Question

Find the form of all the matrices which commute with the matrix A = $=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$

Answer 1

If B be the required matrix then both Ab , and Ba should be defined and also AB = BA
$A_{2\times 2}\therefore Bshouldbe2\times p$
$\therefore AB=2\times p$v
$B_{2\times p}{A}_{2\times 2}$ will be defined if p = 2
Hence ,B must be $2\times 2$ matrix and let it be$=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$
$AB=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$$=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$$=\left[\begin{array}{cc}p=3r& q+3s\\ 2p+4r& 2q+4s\end{array}\right]$
$=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$$=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$$=\left[\begin{array}{cc}p+2q& 3p+4q\\ r+2s& 3r+4s\end{array}\right]$
By definition of equality of matrix we have
p + 3r = p + 2q $\therefore r=\frac{2q}{3}$
q + 3s = 3p + 4q $\therefore$ s = p + q
2p + 4r = r + 2s = r + 2 (p + q) $\therefore r=\frac{2q}{3}$
2p + 4s = 3r + 4s $\therefore r=\frac{2q}{3}$
Let us put $\frac{q}{3}=\lambda$
$\therefore q=3\lambda .r=2\lambda ,s=p+3\lambda$
$\therefore$ The family of matrix which commute with the given matrix are to the types
$=\left[\begin{array}{cc}p& 3\lambda \\ 2\lambda & p+3\lambda \end{array}\right]$
Where p and $\lambda$ are arbitrary constants