Solving Identity Matrices Homework Problems

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Homework Statement


0vzUmzv.jpg


Homework Equations


ImA = AIn = A.
(A−1)−1 = A
(AB)−1 = B−1A−1

The Attempt at a Solution


Determinates:
Det(A) = 3 – 0 = 3
Det (2A+BT) = 4 – 8 = -2

Matrices
B^T = 2 -2
0 -5

(2A + B^T)^-1 = -8 -4
-8 -2

So I've kinda figured out those sides, but I'm having real issues finding the identity matrix for this question. I'm not really sure exactly how to find it. Then for the second question, I just have no idea what it's asking me to do. Thanks for any help.
 
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Hey Siann122 and welcome to the forums.

The identity matrix for most purposes is just an nxn matrix with 1's in the diagonal position and 0's everywhere else.

The second question is asking to prove a statement that LHS = RHS. One hint for this would be to use that fact that A * A^(-1) = I and use other properties of matrices to get your answer.
 
chiro said:
Hey Siann122 and welcome to the forums.

The identity matrix for most purposes is just an nxn matrix with 1's in the diagonal position and 0's everywhere else.

So the 2I would be a matrix of: 2 0
2 2

Because it would have 1 on the diagonals and 0 in the corner as it's a 2*2 matrix?
 
No.
##
\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I##
So ##2I = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right]##Really, I should be saying ##I_n## (##I_2## in this case), but generally the dimension of the identity matrix is suppressed and you make it as big as whatever it needs to be; it's usually obvious.
 
Alright so using the identity matrix I've come up with this result:

(2A + B^T)^-1 = -

2 -1
-2 -1/2

(A-1 * B^T + 2I) =

(2 2/3) (-4)
(1 1/3) (-4 1/3)Det(A^-1 * B^T + 2I) = - 44/9 – ( - 16/3) = 4/9

(A^-1 * B^T + 2I)^-1 =

6 -9
3 -9 3/4But then (A^-1 * B^T + 2I)^-1 =/= (2A + B^T)^-1 , so what am I doing wrong?

PS: How did you do the matrix formatting :P
 
Siann122 said:
Alright so using the identity matrix I've come up with this result:

(2A + B^T)^-1 = -

2 -1
-2 -1/2

(A-1 * B^T + 2I) =

(2 2/3) (-4)
(1 1/3) (-4 1/3)Det(A^-1 * B^T + 2I) = - 44/9 – ( - 16/3) = 4/9

(A^-1 * B^T + 2I)^-1 =

6 -9
3 -9 3/4But then (A^-1 * B^T + 2I)^-1 =/= (2A + B^T)^-1 , so what am I doing wrong?

PS: How did you do the matrix formatting :P

I wouldn't worry too much about the matrix formatting at this point. I write a matrix like A as [[1,2],[2,3]]. It's quick and it gets the point across. You can worry about the formatting later. Can you show more of what you are doing? Before you say (2A+B^T)^(-1)=[[2,-1],[-2,-1/2]], can you say what you got for (2A+B^T)?
 
Dick said:
I wouldn't worry too much about the matrix formatting at this point. I write a matrix like A as [[1,2],[2,3]]. It's quick and it gets the point across. You can worry about the formatting later. Can you show more of what you are doing? Before you say (2A+B^T)^(-1)=[[2,-1],[-2,-1/2]], can you say what you got for (2A+B^T)?

Here's my full working out.

A =
1 2
2 3
Det(A) = 3 – 0 = 3

A-1 =
1/3 2/3
2/3 1

2A+B^T =
4 2
4 1
 
Siann122 said:
Here's my full working out.

A =
1 2
2 3
Det(A) = 3 – 0 = 3

A-1 =
1/3 2/3
2/3 1

2A+B^T =
4 2
4 1

A^(-1) is wrong and Det(A) is wrong. 2A+B^T looks ok. Can you maybe show a little more working? Why do you think Det(A)=3?
 
Dick said:
A^(-1) is wrong and Det(A) is wrong. 2A+B^T looks ok. Can you maybe show a little more working? Why do you think Det(A)=3?

Sorry, Det(A) should be -1 as (3x1)-(2x2), which would make A^-1 wrong.

So A^-1 =
-1 -2
-2 -3

Which means A^-1 * B^T =
(-1)(2)+(-2)(0) (-1)(-2)+(-2)(-5)
(-2)(2)+(-3)(0) (-2)(-2)+(-3)(-5)
= -2 12
-4 19

That means (A-1 * BT + 2I) =
(0) (12)
(-4) (21)

Det(A-1 * BT + 2I) = - 0 – -48 = 48

(A-1 * BT + 2I)-1 =
0 1/4
-1/12 7/16

This is still wrong though...
 
