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Section 5.3 Concepts

Subsection 5.3.1 The identity matrix

The number one plays a special role with respect to multiplication of numbers: it is the only number that has no effect when it is multiplied against another number. In multiplication of matrices, there is only one kind of matrix that has the same effect: a square matrix with all ones down the main diagonal and zeros in every other entry. We call this matrix the identity matrix, and write \(I\) to represent it.

The identity matrix is to multiplication what the zero matrix is to addition, and it will allow us to (sometimes) do the matrix version of the algebra in the preamble to Discovery 5.1. Except there is one wrinkle that we will explore in this chapter and next: while we can always “cancel” a matrix to the zero matrix by subtracting, unfortunately we will not always be able to “cancel” a matrix to the identity by “dividing.”

Subsection 5.3.2 Inverse matrices

If \(a\) is a nonzero number, we can use the inverse \(\inv{a} = 1/a\) to multiply \(a\) to \(1\text{.}\) In algebra, we often use this fact to “cancel” a number from an algebraic expression. In matrix algebra, we can attempt to do the same thing for square matrices. For a square matrix \(A\text{,}\) we would like to find a square matrix of the same size that multiplies \(A\) to the identity \(I\) (where the identity is the matrix version of the number \(1\)). If we can find such an inverse for \(A\text{,}\) we write \(\inv{A}\) for it. Note that we need this inverse to multiply \(A\) to \(I\) from both sides, because order of multiplication matters. That is, we need to be sure that both \(\inv{A} A = I\) and \(A \inv{A} = I\text{.}\)

The only number that doesn't have an inverse is \(0\text{.}\) However, we saw in Discovery 5.2 that some nonzero matrices do not have inverses (i.e. are singular). While the singular example in Discovery 5.2.b only had one nonzero entry, it is possible to come up with examples of singular matrices that have no entries that are zero — see Subsection 5.4.1 for one example.

A look ahead.
  • We will see in Section 5.5 that a square matrix can have only one inverse matrix (Theorem 5.5.2), so writing \(\inv{A}\) to mean the inverse of an invertible matrix \(A\) is unambiguous. We will also see in Section 6.5 that it is enough to check only one of \(BA=I\) and \(AB=I\) in order to know that \(B = \inv{A}\) (Proposition 6.5.4 and Proposition 6.5.6).

  • We will also see that a square matrix is singular when it has some relationship to another matrix that has too many zero entries to be invertible.

    • In Chapter 6, we learn that a square matrix is singular precisely when its RREF has too many zeros (Theorem 6.5.2).
    • In Part III, we will see that a matrix is singular precisely when it is similar to one that has too many zeros (a fact closely related to Theorem 24.6.3).

Subsection 5.3.3 Matrix division

In Chapter 4, we defined the operations of addition, subtraction, and multiplication of matrices (as well as scalar multiplication), but we did not define division of matrices. Matrix inverses are similar to division in that they can be used to cancel a square matrix to \(I\text{,}\) the matrix version of \(1\text{.}\) But we need to be careful with this analogy, because order of matrix multiplication matters. With numbers, when we write division as a fraction \(a/b\text{,}\) it doesn't matter if we mean \(a\inv{b}\) or \(\inv{b}a\) because both orders of multiplication yield the same result. For matrices, it is ambiguous to write a fraction \(A/B\) because it is not clear whether \(A\inv{B}\) or \(\inv{B}A\) should be meant, and it matters because \(A\inv{B}\) and \(\inv{B}A\) might not yield the same result.

Warning 5.3.1.

There is no such operation as division of matrices. Never write fractions of matrices in matrix algebra — use matrix inverses instead.

Subsection 5.3.4 Cancellation

In the algebra of numbers, if we have an equation \(ab=ac\text{,}\) we would usually cancel the \(a\) from both sides to conclude that \(b=c\text{.}\) In Discovery 5.4, we see that this doesn't always work for matrices. When we cancel \(a\) from both sides of \(ab=ac\text{,}\) what we are really doing algebraically is to divide both sides by \(a\text{.}\) However, we cannot do this if \(a\) is \(0\text{,}\) and similarly we cannot cancel the \(A\) from \(AB=AC\) if \(A\) is not invertible. If it is invertible, however, then we may cancel \(A\) by applying \(\inv{A}\) to both sides:

\begin{align*} AB \amp= AC \\ \inv{A}AB \amp= \inv{A}AC \\ IB \amp= IC \\ B \amp= C. \end{align*}

When we apply algebraic manipulations to an equation, we need to make sure we perform the exact same operation on both sides of the equals sign. If we are introducing a new matrix into an equation by multiplication, we need to make sure we multiply the new matrix from the same side on both sides of the equation, because order of matrix multiplication matters! So, when we were faced with \(AB=CA\) in Task c of Discovery 5.4, it would be incorrect to cancel \(A\) here even if \(A\) is invertible, because to cancel it on the left-hand side of the equation we need to multiply by \(\inv{A}\) on the left, and to cancel on the right-hand side we need to multiply \(\inv{A}\) on the right. These are different operations, and doing one on one side of the equation and the other on the other side would violate the equals sign. If we try to do both, we just go in circles:

\begin{align*} AB \amp= CA \\ \inv{A}AB \amp= \inv{A}CA \amp \amp\text{(i)}\\ IB \amp= \inv{A}CA \amp \amp\text{(ii)}\\ B \amp= \inv{A}CA \\ B\inv{A} \amp= \inv{A}CA\inv{A} \amp \amp\text{(iii)}\\ B\inv{A} \amp= \inv{A}CI \\ B\inv{A} \amp= \inv{A}C. \end{align*}

In the above steps:

  1. the same operation (i.e. multiplication by \(\inv{A}\) on the left) must be applied to both sides;
  2. the \(\inv{A}\) and \(A\) on the right do not cancel to \(I\text{;}\) and
  3. the same operation (i.e. multiplication by \(\inv{A}\) on the right) must be applied to both sides.

In more detail, when we have \(\inv{A}CA\) on the right above, unfortunately we cannot cancel the \(\inv{A}\) and \(A\) to \(I\) because we don't have \(\inv{A}A = I\text{,}\) we have a \(C\) between \(A\) and its inverse, and order of matrix multiplication matters! And notice that after all this algebra, in the last line we are no further ahead than when we began.

Subsection 5.3.5 Solving systems using inverses

As we explored in Discovery 5.5, when the coefficient matrix of a linear system is square and invertible, we can solve the system by matrix algebra instead of row reducing. In this case, we can use the inverse to cancel the coefficient matrix from the left-hand side of the system equation \(A\uvec{x}=\uvec{b}\text{:}\)

\begin{align*} A\uvec{x} \amp= \uvec{b} \\ \inv{A}A\uvec{x} \amp= \inv{A}\uvec{b} \\ I\uvec{x} \amp= \inv{A}\uvec{b} \\ \uvec{x} \amp= \inv{A}\uvec{b}. \end{align*}

We will see in the next chapter that inverting a matrix is the same amount of work as row reducing it, so solving a system this way is not a shortcut method. But it can be faster if you want to solve several systems with the same coefficient matrix \(A\) but different vectors of constants \(\uvec{b}\text{,}\) as we had in Discovery 5.5, so that you only do the row reducing work (in computing \(\inv{A}\)) once.