Question

Select all statements below which are true for all invertible $n imes n$ matrices $A$ and $B$ A. $(I_n + A)(I_n + A^{-1}) = 2I_n + A + A^{-1}$ B. $A + I_n$ is invertible C. $A^7B^4$ is invertible D. $ABA^{-1} = B$ E. $(A + B)^2 = A^2 + B^2 + 2AB$ F. $(AB)^{-1} = A^{-1}B^{-1}$

          Select all statements below which are true for all invertible $n 	imes n$ matrices $A$ and $B$
A. $(I_n + A)(I_n + A^{-1}) = 2I_n + A + A^{-1}$
B. $A + I_n$ is invertible
C. $A^7B^4$ is invertible
D. $ABA^{-1} = B$
E. $(A + B)^2 = A^2 + B^2 + 2AB$
F. $(AB)^{-1} = A^{-1}B^{-1}$

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Select all statements below which are true for all invertible n imes n matrices A and B
A. (In + A)(In + A^-1) = 2In + A + A^-1
B. A + In is invertible
C. A^7B^4 is invertible
D. ABA^-1 = B
E. (A + B)^2 = A^2 + B^2 + 2AB
F. (AB)^-1 = A^-1B^-1

Added by Christopher T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Joseph David

10/27/2022

Step 1

**Step 3:** A^7B^4 is invertible because both A and B are invertible. **Step 4:** AB(A^-1) = B, which means ABA^-1 = B. **Step 5:** (A + B)^2 = A^2 + B^2 + 2AB is not true because matrices do not commute. **Step 6:** (AB)^-1 = A^-1B^-1 is not true. The Show more…

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Select all statements below which are true for all invertible n x n matrices A and B A. (In + A)(In + A-1) = 2In + A + A-1 B. A + In is invertible C. A7 B4 is invertible D. ABA-1 = B E. (A + B)2 = A2 + B2 + 2AB F. (AB)-1 = A-1B-1

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Transcript

0:00 Hi.

00:01 In this question, we're given that a and b are n by n invertible matrices.

00:09 N by n, and they're invertible.

00:13 And we want to say which of these facts are true.

00:18 So the first one is we have i plus a, or sorry, in, but it doesn't really matter, and then in plus a inverse.

00:31 And so if we want to see if it equals the right -hand side, let's just multiply it out.

00:37 And what do we get? we get i .n times i .n plus i .n times a inverse.

00:45 And be careful that you get the ordering correct, because matrices in general don't commute, even if they're invertible.

00:52 Plus a times i .n.

00:55 Plus a times a inverse.

01:00 Now i n times i n is just i n.

01:04 And then i .n times a inverse is just a inverse.

01:07 A times i .n is just a.

01:10 And then a plus a...

01:12 And then a times a inverse is in.

01:16 So we get 2in plus a inverse plus a.

01:20 So the first one is correct.

01:27 Then b, a plus in is invertible.

01:31 So one way we could do this is to try to find an inverse.

01:37 I'm not sure if we can do that.

01:42 Let's see.

01:45 So maybe we can try a plus i n times, let's see...

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