Inverse Matrix
Key Questions
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The multiplicative inverse of a matrix
A is a matrix (indicated asA^-1 ) such that:
A*A^-1=A^-1*A=I
WhereI is the identity matrix (made up of all zeros except on the main diagonal which contains all1 ).
For example:
if:A=
[4 3]
[3 2]A^-1=
[-2 3]
[3 -4]Try to multiply them and you'll find the identity matrix:
[1 0]
[0 1] -
First, you must be able to write your system in Standard form, before you write your matrix equation.
Ex: 2x + 3y = 7
-x + 5y = 3As you know from other operations, the Identity produces itself (adding 0, multiplying by 1), leaving you with the variables alone on the left side, and your answers on the right! All you have to do is multiply matrix
A^-1 times matrix B. -
Answer:
((a,b),(c,d))^(-1) = 1/abs((a,b),(c,d)) ((d, -b), (-c, a)) Explanation:
Let's have a go a this without simply plugging in a remembered formula.
Given a matrix
((a, b), (c, d)) , let's try multiplying it by((d, -b),(-c,a)) and see what we get:((a, b),(c, d))((d, -b),(-c, a)) = ((ad-bc, 0), (0, ad-bc)) So if we multiply by
1/(ad-bc) = 1/abs((a,b),(c,d)) then we find the inverse matrix:1/abs((a,b),(c,d)) ((d, -b),(-c, a)) = ((d/abs((a,b),(c,d)), -b/abs((a,b),(c,d))), (-c/abs((a,b),(c,d)), a/abs((a,b),(c,d))))