Question

Slove the system of equatins using Gauss-Jordan method X+Y+Z=3, 2X-Y+3Z=16
,3X+Y-Z=-3​

Answer 1

Answer:

sorry

Step-by-step explanation:

x=3xyz

Answer 2

Answer:

Gauss Jordan method

Step-by-step explanation:

Explanation:

Given ,

X+Y+Z=3

2X-Y+3Z=16

3X+Y-Z=-3

The augmented matrix for this given statement ,

C = [A:B]

$\left[\begin{array}{ccc}1&1&1:3\\2&-1&3:16\\3&1&-1:-3\end{array}\right]$

Step1:

Apply,

$R_{2}$ →$R_{2}$ -2$R_{1}$

$R_{3}$→$R_{3}$-3$R_{1}$

$\left[\begin{array}{ccc}1&1&1:3\\0&-3&1:10\\0&-2&-4:-12\end{array}\right]$

Step 2:

$R_{2}$→$R_{2}$/-3

$\left[\begin{array}{ccc}1&1&1:3\\0&1&-1/3:-10/3\\0&-2&-4:-12\end{array}\right]$

on applying,

$R_{3}$→$R_{3}$+2$R_{2}$

$\left[\begin{array}{ccc}1&1&1:3\\0&1&-1/3:-10/3\\0&0&-14/3:-56/3\end{array}\right]$

again apply,

$R_{3}$→$R_{3}$/(-14/3)

$\left[\begin{array}{ccc}1&1&1:3\\0&1&-1/3:-10/3\\0&0&1:4\end{array}\right]$

Step3:

$R_{2}$→$R_{2}$+1/3$R_{3}$

$R_{1}$→$R_{1}$-$R_{3}$

$\left[\begin{array}{ccc}1&1&0:-1\\0&1&0:-3\\0&0&1:4\end{array}\right]$

Step 4:

$R_{1}$→$R_{1}$-$R_{2}$

$\left[\begin{array}{ccc}1&0&0:2\\0&1&0:-3\\0&0&1:4\end{array}\right]$

Final answer :

Hence ,the value of X = 2, Y = -3 and Z= 4 is the final answer of this given question.