1= let p and q be 3 by 3 matrices p3= q3 and p2q= q2p , p not equal to q find determinant of p2+q2

2- no of 3 by 3 non sing matrices with 4 enteries as 1 and all other enteries is 0 is

P ³ = Q³  (1)
P²Q = Q²P (2)

Subtracting (2) from (1), we have
P³ -P²Q = Q³ -Q²P
P²( P-Q ) = Q² (Q-P)
P²(P-Q) +Q² (P-Q) = 0
(P²+Q²) (P-Q) = 0
As P is not equal to Q .
So (P²+Q²) = 0

2)In this case there is no particular method, you have to check for all the matrices for singularity .
So for the case of 4 ones and 5 zeros , we have 9! /(4!*5!) matrices possible.

Here are some of the matrices , which are non -singular .
$$\begin{gathered} Put the diagonal element as 1. \\ \left|\begin{array}{ccc}1& & \\ & 1& \\ & & 1\end{array}\right| , in this case there are six places left , and we have to fill the by 5 zeros and one 1. \\ Hence the number of arrangement is 6! /(5!) = 6 \\ \left|\begin{array}{ccc}& & 1\\ & 1& \\ 1& & \end{array}\right| , similar to above . it also has 6 cases. \\ Some more cases are \\ \left|\begin{array}{ccc}0& 0& 1\\ 1& 0& 1\\ 0& 1& 0\end{array}\right|, in this matrix , just interchange the rows to get two more matrix .3 cases are there \\ \left|\begin{array}{ccc}1& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right| , in this matrix , just interchange the rows to get two more matrix .3 cases are there \\ \left|\begin{array}{ccc}0& 1& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right| , three cases. \\ \left|\begin{array}{ccc}0& 1& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right| , three cases. \\ \left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right| , three cases . \\ So 12 + 15 = 27 cases \\ The key point here is do not put any row or column as zero . \end{gathered}$$