Question

Can someone check if my answers are correct. True and False questions.

Answer 1

(a) For an invertible matrix M, if M = M^−1, then |M| = 1. False
(b) Let A be any 3 × 3 real matrix such that |A| does not equal 0. Then |2A|=2|A|. False
(c) If A is an n × n matrix and b ∈ R^n such that Ax = b has no solution, then Ax=c has no solution for any other nonzero vector c∈ R^n. True
(d) If A is a square matrix and Ax = 0 has a unique solution, then A is invertible. False
(e) The set of functions {at^4 + bt + c | a, b, c ∈ R} is a vector space. False

Answer 2

so for A you were able to provide a counterexample?

Answer 3

never mind i checked again and a) is true

Answer 4

for part a), you should have

M = M^(-1)

M*M = M^(-1)*M

M^2 = I

then take the determinant of both sides

det(M^2) = det(I)

det(M*M) = 1

det(M)*det(M) = 1

( det(M) )^2 = 1

sqrt[ ( det(M) )^2 ] = sqrt(1)

|det(M)| = 1

Answer 5

there may be another way to do part a)

Answer 6

I like the M^2=I way

Answer 7

But I think that is the only way I know too

Answer 8

(a) is false

Answer 9

so it is false then

Answer 10

let \[M=\left[\begin{array}{cc} 1 & 0 \\[0.em] 0 & -1 \\[0.em] \end{array}\right]\]

Answer 11

\[M=M^{-1}\]

\[|M|=-1\]

Answer 12

the determinant of M is -1, but the absolute value bars force det(M) to be +1

Answer 13

so that's why I put it as |det(M)|

Answer 14

then again, the initial |M| could just mean determinant and not |det(M)|

Answer 15

they asked for |M|

\[|M|=Det(M)\]

Answer 16

hmm good point, so I'm not sure now, but |det(M)| = 1 is definitely true

Answer 17

\[(\det(M))^2=1 \\ means \det(M)=1 \text{ or } \det(M)=-1 \]

Answer 18

so it is true then since it can be both

Answer 19

well that or doesn't mean it is both

Answer 20

it just means if we have M^2=I
we have two possibilities that can happen for the determinant

Answer 21

you know depending on the matrix we choose for M

Answer 22

Like zarkon chosen a matrix such that M^2=I and det(M)=-1

Answer 23

I could find a matrix M such that M^2=I and also det(M)=1
but the point is the det(M) isn't aways 1 for when M^2=I

Answer 24

I think Zarkon is right. He picked a matrix M such that M = M^(-1) is true, but |M| = -1

Answer 25

zarkon is right

Answer 26

you have others that are incorrect in your list

Answer 27

is it d. i think it might be true

Answer 28

yes..that is one of them

Answer 29

is e. another one is it true

Answer 30

you tell me...why did you think it was false...why do you now think it is now true?

Answer 31

the set of all functions, with the standard definitions of "sum of two functions" and "product of a function and a number", is a vector space because for any two functions

Answer 32

@Zarkon

Answer 33

e is true

Answer 34

ok are the rest correct?

Answer 35

no

Answer 36

okay so i am left with b and c. is c false?

Answer 37

c is false...why?

Answer 38

dont know thats just a guess

Answer 39

how many solutions does the following have


\[\left[\begin{array}{cc} 1 & 0 \\[0.em] 0 & 0 \\[0.em] \end{array}\right]x=\left[\begin{array}{c} 0 \\[0.em] 1 \\[0.em] \end{array}\right]\]

Answer 40

many solutions

Answer 41

give me one

Answer 42

im not quite sure what you mean

Answer 43

you said it has many solutions...give be one vector x that satisfies the matrix equation above

Answer 44

well i think it has many solutions since the last row is all 0 but i dont know what you mean still

Answer 45

write it like this


\[\left[\begin{array}{cc} 1 & 0 \\[0.em] 0 & 0 \\[0.em] \end{array}\right] \left[\begin{array}{c} x_1 \\[0.em] x_2 \\[0.em] \end{array}\right] =\left[\begin{array}{c} x_1 \\[0.em] 0\\[0.em] \end{array}\right]\]

Answer 46

is it possible for

\[\left[\begin{array}{c} x_1 \\[0.em] 0\\[0.em] \end{array}\right]=\left[\begin{array}{c} 0\\[0.em] 1\\[0.em] \end{array}\right]\]

Answer 47

are you away from your computer or are you thinking?

Answer 48

im here

Answer 49

just trying to figure out what you're trying to tell me..im confuse

Answer 50

is b. wrong too is it true

Answer 51

you might want to look over your book and notes again to get a better grasp of the fundamentals

Answer 52

why do you think (b) is true?

Answer 53

idk apparently i had almost all of them wrong

Answer 54

yes you did....hence why I think you should take some time to reread the material and try and do as many numerical problems as you can then come back to these