Applied Discrete Structures

We have already investigated, in exercises in the previous section, one special type of matrix. That was the zero matrix, and found that it behaves in matrix algebra in an analogous fashion to the real number 0; that is, as the additive identity. We will now investigate the properties of a few other special matrices.

In the example above, the \(3\times 3\) diagonal matrix \(I\) whose diagonal entries are all 1’s has the distinctive property that for any other \(3\times 3\) matrix \(A\) we have \(A I = I A = A\text{.}\) For example:

In other words, the matrix \(I\) behaves in matrix algebra like the real number 1; that is, as a multiplicative identity. In matrix algebra, the matrix \(I\) is called simply the identity matrix. Convince yourself that if \(A\) is any \(n\times n\) matrix \(A I = I A = A\text{.}\)

In the set of real numbers we recall that, given a nonzero real number \(x\text{,}\) there exists a real number \(y\) such that \(x y = y x =1\text{.}\) We know that real numbers commute under multiplication so that the two equations can be summarized as \(x y = 1\text{.}\) Further we know that \(y =x^{-1}= \frac{1}{x}\text{.}\) Do we have an analogous situation in \(M_{n\times n}(\mathbb{R})\text{?}\) Can we define the multiplicative inverse of an \(n\times n\) matrix \(A\text{?}\) It seems natural to imitate the definition of multiplicative inverse in the real numbers.

When we are doing computations involving matrices, it would be helpful to know that when we find \(A^{-1}\text{,}\) the answer we obtain is the only inverse of the given matrix. This would let us refer to the inverse of a matrix. We refrained from saying that in the definition, but the theorem below justifies it.

Remark: Those unfamiliar with the laws of matrix algebra should return to the following proof after they have familiarized themselves with the Laws of Matrix Algebra in Section 5.5.

Let \(A =\left( \begin{array}{cc} 2 & 0 \\ 0 & 3 \\ \end{array} \right)\) . What is \(A^{-1}\) ? Without too much difficulty, by trial and error, we determine that \(A^{-1}= \left( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \\ \end{array} \right)\) . This might lead us to guess that the inverse is found by taking the reciprocal of all nonzero entries of a matrix. Alas, it isn’t that easy!

If \(A =\left( \begin{array}{cc} 1 & 2 \\ -3 & 5 \\ \end{array} \right)\) , the “reciprocal rule” would tell us that the inverse of \(A\) is \(B=\left( \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{-1}{3} & \frac{1}{5} \\ \end{array} \right)\text{.}\) Try computing \(A B\) and you will see that you don’t get the identity matrix. So, what is \(A^{-1}\text{?}\) In order to understand more completely the notion of the inverse of a matrix, it would be beneficial to have a formula that would enable us to compute the inverse of at least a \(2\times 2\) matrix. To do this, we introduce the definition of the determinant of a \(2\times 2\) matrix.

In addition to \(\det A\text{,}\) common notation for the determinant of matrix \(A\) is \(\lvert A \rvert\text{.}\) This is particularly common when writing out the whole matrix, which case we would write \(\left| \begin{array}{cc} a & b \\ c & d \\ \end{array} \right|\) for the determinant of the general \(2 \times 2\) matrix.