Select all statements below which are true for all invertible n×nn×n matrices aa and bba. (ab)−1=a−1b−1(ab)−1=a−1b−1b. aba−1=baba−1=bc. (in+a)(in+a−1)=2in+a+a−1(in+a)(in+a−1)=2in+a+a−1d. a2b7a2b7 is invertiblee. a+ba+b is invertible f. (a+b)(a−b)=a2−b2(a+b)(a−b)=a2−b2

Answer:The statements that are correct are:c) [tex](in+a)(in+a^{-1})=2in+a+a^{-1}[/tex]d) [tex]a^2b^7[/tex] is invertible.e) [tex]a+b[/tex] is invertible.Step-by-step explanation:We are given that:a and b are invertible n×n matrices.We have to tell which of the following statements are true.a)[tex](ab)^{-1}=a^{-1}b^{-1}[/tex]This statement is false.Since:[tex](ab)^{-1}=b^{-1}a^{-1}[/tex] and it may not be equal to the term [tex]a^{-1}b^{-1}[/tex]b)[tex]aba{-1}=b[/tex]This expression could also be written as:[tex]ab=ba[/tex]Since on Post multiplying by a on both the sides.But here we don't know whether the matrices are commutative or not.Hence, the statement is false.c)[tex](in+a)(in+a^{-1})=2in+a+a^{-1}[/tex]This statement is true.since,[tex](in+a)(in+a^{-1})=in(in+a^{-1})+a(in+a^{-1})\\\\=in^2+in.a^{-1}+a.in+aa^{-1}\\\\=in+a^{-1}+a+in\\\\=in+a^{-1}+a[/tex]where in denote the identity matrix.and we know that:[tex]in^2=in[/tex]d)[tex]a^2b^7[/tex] is invertible.This statement is true.Since we know that prodct of invertible matrices is also invertible.As [tex]a[/tex] is invertible so is [tex]a^2[/tex].Also [tex]b[/tex] is invertible so is [tex]b^7[/tex].Hence Product of  [tex]a^2[/tex] and  [tex]b^7[/tex] is also invertible.i.e. [tex]a^2b^7[/tex] is invertible.e)[tex]a+b[/tex] is invertible.This statement is true as sum of two invertible matrices is invertible.f)[tex](a+b).(a-b)=a^2-b^2[/tex]This statement is false.Since,[tex](a+b).(a-b)=a(a-b)+b(a-b)\\\\=a.a-a.b+b.a-b.b\\\\=a^2-ab+ba-b^2[/tex]Now as we are not given that:[tex]ab=ba[/tex]Hence, we could not say that:[tex](a+b).(a-b)=a^2-b^2[/tex]