More Problems on Linear Systems and Matrices

Set

$$B = \left(\begin{array}{rrrr} a & b \\ c & d\\ \end{array} \right) \;.$$

Then we have

$$A\; B = \left(\begin{array}{rrrr} a-c & b-d \\ 2a+c &2b+d\\ \end{array} \right)$$

and

$$B\; A = \left(\begin{array}{rrrr} a+2b & -a+b \\ c+2d &-c+d\\ \end{array} \right) \;.$$

So the equation $A\;B = B\; A$ implies

$$\left\{\begin{array}{lcl} a-c &=& a+2b \\ b-d &=& -a+b \\ 2a+c &=& c+2d \\ 2b+d &=& -c+d \\ \end{array} \right.$$

This clearly implies

$$c = -2b \;\;\mbox{and}\;\; a=d$$

So

$$B = \left(\begin{array}{rrrr} a & b \\ -2b & a\\ \end{array} \right)$$

where $a$ and $b$ are two arbitrary numbers. Consider the matrix

$$C = \left(\begin{array}{rrrr} 1 & 1 \\ 1 & 1\\ \end{array} \right)\;.$$

Since $C$ is not of the form found before, then we must have

$$A\;C \neq C\; A\;.$$

The reader may want to check that is the case.

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