# Determine which of the formulas hold for all invertible nxn matrices A and BA. (A+B)^2 = A^2+B^2+2ABB. ABA^-1=BC. (I_n-A)(I_n+A)=I_n-A^2D. 8A is invertibleE. (AB)^-1=A^-1B^-1F. A+A^-1 is invertible

### 2 Answers | Add Yours

`(A+B)^2=A^2+AB+BA+B^2,` which is usually not equal to `A^2+2AB+B^2.`so **choice A is not true** **for arbitrary invertible matrices**.

B. If `ABA^(-1)=B,` then `AB=BA.` This is usually not true, so **choice B is not true for arbitrary invertible matrices**.

C. `(I_n-A)(I_n+A)=I_n^2+I_nA-AI_n-A^2=I_n-A^2,` so **choice C is true**.

D. If `A` is invertible, then `det A!=0,` and `det 8A=8detA!=0,` so `8A` is invertible. **Choice D is true.**

` ` E. `(AB)^(-1)=B^(-1)A^(-1).` In general, this will not equal `A^(-1)B^(-1),` so **choice E is not true for arbitrary invertible matrices**.

F. It is easily checked that `A=[[0,1],[-1,0]]` is invertible, and `A+A^(-1)=0,` which is not invertible. **Choice F is not true for arbitrary invertible matrices.**

B,C, D, and F

explanation for F

Let `A=[[1,2],[0,1]]`

`A^(-1)=[[1,-2],[0,1]]`

`A+A^(-1)=[[2,0],[0,2]]=C (say)`

`C^(-1)=[[2,0],[0,2]]`

`` similarly otherpart can be explain.