Quandaries
and Queries
I have to find the determinant of the following matrix
| -2 | 3 | 1 | 2 | 4 |
| -3 | 0 | -2 | 5 | 1 |
| 4 | 2 | 1 | -3 | 5 |
| 2 | 3 | 4 | -1 | 2 |
| 6 | 0 | 3 | 2 | -4 |
Hi,
(I am displaying the matrix as a table since it's easier for me to do that.)
When I looked at this matrix I decided to work with the second column since it contains two zeros. I performed three operations
- Multiply the first row by -1 and add it to the fourth row.
- Multiply the third row by -1 and add it to the first row.
- Multiply the first row by -2 and add it to the third row.
This gave me
| -6 | 1 | 0 | 5 | -1 |
| -3 | 0 | -2 | 5 | 1 |
| 16 | 0 | 1 | -13 | 7 |
| 4 | 0 | 3 | -3 | -2 |
| 6 | 0 | 3 | 2 | -4 |
which is -1 times
| -3 | -2 | 5 | 1 |
| 16 | 1 | -13 | 7 |
| 4 | 3 | -3 | -2 |
| 6 | 3 | 2 | -4 |
Here again I decided to work with the second column. I performed the three operations
- Multiply the second row by 2 and add it to the first row.
- Multiply the third row by -1 and add it to the fourth row.
- Multiply the second row by -3 and add it to the third row.
This gave me
| 29 | 0 | -21 | 15 |
| 16 | 1 | -13 | 7 |
| -44 | 0 | 36 | -23 |
| 2 | 0 | 5 | -2 |
Thus I have -1 times
| 29 | -21 | 15 |
| -44 | 36 | -23 |
| 2 | 5 | -2 |
Looking at the third row I decided to add the first column to the third column to get
| 29 | -21 | 44 |
| -44 | 36 | -67 |
| 2 | 5 | 0 |
At this point I decided to expand along the third row and I finally got the value of the determinant to be 319.
Cheers,
Penny