Thread: Matrix- Determinan and Inverse

Hi,
Remember elementary row transformations can be accomplished by left multiplication of the corresponding elementary row matrix. The elementary row matrix corresponding to a row transformation is the row transformation applied to the identity matrix. Use these ideas for a solution:

Let A be an n x n matrix. Remember these effects of elementary row operations on the determinant.

If B is the matrix obtained from A by interchanging two rows, then det(B) = -det(A)
If B is the matrix obtained from A by adding a mutliple of one row to another, then det(B) = det(A)
If B is the matrix obtained from A by multiplying a row by a scalar c, then det(B) = c det(A)

Let B1 be the matrix obtained from A by L1 <--> L3. Then det(B1) = -det(A).
Let B2 be the matrix obtained from B1 by L3 --> L3 - 2 L2. Then det(B2) = det(B1)
Let B3 be the matrix obtained from B2 by L3 --> 1/2 L3. The det(B3) = (1/2) det(B2)

Since B3 is the final elementary row operation, then we have the identity matrix. Thus, B3 = In. Then det(B3) = 1

det(B3) = (1/2) det(B2)
det(B3) = (1/2) det(B1)
det(B3) = (1/2) - det(A)
1 = -1/2 det(A)
det(A) = -2

Use augmented matrix.

Let A =
[a b c | 1 0 0]
[d e f | 0 1 0]
[g h i | 0 0 1]

Then B1 =
[g h i | 0 0 1]
[d e f | 0 1 0]
[a b c | 1 0 0]

Then B2 =
[g h i | 0 0 1]
[d e f | 0 1 0]
[a-2d b-2e c-2f | 1 -2 0

Then B3 =
[g h i | 0 0 1]
[d e f | 0 1 0]
[.5a-d, .5b-2, .5c-2 | .5 -1 0]

Since B3 = I3, we see that A^-1 =

[0 0 1]
[0 1 0]
[.5 -1 0]

I got it, thank you