How can I find A^-1?

0 ⤊

How can I find A^-1?

A= [-2,4;0,3] B=[-6,2;4,0]

2006-07-04 06:15:01 · 4 answers · asked by Nina E 1 in Science & Mathematics ➔ Mathematics

4 answers

let A = [a11, a12; a21, a22]
then inverse(A) = [a22, -a12; -a21, a11]/det
determinant det = (a11a22 - a12a21)

so in your case
det = -6
inverse(A) = [-3/6, -4/6 ; 0 , 2/6] = [-1/2, -2/3; 0, 1/3]

Check
A*inverse(A) = [-2/-2, -4/3 + 4/3; 0, 3/3] = [1,0;0,1]
correct

2006-07-04 17:02:49 · answer #1 · answered by none2perdy 4 · 0⤊ 0⤋

See the link

2006-07-04 06:39:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋

A^-1 = 1/A

Making this change, solve... much easier than using exponents.

2006-07-04 06:25:31 · answer #3 · answered by CosmikAlex 2 · 0⤊ 0⤋

whats that B for?

2006-07-04 06:19:08 · answer #4 · answered by kalkmat 3 · 0⤊ 0⤋