Subsection5.4.1Dissimilarities with elementary algebra
We have seen that matrix algebra is similar in many ways to elementary algebra. Indeed, if we want to solve the matrix equation \(A X = B\) for the unknown \(X\text{,}\) we imitate the procedure used in elementary algebra for solving the equation \(a x = b\text{.}\) One assumption we need is that \(A\) is a square matrix that has an inverse. Notice how exactly the same properties are used in the following detailed solutions of both equations.
The solution of \(x a = b\) is \(x = b a^{-1} = a^{-1}b\text{.}\) In fact, we usually write the solution of both equations as \(x =\frac{b}{a}\text{.}\) In matrix algebra, the solution of \(X A = B\) is \(X = B A^{-1}\) , which is not necessarily equal to \(A^{-1} B\text{.}\) So in matrix algebra, since the commutative law (under multiplication) is not true, we have to be more careful in the methods we use to solve equations.
It is clear from the above that if we wrote the solution of \(A X = B\) as \(X=\frac{B}{A}\text{,}\) we would not know how to interpret \(\frac{B}{A}\text{.}\) Does it mean \(A^{-1} B\) or \(B A^{-1}\text{?}\) Because of this, \(A^{-1}\) is never written as \(\frac{I}{A}\text{.}\)
There exist matrices \(A\) and \(B\) such that \(AB = \pmb{0}\text{,}\) yet \(A\neq \pmb{0}\)and \(B\neq \pmb{0}\text{.}\) In elementary algebra, the only way \(ab = 0\) is if either \(a\) or \(b\) is zero. There are no exceptions.
There exist matrices \(A^2=A\text{.}\) where \(A\neq \pmb{0}\) and \(A\neq I\text{.}\) In elementary algebra, \(a^2=a\Leftrightarrow a=0 \textrm{ or } 1\text{.}\)
There exist matrices \(A\) where \(A^2=I\) but \(A\neq I\) and \(A\neq -I\text{.}\) In elementary algebra, \(a^2=1\Leftrightarrow a=1\textrm{ or }-1\text{.}\)
Let \(M_{n\times n}(\mathbb{R})\) be the set of real \(n\times n\) matrices. Let \(P \subseteq M_{n\times n}(\mathbb{R})\) be the subset of matrices defined by \(A \in P\) if and only if \(A^2 = A\text{.}\) Let \(Q \subseteq P\) be defined by \(A\in Q\) if and only if \(\det A \neq 0\text{.}\)
Consider the special case \(n = 2\) and prove that a sufficient condition for \(A \in P \subseteq M_{2\times 2}(\mathbb{R})\) is that \(A\) has a zero determinant (i.e., \(A\) is singular) and \(tr(A) = 1\) where \(tr(A) = a_{11}+ a _{22}\) is the sum of the main diagonal elements of \(A\text{.}\)
The matrix of coefficients for this system has a zero determinant; therefore, it has no inverse. The system cannot be solved by this method. In fact, the system has no solution.
