Question

Solve the system of linear equations using the Gauss-Jordan elimination method. 2x + 3y 6212 3x + (x, y. z)

Answer 1

Answer:

The solution of the system of linear equations is $x=3, y=4, z=1$

Step-by-step explanation:

We have the system of linear equations:

$2x+3y-6z=12\\x-2y+3z=-2\\3x+y=13$

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.

The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).

From the system of linear equations that we have, the coefficient matrix is

$\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right]$

the variable matrix is

$\left[\begin{array}{c}x&y&z\end{array}\right]$

and the constant matrix is

$\left[\begin{array}{c}12&-2&13\end{array}\right]$

We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so

$\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:

• multiply the 1st row by 1/2

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

• add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]$

• add -3 times the 1st row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]$

• multiply the 2nd row by -2/7

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]$

• add 7/2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]$

• multiply the 3rd row by 1/3

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right]$

• add 12/7 times the 3rd row to the 2nd row

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

• add 3 times the 3rd row to the 1st row

$\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right]$

• add -3/2 times the 2nd row to the 1st row

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

From the reduced row echelon form we have that

$x=3\\y=4\\z=1$

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.