -2 \\6 & -3 & 8\end{array}\right|=100 \text { (Expanding } \Delta_{1} about the first column \\& \Delta_{2}=\left|\begin{array}{ccc}5 & 4 & -3 \\-5 & -10 & -2 \\-3 & 6 & 8\end{array}\right|=24 ; \quad \Delta_{3}=\left|\begin{array}{ccc}5 & -2 & 4 \\-5 & 7 & -10 \\-3 & -3 & 6\end{array}\right|=84 \\& x_{1}=\frac{\Delta_{1}}{\Delta}=2 ; \quad x_{2}=\frac{\Delta_{2}}{\Delta}=0.48 ; \quad x_{3}=\frac{\Delta_{3}}{\Delta}=1.68\end{aligned}$$
By matrix method
$\begin{aligned}& A X=Y ; \quad A^{-1} A X=A^{-1} Y ; \quad X=A^{-1} Y \text { where } A^{-1}=\frac{1}{50}\left[\begin{array}{ccc}50 & 25 & 25 \\46 & 31 & 25 \\36 & 21 & 25\end{array}\right] \\& x_{1}=2 ; \quad x_{2}=0.48 ; \quad x_{3}=1.68\end{aligned}$