Janelle S. answered • 09/24/20

Penn State Grad for ME, Math & Test Prep Tutoring (10+ yrs experience)

**1) ABA**^{-1}** = B => FALSE**

Multiply both sides by A: ABA^{-1}A = 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}** = A**^{9}** A**^{-9}** => FALSE**

Using binomial expansion:

(A + A^{-1})^{9} = A^{9} + 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^{m}A^{n} = A^{m+n}:

= A^{9} + 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}

= A^{9} + 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}** - A**^{2}** => TRUE**

(I_{n} - A) (I_{n} + A) = I_{n}^{2} + I_{n}A - AI_{n} - A^{2} = I_{n} + A - A - A^{2} = I_{n} - A^{2}

Since I_{n} is the identity matrix, I_{n}^{2} = I_{n}, AI_{n} = A, and I_{n}A = A.

**5) (A + B)**^{2}** = A**^{2}** + B**^{2}** + 2AB => FALSE**

(A + B)^{2} = (A + B) (A + B) = A^{2} + AB + BA + B^{2}

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) A**^{5}** is invertible => TRUE**

A^{5} = A^{4} * A = A^{3} * A * A = A^{2} * A * A *A = A * A * A * A * A

The product of invertible matrices is always invertible.

Greg K.

I meant #3****, sorry.12/15/21

Greg K.

#4 is not true, A+B is not always invertible. Take for example the 2x2 invertible matrices (1, 0)(0, 4) and (1,0)(0, -4) (format is (row1)(row2)). The addition of these matrices makes (2,0)(0,0), a matrix that is not invertible.12/15/21