  • #10
Siann122 said:
Sorry, Det(A) should be -1 as (3x1)-(2x2), which would make A^-1 wrong.

So A^-1 =
-1 -2
-2 -3

Which means A^-1 * B^T =
(-1)(2)+(-2)(0) (-1)(-2)+(-2)(-5)
(-2)(2)+(-3)(0) (-2)(-2)+(-3)(-5)
= -2 12
-4 19

That means (A-1 * BT + 2I) =
(0) (12)
(-4) (21)

Det(A-1 * BT + 2I) = - 0 – -48 = 48

(A-1 * BT + 2I)-1 =
0 1/4
-1/12 7/16

This is still wrong though...

Now Det(A) is good. But A^(-1) is still wrong. You can check this by multiplying A*A^(-1) and you won't get I. You don't just divide A or A^T by Det(A). Can you show your working there?
 
  • #11
Dick said:
Now Det(A) is good. But A^(-1) is still wrong. You can check this by multiplying A*A^(-1) and you won't get I. Can you show your working there?

Yep so 1/-1 = -1, and then A^-1 =

(-1)(1) (-1)(2)
(-1)(2) (-1)(3)

As it's only a scalar amount that is being multiplied in.
 
  • #12
Siann122 said:
Yep so 1/-1 = -1, and then A^-1 =

(-1)(1) (-1)(2)
(-1)(2) (-1)(3)

As it's only a scalar amount that is being multiplied in.

Ok, we are getting down to where your problem lies. To get the inverse you divide the transpose of the cofactor matrix of A by Det(A). The cofactor matrix of [[a,b],[c,d]] is [[d,-c],[-b,a]]. Transpose that and divide by Det(A). And go back and review how to find inverses! And as I said before you can easily check whether you have the correct inverse by multiplying A by A^(-1). Do you get I?
 
  • #13
Dick said:
Ok, we are getting down to where your problem lies. To get the inverse you divide the transpose of the cofactor matrix of A by Det(A). The cofactor matrix of [[a,b],[c,d]] is [[d,-c],[-b,a]]. Transpose that and divide by Det(A). And go back and review how to find inverses! And as I said before you can easily check whether you have the correct inverse by multiplying A by A^(-1). Do you get I?

Ah okay, so
A-1 =
-3 2
2 -1 ??
 
  • #14
Dick said:
Did you check it like I advised? There's still a problem with the off diagonal elements. Can you fix it?

I edited it while you were posting so you must have missed it, to reiterate it should be:

A-1 =
-3 2
2 -1 ??
 
  • #15
Siann122 said:
Ah okay, so
A-1 =
-3 2
2 -1 ??

That's more like it. You can check it yourself by doing the multiplication. I recommend that. So continue from there.
 
  • #16
Dick said:
That's more like it. You can check it yourself by doing the multiplication. I recommend that. So continue from there.

A-1 * BT =
(-3)(2)+(2)(0) (-3)(-2)+(2)(-5)
(2)(2)+(-1)(0) (2)(-2)+(-1)(-5)
= -6+0 6+(-10)
(-4)+0 (-4)+(5)
=-6 -4
4 1

(A-1 * BT + 2I) =
(-4) (-4)
(4) (3)

Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 -1
4 -1Is this correct?
 
  • #17
Siann122 said:
A-1 * BT =
(-3)(2)+(2)(0) (-3)(-2)+(2)(-5)
(2)(2)+(-1)(0) (2)(-2)+(-1)(-5)
= -6+0 6+(-10)
(-4)+0 (-4)+(5)
=-6 -4
4 1

(A-1 * BT + 2I) =
(-4) (-4)
(4) (3)

Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 -1
4 -1Is this correct?

Almost. If (A-1 * BT + 2I)=[[-4,-4],[4,3]] (which I agree with) then did you really get the right inverse finally? Check it. I don't think you did. Try checking yourself more. I'm signing off soon so I can't keep this up much longer. But I think you are getting the hang of it.
 
  • #18
Dick said:
Almost. If (A-1 * BT + 2I)=[[-4,-4],[4,3]] (which I agree with) then did you really get the right inverse finally? Check it. I don't think you did. Try checking yourself more. I'm signing off soon so I can't keep this up much longer.

Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 1
-4 -1

It's this last bit, the -1 should have been 1, and the 4 should have been -4 (I think).
 
  • #19
Siann122 said:
Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 1
-4 -1

It's this last bit, the -1 should have been 1, and the 4 should have been -4 (I think).

(I think)!? You should know. You have to learn how to correct yourself at some point. This would be a good time to start. I'll give the hint that it's not correct. And I think you are perfectly capable of doing it correctly if you do it more carefully.
 
