If`D=d i ag[d_1, d_2, d_n]` , then prove that `f(D)=d i ag[f(d_1),f(d_2), ,f(d_n)],w h e r ef(x)` is a polynomial with scalar coefficient.

If D=diag[d_(1),d_(2),...d_(n)], then prove that f(D)=diag[f(d_(1)),f(d_(2)),...,f(d_(n))], where f(x) is a polynomial with scalar coefficient.

If D=diag(d_1,d_2,d_3,…,d_n)" where "d ne 0" for all " I = 1,2,…,n," then " D^(-1) is equal to

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 such that R={(a d) (c g) (d e) (d f) (g f)} then (ROR^(1))^(-1) is

Find sum f_(i)d_(i) , if bar(d) = 1.2 and sum f_(i) = 25 .

Let f(x)=(ax+b)/(cx+d) , x!=-d/c . If d=-a, show that f{(x)}=x is an identity.

Instead of the usual definition of derivative Df(x), if we define a new kind of derivative D^*F(x) by the formula D*f(x)=lim_(h->0)(f^2(x+h)-f^2(x))/h ,w h e r ef^2(x) mean [f(x)]^2 and if f(x)=xlogx ,then D^*f(x)|_(x=e) has the value (A)e (B) 2e (c) 4e (d) none of these

If (d(f(x))/(dx)=(1)/(1+x^(2)) then (d)/(dx){f(x^(3))} is