Let us solve the system of following equation by Guass elimination method:
{ 3 x 1 + 2 x 2 + 6 x 3 = 2 6 x 1 − 7 x 2 − 11 x 3 = 8 − 5 x 1 + 9 x 2 + 3 x 3 = 10 \begin{cases}3x_1+2x_2+6x_3=2\\
6x_1-7x_2-11x_3=8\\
-5x_1+9x_2+3x_3=10
\end{cases} ⎩ ⎨ ⎧ 3 x 1 + 2 x 2 + 6 x 3 = 2 6 x 1 − 7 x 2 − 11 x 3 = 8 − 5 x 1 + 9 x 2 + 3 x 3 = 10
The augmented matrix of the systems is the following:
[ 3 2 6 2 6 − 7 − 11 8 − 5 9 3 10 ] ( r 2 → r 2 − 2 r 1 r 3 → r 3 + 2 r 1 ) [ 3 2 6 2 0 − 11 − 23 4 1 13 15 14 ] r 1 → r 1 − 3 r 3 \left[
\begin{array}{cccc}
3 & 2 & 6 & 2\\
6 & -7 & -11 & 8\\
-5 & 9 & 3 & 10
\end{array}
\right]
{r_2\to r_2-2r_1\choose r_3\to r_3+2r_1}
\left[
\begin{array}{cccc}
3 & 2 & 6 & 2\\
0 & -11 & -23 & 4\\
1 & 13 & 15 & 14
\end{array}
\right]
{r_1\to r_1-3r_3} ⎣ ⎡ 3 6 − 5 2 − 7 9 6 − 11 3 2 8 10 ⎦ ⎤ ( r 3 → r 3 + 2 r 1 r 2 → r 2 − 2 r 1 ) ⎣ ⎡ 3 0 1 2 − 11 13 6 − 23 15 2 4 14 ⎦ ⎤ r 1 → r 1 − 3 r 3
[ 0 − 37 − 39 − 40 0 − 11 − 23 4 1 13 15 14 ] r 1 → r 1 − 3 r 2 [ 0 − 4 30 − 52 0 − 11 − 23 4 1 13 15 14 ] r 2 → r 2 − 3 r 1 \left[
\begin{array}{cccc}
0 & -37 & -39 & -40\\
0 & -11 & -23 & 4\\
1 & 13 & 15 & 14
\end{array}
\right]
{r_1\to r_1-3r_2}
\left[
\begin{array}{cccc}
0 & -4 & 30 & -52\\
0 & -11 & -23 & 4\\
1 & 13 & 15 & 14
\end{array}
\right]
{r_2\to r_2-3r_1} ⎣ ⎡ 0 0 1 − 37 − 11 13 − 39 − 23 15 − 40 4 14 ⎦ ⎤ r 1 → r 1 − 3 r 2 ⎣ ⎡ 0 0 1 − 4 − 11 13 30 − 23 15 − 52 4 14 ⎦ ⎤ r 2 → r 2 − 3 r 1
[ 0 − 4 30 − 52 0 1 − 113 160 1 13 15 14 ] r 1 → r 1 + 4 r 2 [ 0 0 − 422 588 0 1 − 113 160 1 13 15 14 ] \left[
\begin{array}{cccc}
0 & -4 & 30 & -52\\
0 & 1 & -113 & 160\\
1 & 13 & 15 & 14
\end{array}
\right]
{r_1\to r_1+4r_2}
\left[
\begin{array}{cccc}
0 & 0 & -422 & 588\\
0 & 1 & -113 & 160\\
1 & 13 & 15 & 14
\end{array}
\right] ⎣ ⎡ 0 0 1 − 4 1 13 30 − 113 15 − 52 160 14 ⎦ ⎤ r 1 → r 1 + 4 r 2 ⎣ ⎡ 0 0 1 0 1 13 − 422 − 113 15 588 160 14 ⎦ ⎤
Consequently, we have the following system:
{ − 422 x 3 = 588 x 2 − 113 x 3 = 160 x 1 + 13 x 2 + 15 x 3 = 14 \begin{cases}-422x_3=588\\
x_2-113x_3=160\\
x_1+13x_2+15x_3=14
\end{cases} ⎩ ⎨ ⎧ − 422 x 3 = 588 x 2 − 113 x 3 = 160 x 1 + 13 x 2 + 15 x 3 = 14
{ x 3 = − 294 211 x 2 = 160 + 113 ( − 294 211 ) x 1 = 14 − 13 x 2 − 15 ( − 294 211 ) \begin{cases}x_3=-\frac{294}{211}\\
x_2=160+113(-\frac{294}{211})\\
x_1=14-13x_2-15(-\frac{294}{211})
\end{cases} ⎩ ⎨ ⎧ x 3 = − 211 294 x 2 = 160 + 113 ( − 211 294 ) x 1 = 14 − 13 x 2 − 15 ( − 211 294 )
{ x 3 = − 294 211 x 2 = 538 211 x 1 = 14 − 13 ( 538 211 ) − 15 ( − 294 211 ) \begin{cases}x_3=-\frac{294}{211}\\
x_2=\frac{538}{211}\\
x_1=14-13(\frac{538}{211})-15(-\frac{294}{211})
\end{cases} ⎩ ⎨ ⎧ x 3 = − 211 294 x 2 = 211 538 x 1 = 14 − 13 ( 211 538 ) − 15 ( − 211 294 )
{ x 3 = − 294 211 x 2 = 538 211 x 1 = 370 211 \begin{cases}x_3=-\frac{294}{211}\\
x_2=\frac{538}{211}\\
x_1=\frac{370}{211}
\end{cases} ⎩ ⎨ ⎧ x 3 = − 211 294 x 2 = 211 538 x 1 = 211 370
#339914 on Dec 2023
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