  • #20
Dick said:
(I think)!? You should know. You have to learn how to correct yourself at some point. This would be a good time to start. I'll give the hint that it's not correct. And I think you are perfectly capable of doing it correctly if you do it more carefully.

Haha, basic arithmetic fail, totally overlooked that 1/4 * 4 = 1, so it should be:

(A-1 * BT + 2I)-1 =
3/4 1
-1 -1
 
  • #21
Siann122 said:
Haha, basic arithmetic fail, totally overlooked that 1/4 * 4 = 1, so it should be:

(A-1 * BT + 2I)-1 =
3/4 1
-1 -1

Indeed it is. Did you check it like I suggested?
 
  • #22
Dick said:
Indeed it is. Did you check it like I suggested?

Yep, here's my total working out (Which appears to be sound).

A-1 =
-3 2
2 - 1

2I =
2 0
0 2

A-1 * BT =
(-3)(2)+(2)(0) (-3)(-2)+(2)(-5)
(2)(2)+(-1)(0) (2)(-2)+(-1)(-5)
= -6+0 6+(-10)
(-4)+0 (-4)+(5)
=-6 -4
4 1

(A-1 * BT + 2I) =
(-4) (-4)
(4) (3)

Det(A-1 * BT + 2I) = -12 - -16 = 4

(A-1 * BT + 2I)-1 =
3/4 1
-1 -1

After all that though, it still doesn't match this result:

2A+BT =
4 2
4 1

Det (2A+BT) = 4 – 8 = -2

(2A + BT)-1 =
-1/2 1
2 -2

(2A + BT)^-1 A =
(-1/2)(1)+(1)(2) (-1/2)(2)+(1)(3)
(2)(1)+(-2)(2) (2)(2)+(-2)(3)
=(1.5) (2)
(-2) (-2)

This was the original aim of the question.
 
  • #23
I've had a few more goes at trying to fix it but it just seems to be right, yet I don't get a "correct answer" here. What am I doing wrong?
 
  • #24
Siann122 said:
Yep, here's my total working out (Which appears to be sound).

A-1 =
-3 2
2 - 1

2I =
2 0
0 2

A-1 * BT =
(-3)(2)+(2)(0) (-3)(-2)+(2)(-5)
(2)(2)+(-1)(0) (2)(-2)+(-1)(-5)
= -6+0 6+(-10)
(-4)+0 (-4)+(5)
=-6 -4
4 1

(A-1 * BT + 2I) =
(-4) (-4)
(4) (3)

Det(A-1 * BT + 2I) = -12 - -16 = 4

(A-1 * BT + 2I)-1 =
3/4 1
-1 -1

After all that though, it still doesn't match this result:

2A+BT =
4 2
4 1

Det (2A+BT) = 4 – 8 = -2

(2A + BT)-1 =
-1/2 1
2 -2

(2A + BT)^-1 A =
(-1/2)(1)+(1)(2) (-1/2)(2)+(1)(3)
(2)(1)+(-2)(2) (2)(2)+(-2)(3)
=(1.5) (2)
(-2) (-2)

This was the original aim of the question.

Really? I can't say much for your fixing ability if you still think 4-8=(-2).
 
  • #25
Dick said:
Really? I can't say much for your fixing ability if you still think 4-8=(-2).

Yeah I picked this one up. Updated:

2A+BT =
4 2
4 1

Det (2A+BT) = 4 – 8 = -4

(2A + BT)-1 =
5/4 2
-1 -1

(2A + BT)-1A =
(-1/2)(1)+(1)(2) (-1/2)(2)+(1)(3)
(2)(1)+(-2)(2) (2)(2)+(-2)(3)
=(21/4) (17/2)
(-3) (-5)Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 1
-1 -1
 
  • #26
Siann122 said:
Yeah I picked this one up. Updated:

2A+BT =
4 2
4 1

Det (2A+BT) = 4 – 8 = -4

(2A + BT)-1 =
5/4 2
-1 -1

(2A + BT)-1A =
(-1/2)(1)+(1)(2) (-1/2)(2)+(1)(3)
(2)(1)+(-2)(2) (2)(2)+(-2)(3)
=(21/4) (17/2)
(-3) (-5)


Det(A-1 * BT + 2I) = -12 - -16 = 4
(A-1 * BT + 2I)-1 =
3/4 1
-1 -1

Check (2A+B^T)^(-1). By which I mean check it. Not just repeat the mistake you made again.
 
  • #27
Dick said:
Check (2A+B^T)^(-1). By which I mean check it. Not just repeat the mistake you made again.

Thanks for sticking with me, I screwed up a few times with some silly errors :P.

(2A + BT)-1 =
1/4 ½
1 -1
(2A + BT)-1A =
¾ 1
-1 -1


Which is correct! Thanks!
 
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