Question

solve by gauss elimination method 10x + y + 2z = 13, 3x+10y+z = 14, 2x + 3y + 10z = 15​

Answer 1

Answer:

(10x+y+2z=13)

(3x+10y+z=14)×2

subtract both outcomes become

4x-19y=-15 be equation 1st

another second one and third one

(3x+10y+z=14)×10

2x+3y+10z=15

subtract both then become

28x+97y=125

be second equation

from equation 1st and 2nd

28x+97y=125

(4x+19y=-15)×7

subtract both then become y=1

in equation 1st use y=1

finally become x=1

in given equation used value of x and y then become z=1 the value of x and y and z become 1 ...

Explanation:

Answer 2

Answer:

The value of x, y, z is 1,1,1 respectively by gauss elimination method.

Explanation:

Given system is

$10x+y+2z=13\\3x+10y+z=14\\2x+3y+10z=15$

Now rewrite the system in matrix form and solve it by gaussian elimination:

$\begin{bmatrix}10 & 1 &2 \\ 3 &10 &1 \\ 2 &3 &10 \end{bmatrix}=\begin{bmatrix}13\\14 \\15 \end{bmatrix}$

Now divide the row by $10$

$\begin{bmatrix}1 &0.1 &0.2 \\ 3 &10 &1\\ 2 &3 & 10\end{bmatrix}=\begin{bmatrix}1.3\\ 14\\ 15\end{bmatrix}$

Now multiply row $1$ by $3$ and subtract it from row $2$ and also multiply row $1$by  $2$ and subtract it from row $3$, we get

$\begin{bmatrix}1 &0.1 &0.2 \\ 0 &9.7 &0.4 \\ 0&2.8 &9.6 \end{bmatrix}=\begin{bmatrix}1.3\\ 10.1\\ 12.4\end{bmatrix}$

Now divide the row $2$ by $9.7$, we get

$\begin{bmatrix}1 & 0.1&0.2 \\ 0 &1 &\frac{4}{97} \\ 0 &2.8 & 9.6\end{bmatrix}=\begin{bmatrix}1.3\\ \frac{101}{97}\\ 12.4\end{bmatrix}$

Now multiply row $2$ by $0.1$ and subtract it from row $1$ and multiply row $2$ by $2.8$ and subtract it from row $3$, we get

$\begin{bmatrix} 1& 0 &\frac{19}{97} \\ 0& 1& \frac{4}{97} \\ 0 & 0& \frac{920}{97}\end{bmatrix}=\begin{bmatrix}\frac{116}{97} \\ \frac{101}{97} \\ \frac{920}{97} \end{bmatrix}$

Now divide row $3$ by $\frac{920}{97}$, we get

$\begin{bmatrix} 1& 0 &\frac{19}{97} \\ 0 & 1 & \frac{4}{97} \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\frac{116}{97} \\ \frac{101}{97} \\ 1\end{bmatrix}$

Now multiply row $3$ by $\frac{19}{97}$ and subtract it from row $1$ and also multiply row $3$ by $\frac{4}{97}$ and subtract it from row $2$, we get

$\begin{bmatrix} 1& 0 &0 \\ 0& 1& 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$

Therefore, the value of x, y, z is 1,1,1 respectively by gauss elimination method.

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