4 5 6
5 6 4
6 4 5
Solution: Applying R3----> R3-R2 and R2----->R2-R1, we obtain
4 5 6
5 6 4
6 4 5
Applying R2---->R2-R2, we obtain
0 13 2
0 3 -3
1 -2 1
Expanding along C1, we get
det= (-1)3+1 (1) | 13 2 |
| 3 -3|
=-39-6 = -45
Dear sir/mam
I tried to upload photo of the determinant which i got from my book but there was some error, so i have written it. Now i know the expansion method to solve the det but i am unable to understand the above method, that means (Applying R3----> R3-R2 and R2----->R2-R1 and Expanding C1). From which formula or property this belongs.How can i know when to deduct which row from which or when to expand which column. Actually i am facing a great mess here. So i request you to explain me as nicely as possible.
Applying row operations has the following effects on the determinant:
Swapping 2 rows changes the sign of the determinant (i.e. multiplication by $-1$).
Adding a multiple of one row to another does not change the determinant.
Multiplying a row by a scalar changes the determinant by the reciprocal of that factor (i.e. multiplying a row by $2$ changes the determinant by $\frac{1}2$.
So one can use row operations to reduce the matrix to a triangular one, or to just a matrix with more $0$'s so the determinant computation is simplified. The steps in your particular problem are confusing, as the second matrix is equal to the first one, so there appears to be no row operations done on the matrix at the first step. Also, you cannot subtract a row from itself as that would be equivalent to multiplying the row by $0$ which is not allowed (so $R_2-R_2\to R_2$ makes no sense).