Question

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

Answer 1

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

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}c2&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}c1&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}ccc1&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}c1&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}c1&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}c1&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}ccc1&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}ccc1&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}ccc1&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}ccc1&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.