Question

solve this by using Gauss elimination method:

4x+y+z=4,
x+4y-2z=4,
3x+2y-4z=6
​

Answer 1

Step-by-step explanation:

hopes it helps u.........

Answer 2

The value of x, y and z is 1, $\frac{1}{2}$ and $\frac{-1}{2}$ respectively.

Given:

Three equations i.e.

4x+y+z = 4

x+4y-2z = 4

3x+2y-4z = 6

To Find:

To solve the above equations by the gauss elimination method

Solution:

We can simply solve this problem by using the following mathematical process.

The matrix form of the above equations is

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

The augment matrix (A, B) is

$\left[\begin{array}{cccc}4&1&1&4\\1&4&-2&4\\3&2&-4&6\end{array}\right]$

$R_{1}$ ↔ $R_{2}$

$\left[\begin{array}{cccc}1&4&-2&4\\4&1&1&4\\3&2&-4&6\end{array}\right]$

$R_{2}$ → $R_{2} -4R_{1}$

$R_{3}$ → $R_{3} -3R_{1}$

$\left[\begin{array}{cccc}1&4&-2&4\\0&-15&9&-12\\0&-10&2&-6\end{array}\right]$

$R_{3}$ → $R_{3} -\frac{2}{3} R_{2}$

$\left[\begin{array}{cccc}1&4&-2&4\\0&-15&9&-12\\0&0&-4&2\end{array}\right]$

The above matrix is in echelon form.  

Now writing the equivalent equations

$\left[\begin{array}{ccc}1&4&-2\\0&-15&9\\0&0&-4\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}4\\-12\\2\end{array}\right]$

This gives us

-4z = 2

$z=\frac{-1}{2}$

And

-15y + 9z = -12

Putting the value of z we get,

$y=\frac{1}{2}$

Now

x + 4y + -2z = 4

Putting the values of y and z we have,

x = 1

Hence, The value of x, y and z is 1, $\frac{1}{2}$ and $\frac{-1}{2}$ respectively.

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