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.

Question

Quizplus · Accepted Answer

The Answer is

\left[\begin{array}{ccc}5 & -2 & -3 \-5 & 7 & -2 \-3 & -3 & 8\end{array} ight]\left[\begin{array}{l}x_{1} \x_{2} \x_{3}\end{array} ight]=\left[\begin{array}{c}4 \-10 \6\end{array} ight]

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} ight]\left[\begin{array}{l}x_{1} \x_{2} \x_{3}\end{array} ight]=\left[\begin{array}{c}4 \-6 \\frac{42}{5}\end{array} ight]

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} ight]\left[\begin{array}{l}x_{1} \x_{2} \x_{3}\end{array} ight]=\left[\begin{array}{c}4 \-6 \\frac{84}{25}\end{array} ight]

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} ight|=50 ; \quad \Delta_{1}=\left|\begin{array}{ccc}4 & -2 & -3 \-10 & 7 & -2 \6 & -3 & 8\end{array} ight|=100 ext { (Expanding } \Delta_{1} about the first column \& \Delta_{2}=\left|\begin{array}{ccc}5 & 4 & -3 \-5 & -10 & -2 \-3 & 6 & 8\end{array} ight|=24 ; \quad \Delta_{3}=\left|\begin{array}{ccc}5 & -2 & 4 \-5 & 7 & -10 \-3 & -3 & 6\end{array} ight|=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 ext { where } A^{-1}=\frac{1}{50}\left[\begin{array}{ccc}50 & 25 & 25 \46 & 31 & 25 \36 & 21 & 25\end{array} ight] \& x_{1}=2 ; \quad x_{2}=0.48 ; \quad x_{3}=1.68\end{aligned}