We develop a method for finding the inverse of a square matrix, discuss when the inverse does not exist, and use matrix inverses to solve matrix equations.
Consider the equation . It takes little time to recognize that the solution to this equation is . In fact, the solution is so obvious that we do not think about the algebraic steps necessary to find it. Let’s take a look at these steps in detail. This process utilizes many properties of real-number multiplication. In particular, we make use of the existence of multiplicative inverses. Every non-zero real number has a multiplicative inverse with the property that . We say that is the multiplicative identity because .
Given a matrix equation , we would like follow a process similar to the one above to solve this matrix equation for .
Given the set of all matrices, with standard matrix multiplication, the role of the multiplicative identity is filled by because .
Given an matrix , a multiplicative inverse of would have to be some matrix that satisfies the following property:
Assuming that such an inverse exists, this is what the process of solving the equation would look like:
The following theorem shows that matrix inverses are unique.
- Because is an inverse of , we have: Suppose there exists another matrix such that Then
Now that we know that a matrix cannot have more than one inverse, we can safely refer to the inverse of as .
We now prove several useful properties of matrix inverses.
- Proof of Property item:inverseofproduct:
- We will check to see if is the inverse of . Thus is invertible and .
We now turn to the question of how to find the inverse of a square matrix, or determine that the inverse does not exist.
Given a square matrix , we are looking for a square matrix such that We will start by attempting to satisfy . Let be the columns of , then where each is a standard unit vector of . This gives us a system of equations for each . If each has a unique solution, then finding these solutions will give us the columns of the desired matrix .
First, suppose that , then we can use elementary row operations to carry each to its reduced row-echelon form. Observe that the row operations that carry to will be the same for each . We can, therefore, combine the process of solving systems of equations into a single process Each is a unique solution of , and we conclude that is a solution to .
By Problem prob:elemrowopsreverse of SYS-0010, we can reverse the elementary row operations to obtain But the same row operations would also give us We conclude that , and .
Next, suppose that . Then must contain a row of zeros. Because one of the rows of was completely wiped out by elementary row operations, one of the rows of must be a linear combination of the other rows. Suppose row is a linear combination of the other rows. Then row can be carried to a row of zeros. But then the system is inconsistent. This is because has a as the entry and zeros everywhere else. The in the spot will not be affected by elementary row operations, and the row will eventually look like this This shows that a matrix such that does not exist, and does not have an inverse.
We have just proved the following theorem.
We will conclude this section by discussing the inverse of a nonsingular matrix. Let be a non-singular matrix. We can find by using the row reduction method described above, that is, by computing the reduced row-echelon form of . Row reduction yields the following:
Note that the denominator of each term in the inverse matrix is the same. Factoring it out, gives us the following formula for .
Clearly, the expression for is defined, if and only if . So, what happens when ? In Practice Problem prob:inverseformula you will be asked to fill in the steps of the row reduction procedure that produces this formula, and show that if then does not have an inverse.
Use to solve the equation Answer: