Question-8

\(AB=BA\)

In general, matrix multiplication is not commutative. That is, if \(A\) and \(B\) are two matrices, then \(AB\neq BA\). However, not all pairs of matrices are like this.

  • A simple but trivial example is the case of \(A=B\).

  • Another example is the case of \(B=I\), since \(AI=IA=A\).

  • Two diagonal matrices always commute. This is because the product of two diagonal matrices is another diagonal matrix whose entries are the product of the corresponding diagonal entries:

\[ \begin{aligned} D_{1} & =\begin{bmatrix}2 & 0\\ 0 & 3 \end{bmatrix}\\ D_{2} & =\begin{bmatrix}3 & 0\\ 0 & 5 \end{bmatrix}\\ D_{1}D_{2} & =D_{2}D_{1}=\begin{bmatrix}6 & 0\\ 0 & 15 \end{bmatrix} \end{aligned} \]

\(A^{2}=A\)

In the case of \(A^{2}=A\), we can see that \(A=I\) would satisfy \(A^{2}=A\). Interestingly, \(A=-I\) would not satisfy \(A^{2}=A\). Compare this to the single variable equation \(a^{2}=a\), in which \(a=-1\) is also a solution. To construct a non-trivial example, we can take any \(2\times2\) matrix \(A=\begin{bmatrix}a & b\\ c & d \end{bmatrix}\)

\[ \begin{aligned} \begin{bmatrix}a & b\\ c & d \end{bmatrix}\begin{bmatrix}a & b\\ c & d \end{bmatrix} & =\begin{bmatrix}a^{2}+bc & ab+bd\\ ac+cd & bc+d^{2} \end{bmatrix}\\ & =\begin{bmatrix}a & b\\ c & d \end{bmatrix} \end{aligned} \]

We now have:

\[ \begin{aligned} a^{2}+bc & =a\\ b(a+d) & =b\\ c(a+d) & =c\\ d^{2}+bc & =d \end{aligned} \]

One solution can be \(a=1,b=2,c=0,d=0\) which gives us the matrix \(\begin{bmatrix}1 & 2\\ 0 & 0 \end{bmatrix}\). You can verify that \(A^{2}=A\) for this matrix.

\(A^{2}+A+I=0\)

First consider the corresponding equation in one variable:

\[ a^{2}+a+1=0 \]

This equation does not have any real solutions. Can we expect something similar for the matrix equation? Taking a general matrix, we get:

\[ \begin{aligned} \begin{bmatrix}a^{2}+bc & ab+bd\\ ac+cd & bc+d^{2} \end{bmatrix}+\begin{bmatrix}a & b\\ c & d \end{bmatrix}+\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix} & =\begin{bmatrix}a^{2}+bc+a+1 & ab+bd+b\\ ac+cd+c & bc+d^{2}+d+1 \end{bmatrix} \end{aligned} \]

Setting this to the zero matrix:

\[ \begin{aligned} a^{2}+a+1+bc & =0\\ b(a+d+1) & =0\\ c(a+d+1) & =0\\ d^{2}+d+1+bc & =0 \end{aligned} \]

If \(a+d+1\neq0\), then \(b=c=0\). This would imply that \(a^{2}+a+1=d^{2}+d+1=0\), which is impossible for real numbers \(a,d\). Therefore, \(a+d+1=0\). Equating the first and last equations, we get:

\[ a^{2}+a+1=d^{2}+d+1\implies(a-d)(a+d+1)=0. \]

We see that \(a+d+1=0\) will take care of the second and third equations. We now take up the first equation:

\[ \begin{aligned} a^{2}+a+1+bc & =0\\ \left(a+\frac{1}{2}\right)^{2}+\frac{3}{4}+bc & =0\\ \left(a+\frac{1}{2}\right)^{2} & =-\left(\frac{3}{4}+bc\right) \end{aligned} \]

We can use this to get one set of values:

\[ a=-\frac{1}{2},b=3,c=-\frac{1}{4},d=-\frac{1}{2} \]

The resulting matrix is:

\[ \begin{bmatrix}-\frac{1}{2} & 3\\ \\\frac{-1}{4} & -\frac{1}{2} \end{bmatrix} \]

We can characterize all possible solutions here. This is left as an exercise to the reader.