!!!!!!!!Algebra matrix question plz help!!!!!!!!?

Question

A=
2 -1 (is matrix)
3 6

a) write A as a product of elementary matrices.

b) use your result for a) to write A^-1 as a product of elementary matrices.

c) AB=
-3 5
1 2

solve for B?

Answer 1

a) This is just like doing Gaussian elimination, except that at each step you write down to the left the elementary matrix that corresponds to the _opposite_ of your last operation (since that matrix times the result of your last operation will give you back the original matrix). So we have:

A =
[2, -1]
[3, 6]
=
[2, 0] [1, -1/2]
[0, 1] [3, 6]
=
[2, 0] [1, 0] [1, -1/2]
[0, 1] [3, 1] [0, 15/2]
=
[2, 0] [1, 0] [1, 0] [1, -1/2]
[0, 1] [3, 1] [0, 15/2] [0, 1]

b) the inverse of an elementary matrix is simply the matrix that corresponds to the inverse operation -- so simply invert each matrix and write them in reverse order (reverse order, since (ABCD)⁻¹ = D⁻¹C⁻¹B⁻¹A⁻¹. Thus we have that:

A⁻¹ =
[1, 1/2] [1, 0] [1, 0] [1/2, 0]
[0, 1] [0, 2/15] [-3, 1] [0, 1]

c) Since you have the inverse matrix as a product of elementary matrices, simply multiply it out to get A⁻¹, and then left-multiply AB by A⁻¹ to obtain B:

A⁻¹ =
[1, 1/2] [1, 0] [1, 0] [1/2, 0]
[0, 1] [0, 2/15] [-3, 1] [0, 1]
=
[1, 1/2] [1, 0] [1/2, 0]
[0, 1] [0, 2/15] [-3/2, 1]
=
[1, 1/2] [1/2, 0]
[0, 1] [-1/5, 2/15]
=
[2/5, 1/15]
[-1/5, 2/15]

A⁻¹AB =
[2/5, 1/15] [-3, 5]
[-1/5, 2/15] [1, 2]
=
[-17/15, 32/15]
[11/15, -11/15]

And we are done. That's one ugly matrix.

Answer 2

how many times are you going to write this question

use this it might help