Question

Solve the given system of equation by Gauss Elimination method. 3x + 4y – z = -6; -2y + 10z = -84; y – 2z = -2​

Answer 1

Answer:

-7z =42

Step-by-step explanation:

2x + 4y + 6z = 4

a) 3x - 5y =12

10x + 5y = -25

2x + y - 2z = 10

b) y + 10z = - 28

Answer 2

Answer:

The solution set of the given system of equations is

$\bigg(\dfrac{314}{9},-\dfrac{94}{3},-\dfrac{44}{3}\bigg)$

Step-by-step explanation:

Gaussian  method or Gauss-Jordan method, also known as row reduction method, is an algorithm for solving systems of linear equations.

In this method, a series of operations are performed to reduce the coefficients, thus finally arriving at the solution.

Given the system of equations

$\begin{array}{ccc}3x + 4y~~- z = -6\\~~~~~-2y + 10z = -84\\~~~~~~~~ y ~- 2z =-2 \end{array}$

Rewriting the system of equations in matrix form

$\left[\begin{array}{ccc|c}3 &4&-1&-6\\ 0&-2&10&-84\\0 &1&-2&-2 \end{array}\right]$

Multiply row 3 with 2 and add with row 2, i.e., $\big(2 R_3+R_2 \rightarrow R_3\big)$ and divide row 3 with 3 i.e., $\big(R_3/3 \rightarrow R_3\big)$

$\left[\begin{array}{ccc|c}3 &4&-1&-6\\ 0&-2&10&-84\\0 &0&1&-\dfrac{88}{6} \end{array}\right]$

Thus, the value of

$z=-\dfrac{88}{6} =-\dfrac{44}{3}$

Now using back substitution, substitute z in equation (3),

$y ~- 2z =-2$

$\implies y ~- 2\times-\dfrac{44}{3} =-2$

$\implies y ~=-\dfrac{88}{3} -2=\dfrac{-88-6}{3}=-\dfrac{94}{3}$

Substitute y and z in equation (1),

$3x + 4y- z = -6$

$\implies 3x + 4\bigg(-\dfrac{94}{3}\bigg)- \bigg(-\dfrac{44}{3}\bigg) = -6$

$\implies 3x -\dfrac{376}{3}+\dfrac{44}{3}= -6$  

$\implies 3x=\dfrac{376}{3}-\dfrac{44}{3} -6=\dfrac{376-44-18}{3}=\dfrac{314}{3}$

$\implies x=\dfrac{314}{3\times 3}=\dfrac{314}{9}$

Therefore, the solution set of the given system of equations is

$\bigg(\dfrac{314}{9},-\dfrac{94}{3},-\dfrac{44}{3}\bigg)$

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