Question

Using Gauss elimination, solve the following linear algebraic equations:
$\begin{aligned}
5 x_{1}-2 x_{2}-3 x_{3} & =4 \
-5 x_{1}+7 x_{2}-2 x_{3} & =-10 \
-3 x_{1}-3 x_{2}+8 x_{3} & =6
\end{aligned}$
Find the three unknowns $x_{1}, x_{2}$, and $x_{3}$. Check your answers using Cramer's rule. Also check by using matrix solution of linear equations.

Accepted Answer

$\left[\begin{array}{ccc}
5 & -2 & -3 \
-5 & 7 & -2 \
-3 & -3 & 8
\end{array}\right]\left[\begin{array}{l}
x_{1} \
x_{2} \
x_{3}
\end{array}\right]=\left[\begin{array}{c}
4 \
-10 \
6
\end{array}\right]$
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
$\begin{aligned}
& A X=Y ; \quad A^{-1} A X=A^{-1} Y ; \quad X=A^{-1} Y \text { where } A^{-1}=\frac{1}{50}\left[\begin{array}{ccc}
50 & 25 & 25 \
46 & 31 & 25 \
36 & 21 & 25
\end{array}\right] \
& x_{1}=2 ; \quad x_{2}=0.48 ; \quad x_{3}=1.68
\end{aligned}$

Quiz 5: Power Flows