Inverse Matrix

Key Questions

  • What is the multiplicative inverse of a matrix?

    The multiplicative inverse of a matrix #A# is a matrix (indicated as #A^-1#) such that:
    #A*A^-1=A^-1*A=I#
    Where #I# is the identity matrix (made up of all zeros except on the main diagonal which contains all #1#).
    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]

    Gió · 3 · Feb 18 2015
  • How do I use an inverse matrix to solve a system of equations?

    First, you must be able to write your system in Standard form, before you write your matrix equation.
    Ex: 2x + 3y = 7
    -x + 5y = 3

    my screenshot5

    As 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.

    my screenshot 6

    MrsRudolph221 · 1 · Apr 25 2015
  • How do I find the inverse of a #2xx2# matrix?

    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))))#

    George C. · 1 · Jun 9 2018

Questions

Matrix Algebra