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" If "D=diag[d(1),d(2),d(3)],d(1),d(2),d...

" If "D=diag[d_(1),d_(2),d_(3)],d_(1),d_(2),d_(3)!=0," show that "D^(-1)=diag[d_(1)^(-1),d_(2)^(-1),d_(3)^(-1)]

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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)] .

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 D = diag [d_1,d_2,d_3] where none of d_1,d_2,d_3 is 0. prove that: D^-1=diag [d_1^-1,d_2^-1,d_3^-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.

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]

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