Let A= [ array lll a & x & p y & q & b r & c & z array ] and B= [ array lll 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 array ] where a, b, c, x, y, z, p, q, r are natural numbers. If tr . (A B+A B^3+A B^5+ +A B¹⁹ )=210 , then find number of ordered triplets (p, q, r) . [Note: tr .(P) denotes the trace of matrix P .

Step-by-step solution

aligned & B^2=I & A B= [ array lll a & x & p y & q & b r & c & z array ] [ array lll 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 array ]= [ array lll p & x & a b & q & y z & c & r array ] & A B=A B^3= =A B¹⁹= [ array lll p & x & a b & q & y z & c & r array ] aligned tr (A B+A B^3+ .+A B¹⁹ )=210 gathered 10(p+q+r)=210 p+q+r=21, p, q, r N p^ +q^ +r^ =18, p^ , q^ , r^ W gathered Number of ordered triplets (p, q, r)= ²⁰ C₂= 20 19 2 =190