Select all statements below which are true for all invertible 𝑛×𝑛 matrices 𝐴 and 𝐵

Question

1. ABA^-1=B
2. (A+A^-1)⁹=A⁹+A^-9
3. A+B is invertible
4. (I_n-A)(I_n+A)=I_n-A²
5. (A+B)²=A²+B²+2AB
6. A⁵ is invertible

Answer 1

1) ABA^-1 = B => FALSE

Multiply both sides by A: ABA^-1A = BA => AB = BA

Matrix multiplication is not commutative (you cannot switch the order of the factors), so AB does not equal BA.

2) (A + A^-1)⁹ = A⁹ A^-9 => FALSE

Using binomial expansion:

(A + A^-1)⁹ = A⁹ + 9(A)⁸(A^-1)¹ + 36(A)⁷(A^-1)² + 84(A)⁶(A^-1)³ + 126(A)⁵(A^-1)⁴ + 126(A)⁴(A^-1)⁵ + 84(A)³(A^-1)⁶ + 36(A)²(A^-1)⁷ + 9(A)¹(A^-1)⁸ + (A^-1)⁹

To simplify, use the rules (A^m)ⁿ = A^mn and A^mAⁿ = A^m+n:

= A⁹ + 9(A)^8-1 + 36(A)^7-2 + 84(A)^6-3 + 126(A)^5-4 + 126(A)^4-5 + 84(A)^3-6 + 36(A)^2-7 + 9(A)^1-8 + A^-9

= A⁹ + 9(A)⁷ + 36(A)⁵ + 84(A)³ + 126A + 126(A)^-1 + 84(A)^-3 + 36(A)^-5 + 9(A)^-7 + A^-9

3) A + B is invertible => TRUE

The sum of two invertible matrices is always invertible.

4) (I_n - A) (I_n + A) = I_n - A² => TRUE

(I_n - A) (I_n + A) = I_n² + I_nA - AI_n - A² = I_n + A - A - A² = I_n - A²

Since I_n is the identity matrix, I_n² = I_n, AI_n = A, and I_nA = A.

5) (A + B)² = A² + B² + 2AB => FALSE

(A + B)² = (A + B) (A + B) = A² + AB + BA + B²

Matrix multiplication is not commutative (you cannot switch the order of the factors), so AB does not equal BA and AB + BA does not equal 2AB.

6) A⁵ is invertible => TRUE

A⁵ = A⁴ * A = A³ * A * A = A² * A * A *A = A * A * A * A * A

The product of invertible matrices is always invertible.