`abs((2,0,1),(1,0,0),(3,2,0)).B=[B_I,B_(II),B_(III)]` are column vector if `AB_1=abs((1),(0),(0))AB_(II)=abs((2),(0),(1))AB_(III)=abs((3),(2),(1))` if `alpha`= sum of the diagonal elements of B.
`beta`= det (B) in find `abs(a^3+beta^3)`

Let A=[[2,0,1],[1,1,0],[1,0,1]],B=[B_(1),B_(2),B_(3)] ,where B_(1),B_(2),B_(3) are column matrics,and [AB_(1)=[[1],[0],[0]],AB_(2)=[[2],[3],[0]],AB_(3)=[[3],[2],[1]]] If alpha=|B| and beta is the sum of all the diagonal elements of "B" ,then alpha^(3)+beta^(3) is equal to

A=[(-1,3,0),(-7,2,8)],B=[(-5,0),(0,3),(1,-8)] . then AB

If A=({:(1,2),(2,3):})and B=({:(1,0),(1,0):}) then what is determinant of AB ?

If A=[(1,0),(0,3)] and B=[(2,0),(0,1)] then show AB =BA =[(2,0),(0,3)]

If A= [(3,1),(4,0)] and B=[(4,0),(2,5)] then verify that (AB)^(-1)= B^(-1) A^(-1)

If 9a^(2)-6ab+b^(2)=0 then find a:b

If A = [(-1),(2),(3)] and B = [(2,0,-4)] , then find AB .