" 39.If "A=diag[a,b,c]," show that "A^(n)=diag[a^(n),b^(n),c^(n)]" for all "n in N

If A=diag[a,b,c] then show that A^(n)=diag[a^(n),b^(n),c^(n)] for all n in N

If A=diag(abc), show that A^(n)=diag(a^(n)b^(n)c^(n)) for all positive integer n.

If A=diag(abc), show that A^(n)=diag(a^(n)b^(n)c^(n)) for all positive integer n .

If A = diag [{:(a,b,c):}] , then show that A^(n)="diag"[{:(a^(n),b^(n),c^(n)):}],AAn in N .

If A=d i ag\ (a\ \ b\ \ c) , show that A^n=d i ag\ (a^n\ \ b^n\ \ c^n) for all positive integer n .

If diagA=[a_(1),a_(2),a_(3)] , then for any integer n ge 1 show that A^(n) = diag[a_(1)^(n), a_(2)^(n), a_(3)^(n)]

If a,b,c,d are in G.P.prove that (a^(n)+b^(n)),(b^(n)+c^(n)),(c^(n)+a^(n)) are in G.P.

If a,b,c,d are in G.P.prove that (a^(n)+b^(n)),(b^(n)+c^(n)),(c^(n)+d^(n)) are in G.P.

Statement 1: if D=diag[d_(1),d_(2),,d_(n)], then D^(-1)=diag[d_(1)^(-1),d_(2)^(-1),...,d_(n)^(-1)] Statement 2: if D=diag[d_(1),d_(2),,d_(n)], then D^(n)=diag[d_(1)^(n),d_(2)^(n),...,d_(n)^(n)]

Let R be a relation on N xx N defined by (a,b) R(c, d) hArr a + d= b + c for all (a,b) (c,d) in N xx N show that, (a, b) R (a,b) for all (a, b) in N xx N