An n × n matrix, A, is invertible if there exists an n × n matrix, A-1, called the inverse of A, such that
A-1A = AA-1 = In
where In is the n × n matrix. We will denote the identity matrix simply as I from now on since it will be clear what size I should be in the context of each problem. Remember that I is special because for any other matrix A,
IA = AI = A
It may be worth nothing that given an n × n invertible matrix, A, the following conditions are equivalent (they are either all true, or all false):
- rref(A) = I
- det(A) ≠ 0
- x = 0 is the only solution to Ax = 0, where 0 is the n-dimensional 0-vector
Inverse of a 2 × 2 matrix
The inverse of a 2 × 2 matrix can be calculated using a formula, as shown below.
To confirm that this is true,
Note that (ad - bc) is also the determinant of the given 2 × 2 matrix.
Inverse of a 3 × 3 matrix
Determining the inverse of a 3 × 3 matrix or larger matrix is more involved than determining the inverse of a 2 × 2. Below are some examples.
Example 1: Inverse of a 3 × 3 matrix
Calculate the inverse of
Below is the augmented matrix, [A|I],
which is matrix A coupled with the 3 × 3 identity matrix on its right. We then perform Gaussian elimination on this 3 × 6 augmented matrix to get
where rref([A|I]) stands for the "reduced row echelon form of [A|I]." The left 3 columns of rref([A|I]) form rref(A) which also happens to be the identity matrix, so rref(A) = I. Therefore, we claim that the right 3 columns form the inverse A-1 of A, so
We can then confirm this as
When we calculate rref([A|I]), we are essentially solving the systems Ax1 = e1, Ax2 = e2, and Ax3 = e3, where e1, e2, and e3 are the standard basis vectors, simultaneously. When rref(A) = I, the solution vectors x1, x2 and x3 are uniquely defined and form a new matrix [x1 x2 x3] that appears on the right half of rref([A|I]). [x1 x2 x3] satisfies A[x1 x2 x3] = [e1 e2 e3]. But since [e1 e2 e3] = I, A[x1 x2 x3] = [e1 e2 e3] = I, and by definition of inverse, [x1 x2 x3] = A-1.
Example 2: A singular (noninvertible) matrix
As in Example 1, we form the augmented matrix [B|I],
However, when we calculate rref([B|I]), we get
Notice that the first 3 columns do not form the identity matrix. Instead, they form
which has all 0's on the 3rd row. Therefore, B is not invertible. No matter what we do, we will never find a matrix B-1 that satisfies BB-1 = B-1B = I. A noninvertible matrix is usually called singular.
Note: The form of rref(B) says that the 3rd column of B is 1 times the 1st column of B plus -3 times the 2nd row of B, as shown below.
Determinants and invertibility
It can be proven that if a matrix A is invertible, then det(A) ≠ 0. The converse is also true: if det(A) ≠ 0, then A is invertible. The proof has to do with the property that each row operation we use to get from A to rref(A) can only multiply the determinant by a nonzero number. Though the proof is not provided here, we can see that the above holds for our previous examples.
In Examples 1 and 2 above, we found that
was invertible while
was singular. If we calculate the determinants of A and B, we find that
See also matrix notation, determinants.