We have seen that the number 1 plays a special role in multiplication, because
1
x
=
x
1x=x.
The inverse of a number is defined as the number that multiplies by that number to give 1: bb is the inverse of
a
a if
a
b
=
1
ab=1. Hence, the inverse of
3
3 is
1
3
1
3
; the inverse of
−58−58 size 12{ { {  5} over {8} } } {} ==
−85−85 size 12{ { {  8} over {5} } } {}. Every number except 0 has an inverse.
By analogy, the inverse of a matrix multiplies by that matrix to give the identity matrix.
The inverse of matrix
[
A
]
[A], designated as
[
A
]
1
[A
]
1
, is defined by the property:
[
A
]
[
A
]
1
=
[
A
]
1
[
A
]
=
[
I
]
[A][A
]
1
=[A
]
1
[A]=[I]
The superscript –1 is being used here in a similar way to its use in functions. Recall that
f
–1
(
x
)
f
–1
(x) does not designate an exponent of any kind, but instead, an inverse function. In the same way,
[
A
]
–1
[A
]
–1
does not denote an exponent, but an inverse matrix.
Note that, just as in the definition of the identity matrix, this definition requires commutativity—the multiplication must work the same in either order.
Note also that only square matrices can have an inverse. Why? The definition of an inverse matrix is based on the identity matrix
[
I
]
[I], and we already said that only square matrices even have an identity!
How do you find an inverse matrix? The method comes directly from the definition, with a little algebra.
Table 1
Example: Finding an Inverse Matrix 
Find the inverse of
34563456 size 12{ left [ matrix {
3 {} # 4 {} ##
5 {} # 6{}
} right ]} {} 
The problem

34563456 size 12{ left [ matrix {
3 {} # 4 {} ##
5 {} # 6{}
} right ]} {}abcdabcd size 12{ left [ matrix {
a {} # b {} ##
c {} # d{}
} right ]} {} =
10011001 size 12{ left [ matrix {
1 {} # 0 {} ##
0 {} # 1{}
} right ]} {} 
This is the key step. It establishes
a
b
c
d
a
b
c
d
size 12{ left [ matrix {
a {} # b {} ##
c {} # d{}
} right ]} {}
as the inverse that we are looking for, by asserting that it fills the definition of an inverse matrix: when you multiply this mystery matrix by our original matrix, you get [I]. When we solve for the four variables a, b, c, and d, we will have found our inverse matrix.

3a+4c3b+4d5a+6c5b+6d3a+4c3b+4d5a+6c5b+6d size 12{ left [ matrix {
3a+4c {} # 3b+4d {} ##
5a+6c {} # 5b+6d{}
} right ]} {} =
10011001 size 12{ left [ matrix {
1 {} # 0 {} ##
0 {} # 1{}
} right ]} {} 
Do the multiplication. (You should check this step for yourself, it’s great practice. For instance, you start by multiplying first row x first column, and you get 3a+4c.)

3
a
+
4
c
=
1
3
b
+
4
d
=
0
5
a
+
6
c
=
0
5
b
+
6
d
=
1
3a+4c=13b+4d=05a+6c=05b+6d=1 
Remember what it means for two matrices to be equal: every element in the left must equal its corresponding element on the right. So, for these two matrices to equal each other, all four of these equations must hold.

a
=
–3
b
=
2
c
=
2
1
2
d
=
–1
1
2
a=–3b=2c=2
1
2
d=–1
1
2

Solve the first two equations for a and c by using either elimination or substitution. Solve the second two equations for b
b and
d
d by using either elimination or substitution. (The steps are not shown here.)

So the inverse is:
−32212−112−32212−112 size 12{ left [ matrix {
 3 {} # 2 {} ##
2 { {1} over {2} } {} #  1 { {1} over {2} } {}
} right ]} {} 
Having found the four variables, we have found the inverse.

Did it work? Let’s find out.
Table 2
Testing our Inverse Matrix 
−
3
2
2
1
2
−
1
1
2
−
3
2
2
1
2
−
1
1
2
size 12{ left [ matrix {
 3 {} # 2 {} ##
2 { {1} over {2} } {} #  1 { {1} over {2} } {}
} right ]} {}
3
4
5
6
3
4
5
6
size 12{ left [ matrix {
3 {} # 4 {} ##
5 {} # 6{}
} right ]} {}
1
0
0
1
1
0
0
1
size 12{ left [ matrix {
1 {} # 0 {} ##
0 {} # 1{}
} right ]} {}

The definition of an inverse matrix: if we have indeed found an inverse, then when we multiply it by the original matrix, we should get [I].

(
−
3
)
(
3
)
+
(
2
)
(
5
)
(
−
3
)
(
4
)
+
(
2
)
(
6
)
2
1
2
(
3
)
+
−
1
1
2
(
5
)
2
1
2
(
4
)
+
−
1
1
2
(
6
)
(
−
3
)
(
3
)
+
(
2
)
(
5
)
(
−
3
)
(
4
)
+
(
2
)
(
6
)
2
1
2
(
3
)
+
−
1
1
2
(
5
)
2
1
2
(
4
)
+
−
1
1
2
(
6
)
size 12{ left [ matrix {
\(  3 \) \( 3 \) + \( 2 \) \( 5 \) {} # \(  3 \) \( 4 \) + \( 2 \) \( 6 \) {} ##
left (2 { {1} over {2} } right ) \( 3 \) + left (  1 { {1} over {2} } right ) \( 5 \) {} # left (2 { {1} over {2} } right ) \( 4 \) + left (  1 { {1} over {2} } right ) \( 6 \) {}
} right ]} {}
1
0
0
1
1
0
0
1
size 12{ left [ matrix {
1 {} # 0 {} ##
0 {} # 1{}
} right ]} {}

Do the multiplication.

−9+10−12+12712−71210−9−9+10−12+12712−71210−9 size 12{ left [ matrix {
 9+"10" {} #  "12"+"12" {} ##
7 { {1} over {2} }  7 { {1} over {2} } {} # "10"  9{}
} right ]} {} =
10011001 size 12{ left [ matrix {
1 {} # 0 {} ##
0 {} # 1{}
} right ]} {} 
It works!

Note that, to fully test it, we would have to try the multiplication in both orders. Why? Because, in general, changing the order of a matrix multiplication changes the answer; but the definition of an inverse matrix specifies that it must work both ways! Only one order was shown above, so technically, we have only halftested this inverse.
This process does not have to be memorized: it should make logical sense. Everything we have learned about matrices should make logical sense, except for the very arbitrarylooking definition of matrix multiplication.
"DAISY and BRF versions of this collection are available."