Question

Solve the following equations by Gauss elimination method
x+y+3z=7
x+3y-3z=11
x+2y+2z=-6

Answer 1

Answer:

here it is

eq.2-eq.1

then

3y=4

y=2

then

x+3z=5

and

x+2z=-10

subtract the eq.

then

z=15

x=-40

ok

follow me

guys

Answer 2

Answer:

$\left\{\begin{array}{l}x_1=50 \\ x_2=-20.5 \\ x_3=-7.5\end{array}\right.$

Step-by-step explanation:

Step 1: $\left\{\begin{array}{l}x_1+x_2+3 x_3=7 \\x_1+3 x_2-3 x_3=11 \\x_1+2 x_2+2 x_3=-6\end{array}\right.$

Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)

$\left(\begin{array}{ccc|c}1 & 1 & 3 & 7 \\1 & 3 & -3 & 11 \\1 & 2 & 2 & -6\end{array}\right)$

$\mathrm{R}_2-1 \mathrm{R}_1 \rightarrow \mathrm{R}_2$(multiply 1 row by 1 and subtract it from 2 row); $\mathrm{R}_3-1 \mathrm{R}_1 \rightarrow \mathrm{R}_3$ (multiply 1 row by 1 and subtract it from 3 row)

$\left(\begin{array}{ccc|c}1 & 1 & 3 & 7 \\0 & 2 & -6 & 4 \\0 & 1 & -1 & -13\end{array}\right)$

$\mathrm{R}_2 / 2 \rightarrow \mathrm{R}_2$ (divide the 2 row by 2 )

$\left(\begin{array}{ccc|c}1 & 1 & 3 & 7 \\0 & 1 & -3 & 2 \\0 & 1 & -1 & -13\end{array}\right)$

Step 2:$R_1-1 R_2 \rightarrow R_1$ (multiply 2 row by 1 and subtract it from 1 row); $R_3-1 R_2 \rightarrow R_3$ (multiply 2 row by 1 and subtract it from 3 row)

$\left(\begin{array}{ccc|c}1 & 0 & 6 & 5 \\0 & 1 & -3 & 2 \\0 & 0 & 2 & -15\end{array}\right)$

$\mathrm{R}_3 / 2 \rightarrow \mathrm{R}_3$ (divide the 3 row by 2 )

$\left(\begin{array}{ccc|c}1 & 0 & 6 & 5 \\0 & 1 & -3 & 2 \\0 & 0 & 1 & -7.5\end{array}\right)$

Step 3:$R_1-6 R_3 \rightarrow R_1$ (multiply 3 row by 6 and subtract it from 1 row);$R_2+3 R_3 \rightarrow R_2$ (multiply 3 row by 3 and add it to 2 row)

$\left(\begin{array}{ccc|c}1 & 0 & 0 & 50 \\0 & 1 & 0 & -20.5 \\0 & 0 & 1 & -7.5\end{array}\right)$

$\left\{\begin{array}{l}x_1=50 \\ x_2=-20.5 \\ x_3=-7.5\end{array}\right.$

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