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]`

If D=diag [2, 3, 4] , then D^(-1)=

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.

Find the inverse of each of the matrices given below : Let D= "diag" [d_(1),d_(2),d_(3)] where none of d_(1),d_(2),d_(3) is ), prove that D^(-1)="diag" [d_(1)^(-1),d_(2)^(-1),d_(3)^(-1)] .

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

If (n+1)!=90[(n-1)!], fin d n

Statement-1: The sum of n terms of the series a+(a+d)+(a+2d)+...(a+(n-1)d)=(n)/(2)[2n+(n-1)d] Statement-2:- Mathematical induction is valid only for natural numbers.

If D_(r) = |(r,1,(n(n +1))/(2)),(2r -1,4,n^(2)),(2^(r -1),5,2^(n) -1)| , then the value of sum_(r=1)^(n) D_(r) , is