# Inverse matrix

An n × n matrix, A, is invertible if there exists an n × n matrix, A^{-1}, called the inverse of A, such that

A^{-1}A = AA^{-1} = I_{n}

where I_{n} 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.

If

then

To confirm that this is true,

and

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

and

When we calculate rref([A|I]), we are essentially solving the systems Ax_{1} = e_{1}, Ax_{2} = e_{2}, and Ax_{3} = e_{3}, where e_{1}, e_{2}, and e_{3} are the standard basis vectors, simultaneously. When rref(A) = I, the solution vectors x_{1}, x_{2} and x_{3} are uniquely defined and form a new matrix [x_{1} x_{2} x_{3}] that appears on the right half of rref([A|I]). [x_{1} x_{2} x_{3}] satisfies A[x_{1} x_{2} x_{3}] = [e_{1} e_{2} e_{3}]. But since [e_{1} e_{2} e_{3}] = I, A[x_{1} x_{2} x_{3}] = [e_{1} e_{2} e_{3}] = I, and by definition of inverse, [x_{1} x_{2} x_{3}] = 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 3^{rd} row. Therefore, B is not invertible. No matter what we do, we will never find a matrix B^{-1} that satisfies BB^{-1} = B^{-1}B = I. A noninvertible matrix is usually called singular.

Note: The form of rref(B) says that the 3^{rd} column of B is 1 times the 1^{st} column of B plus -3 times the 2^{nd} 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

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See also matrix notation, determinants.