A linear system is said to be **square** if the number of equations matches the number of unknowns. If the system *A* **x** = **b** is square, then the coefficient matrix, *A*, is square. If *A* has an inverse, then the solution to the system *A* **x** = **b** can be found by multiplying both sides by *A* ^{−1}:

**Theorem D**. If *A* is an invertible *n* by *n* matrix, then the system *A* **x** = **b** has a unique solution for *every n*‐vector **b**, and this solution equals *A* ^{−1} **b**.

Since the determination of *A* ^{−1} typically requires more calculation than performing Gaussian elimination and backsubstitution, this is not necessarily an improved method of solving *A* **x** = **b** (And, of course, if *A* is not square, then it has no inverse, so this method is not even an option for nonsquare systems.) However, if the coefficient matrix *A* is square, and if *A* ^{−1} is known or the solution of *A* **x** = **b** is required for several different **b**'s, then this method is indeed useful, from both a theoretical and a practical point of view. The purpose of this section is to show how the elemenetary row operations that characterize Gauss‐Jordan elimination can be applied to compute the inverse of a square matrix.

First, a definition: If an elementary row operation (the interchange of two rows, the multiplication of a row by a nonzero constant, or the addition of a multiple of one row to another) is applied to the identity matrix, *I*, the result is called an **elementary matrix**. To illustrate, consider the 3 by 3 identity matrix. If the first and third rows are interchanged,

*I* is multiplied by −2,

*A* ^{−1} reads as follows: *If E is the elementary matrix that results when a particular elementary row operation is performed on I, then the product EA is equal to the matrix that would result if that same elementary row operation were applied to A*. In other words, an elementary row operation on a matrix *A* can be performed by multiplying *A* on the left by the corresponding elementary matrix. For example, consider the matrix

Adding −2 times the first row to the second row yields

If this same elementary row operation is applied to *I*,

*EA* should equal *A*′. You may verify that

If *A* is an invertible matrix, then some sequence of elementary row operations will transform *A* into the identity matrix, *I*. Since each of these operations is equivalent to left multiplication by an elementary matrix, the first step in the reduction of *A* to *I* would be given by the product *E* _{1} *A*, the second step would be given by *E* _{2} *E* _{1} *A*, and so on. Thus, there exist elementary matrices *E* _{1}, *E* _{2},…, *E* _{k} such that

But this equation makes it clear that *E* _{k}… *E* _{2} *E* _{1} = *A* ^{−1}:

Since *E* _{k}… *E* _{2} *E* _{1} = *E* _{k}… *E* _{2} *E* _{1} *I*, where the right‐hand side explicitly denotes the elementary row operations applied to the identity matrix *I*, *the same elementary row operations that transform A into I will transform I into A* ^{−1}. For *n* by *n* matrices *A* with *n* > 3, this describes the most efficient method for determining *A* ^{−1}.

**Example 1**: Determine the inverse of the matrix

Since the elementary row operations that will be applied to *A* will be applied to *I* as well, it is convenient here to augment the matrix *A* with the identity matrix *I*:

Then, as *A* is transformed into *I, I* will be transformed into *A* ^{−1}:

Now for a sequence of elementary row operations that will effect this transformation:

Since the transformation [ *A* | *I*] → [ *I* | *A* ^{−1}] reads

*A* is

**Example 2**: What condition must the entries of a general 2 by 2 matrix

*A* to be invertible? What is the inverse of *A* in this case?

The goal is to effect the transformation [ *A* | *I*] → [ *I* | *A* ^{−1}]. First, augment *A* with the 2 by 2 identity matrix:

Now, if *a* = 0, switch the rows. If *c* is also 0, then the proces of reducing *A* to *I* cannot even begin. So, one necessary condition for *A* to be invertible is that the entries *a* and *c* are not both 0. Assume that *a* ≠ 0. Then

Next, *assuming that ad* − *bc* ≠ 0,

Therefore, if *ad* − *bc* ≠ 0, then the matrix *A* is invertible, and its inverse is given by

(The requirement that *a* and *c* are not both 0 is automatically included in the condition *ad* − *bc* ≠ 0.) In words, the inverse is obtained from the given matrix by interchanging the diagonal entries, changing the signs of the off‐diagonal entries, and then dividing by the quantity *ad* − *bc*. *This formula for the inverse of a 2 x 2 matrix should be memorized*.

To illustrate, consider the matrix

Since *ad* − *bc* = (−2)(5) − (−3)(4) = 2 ≠ 0, the matrix is invertible, and its inverse is

You may verify that

*A* ^{−1} *A* = *I* also.

**Example 3**: Let *A* be the matrix

*A* invertible?

No. Row reduction of *A* produces the matrix

The row of zeros signifies that *A* cannot be transformed to the identity matrix by a sequence of elementary row operations; *A* is noninvertible. Another argument for the noninvertibility of *A* follows from the result Theorem D. If *A* were invertible, then Theorem D would guarantee the existence of a solution to *A* **x** = **b** for *every* column vector **b** = ( *b* _{1}, *b* _{2}, *b* _{3}) ^{T}. But *A* **x** = **b** is consistent only for those vectors **b** for which *b* _{1} + 3 *b* _{2} + *b* _{3} = 0. Clearly, then, there exist (infinitely many) vectors **b** for which *A* **x** = **b** is inconsistent; thus, *A* cannot be invertible.

**Example 4**: What can you say about the solutions of the homogeneous system *A* **x** = **0** if the matrix *A* is invertible?

Theorem D guarantees that for an invertible matrix *A*, the system *A* **x** = **b** is consistent for every possible choice of the column vector **b** and that the unique solution is given by *A* ^{−1} **b**. In the case of a homogeneous system, the vector **b** is **0**, so the system has only the trivial solution: **x** = *A* ^{−1} **0** = **0**.

**Example 5**: Solve the matrix equation *AX* = *B*, where

*Solution 1*. Since *A* is 3 x 3 and *B* is 3 x 2, if a matrix *X* exists such that *AX* = *B*, then *X* must be 3 x 2. If *A* is invertible, one way to find *X* is to determine *A* ^{−1} and then to compute *X* = *A* ^{−1} *B*. The algorithm [ *A* | *I*] → [ *I* | *A* ^{−1}] to find *A* ^{−1} yields

Therefore,

*Solution 2*. Let **b** _{1} and **b** _{2} denote, respectively, column 1 and column 2 of the matrix *B*. If the solution to *A* **x** = **b** _{1} is **x** _{1} and the solution to *A* **x** = **b** _{2} is **x** _{2}, then the solution to *AX* = *B* = [ **b** _{1} **b** _{2}] is *X* = [ **x** _{1} **x** _{2}]. That is, the elimination procedure can be performed on the two systems ( *A* **x** = **b** _{1} and *A* **x** = **b** _{2})

simultaneously:

Gauss‐Jordan elimination completes the evaluation of the components of **x** _{1} and **x** _{2}:

It follows immediately from this final augmented matrix that

It is easy to verify that the matrix *X* does indeed satisfy the equation *AX* = *B*:

Note that the transformation in Solution 1 was [ *A* | *I*] → [ *I* | *A* ^{−1}], from which *A* ^{−1} *B* was computed to give *X*. However, the transformation in Solution 2, [ *A* | *B*] → [ *I* | *X*], gave *X* directly.