#### (Solved): My Question Is, If I Do This: 3??R2-R3 Then Replce R3, I Will Get 0 0 2 For R3, Then The Det?A ...  My question is, if I do this:

3??R2-R3 then replce R3, i will get 0 0 2 for R3, then the det?A?will become 14 instead of-14. Why am i wrong?some people told me if i want to replace a row?then that row must be always stay in the most front? and we can't multiple that row?

but see this example, i can still get the right answer by putting the replaced row in the most front: Can anyone explain more for me

Thanks

Theorem 3.3 Elementary Row Operations and Determinants Let A and B be square matrices. 1. When B is obtained from A by interchanging two rows of A, det (B) = -det (A). 2. When B is obtained from A by adding a multiple of a row of A to another row of A, det (B) = det (A). 3. When B is obtained from A by multiplying a row of A by a nonzero constant c, det (B) = c det (A). Using elementary row operations, rewrite A in triangular form as shown below. JO 1 lo -7 2 3 Interchange the first two rows. 14 1 2 -2 = -10 -7 -81 lo 3 112 70 1 JO 3 11 2 = 710 1 lo 0 -2 141 - 8 -21 -21 -8 -2 -2 -21 Factor - 7 out of the second row. Add-3 times the second row to the third row to produce a new third row. The above matrix is triangular, so the determinant is A = 7(1) (1) (-2) = -14. : use row operations to find the deteminant, A=215 14 04 Solution ? A = 1 lo 2 14 ZR, tks -> R2 4 04 Ho 4 lo 2 14 4R-1,7R; oc l2l det[A1 = 3 (1x2x12) =12 solution 3 LH 04 A- ž o 2 14 ZR ?t R, ->R2 14 o al - ž o 2 14 -4R, tR, R. To o 12 | det 1A = 2 (1&2X(-12)) = -12 why solution 1 is wrong?