Question

Use Gaussian elimination to write each system in triangular form

x+ y+z+ w= 1

x+y −w=−1

−x+ y+z+2w= 2

x+2y−z+ w= 0

Answer 1

To see the steps to the diagonal form see the step-by-step explanation. The solution to the system is
`x =  -(1)/(9)`,
`y= -(1)/(9)`,
`z= (4)/(9)` and
`w = (7)/(9)`

Explanation:

Gauss elimination method consists in reducing the matrix to a upper triangular one by using three different types of row operations (this is why the method is also called row reduction method). The three elementary row operations are:

1. Swapping two rows
2. Multiplying a row by a nonzero number
3. Adding a multiple of one row to another row

To solve the system using the Gauss elimination method we need to write the augmented matrix of the system. For the given system, this matrix is:

`\left[\begin{array}c1 & 1 & 1 & 1 & 1 \\1 & 1 & 0 & -1 & -1 \\-1 & 1 & 1 & 2 & 2 \\1 & 2 & -1 & 1 & 0\end{array}\right]`

For this matrix we need to perform the following row operations:

• 
  `R_2 - 1 R_1 \rightarrow R_2` (multiply 1 row by 1 and subtract it from 2 row)
• 
  `R_3 + 1 R_1 \rightarrow R_3` (multiply 1 row by 1 and add it to 3 row)
• 
  `R_4 - 1 R_1 \rightarrow R_4` (multiply 1 row by 1 and subtract it from 4 row)

• 
  `R_2 \leftrightarrow R_3` (interchange the 2 and 3 rows)

• 
  `R_2 / 2 \rightarrow R_2` (divide the 2 row by 2)
• 
  `R_1 - 1 R_2 \rightarrow R_1` (multiply 2 row by 1 and subtract it from 1 row)
• 
  `R_4 - 1 R_2 \rightarrow R_4` (multiply 2 row by 1 and subtract it from 4 row)
• 
  `R_3 \cdot ( -1) \rightarrow R_3` (multiply the 3 row by -1)
• 
  `R_2 - 1 R_3 \rightarrow R_2` (multiply 3 row by 1 and subtract it from 2 row)
• 
  `R_4 + 3 R_3 \rightarrow R_4` (multiply 3 row by 3 and add it to 4 row)
• 
  `R_4 / 4.5 \rightarrow R_4` (divide the 4 row by 4.5)

After this step, the system has an upper triangular form

The triangular matrix looks like:

`\left[\begin{array}c1 & 0 & 0 & -0.5 & -0.5  \\0 & 1 & 0 & -0.5 & -0.5\\0 & 0 & 1 & 2 &  2 \\0 & 0 & 0 & 1 &  (7)/(9)\end{array}\right]`

If you later perform the following operations you can find the solution to the system.

• 
  `R_1 + 0.5 R_4 \rightarrow R_1` (multiply 4 row by 0.5 and add it to 1 row)
• 
  `R_2 + 0.5 R_4 \rightarrow R_2` (multiply 4 row by 0.5 and add it to 2 row)
• 
  `R_3 - 2 R_4 \rightarrow R_3`(multiply 4 row by 2 and subtract it from 3 row)

After this operations, the matrix should look like:

`\left[\begin{array}cccc1 & 0 & 0 & 0 & -(1)/(9)  \\0 & 1 & 0 & 0 &   -(1)/(9)\\0 & 0 & 1 & 0 &  (4)/(9) \\0 & 0 & 0 & 1 &  (7)/(9)\end{array}\right]`

Thus, the solution is:

`x =  -(1)/(9)`,
`y= -(1)/(9)`,
`z= (4)/(9)` and
`w = (7)/(9)`