Find the matrix A such that

A

- [1 0] = [-5 -4]

[-1 4] [7 12]

Hint: Let A =

- [a b]

[c d]

not looking for the answer just an explanation on where to start. Thanks

I am sure that this is an easy one, but I don't know where to start. I'm taking an online course and all the examples in the book are nothing like this one. Here's the question:

Find the matrix A such that

A

Hint: Let A =

not looking for the answer just an explanation on where to start. Thanks

Find the matrix A such that

A

- [1 0] = [-5 -4]

[-1 4] [7 12]

Hint: Let A =

- [a b]

[c d]

not looking for the answer just an explanation on where to start. Thanks

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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littlebu wrote:Find the matrix A such that

A[1 0] = [-5 -4]

[-1 4] [7 12]

Is the "A"off to one side meant to indicate multiplication, so the exercise is as follows?

. . . . .

If so, have you tried using the hint? (If yes, please reply showing your work so far.)

Also, have you yet learned about finding inverse matrices?

Thank you!

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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stapel_eliz wrote:littlebu wrote:Find the matrix A such that

A[1 0] = [-5 -4]

[-1 4] [7 12]

Is the "A"off to one side meant to indicate multiplication, so the exercise is as follows?

. . . . .

If so, have you tried using the hint? (If yes, please reply showing your work so far.)

Also, have you yet learned about finding inverse matrices?

Thank you!

based on the hint I would bet not.

the original poster is probably required to multiply out the left hand side then set up a system of 4 equations an 4 unknowns and solve this system for a,b,c,d.

Last edited by Martingale on Mon Dec 06, 2010 11:45 pm, edited 1 time in total.

stapel_eliz wrote:Is the "A"off to one side meant to indicate multiplication, so the exercise is as follows?

. . . . .

Yes the problem is written as A[....]

The problem I am having is were to start. I know that AX=B so I'm simply solving for A. Sigh I knew I was over thinking it. When I solve it I will post my solution to ensure I did the problem properly.

So not as simple as I thought or is it?

Since AX=B

[a b] [1 0] = [-5 -4]

[c d] [-1 4]= [7 12]

so I get [a b] [-5 -4]

[-a+4c -b+4d][7 12]

Which in equation form is

a = -5 b= -4

-a + 4c = 7 -b +4d = 12

solving for c and d gives

a = -5 b = -4

c = .5 d = d = 2

Is this correct?

Since AX=B

[a b] [1 0] = [-5 -4]

[c d] [-1 4]= [7 12]

so I get [a b] [-5 -4]

[-a+4c -b+4d][7 12]

Which in equation form is

a = -5 b= -4

-a + 4c = 7 -b +4d = 12

solving for c and d gives

a = -5 b = -4

c = .5 d = d = 2

Is this correct?

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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littlebu wrote:Since AX=B

[a b] [1 0] = [-5 -4]

[c d] [-1 4]= [7 12]

so I get [a b] [-5 -4]

[-a+4c -b+4d][7 12]

You might want to check your

Yea I went through and double checked it and came up with

4b = -4

a - b = -5

4d = 12

c - d = 7

but didn't update the post.

4b = -4

a - b = -5

4d = 12

c - d = 7

but didn't update the post.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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littlebu wrote:Yea I went through and double checked it and came up with

4b = -4

a - b = -5

4d = 12

c - d = 7

Please review how to multiply matrices. The method is demonstrated, with a little "movie", in the link provided earlier.

For instance, the top left entry of the product matrix should be the result of multipying the top row entries of A by the left column entries of X, to give you a(1) + b(-1), which you would then set equal to the top left entry of B, giving you a - b = -5.

stapel_eliz wrote:Please review how to multiply matrices. The method is demonstrated, with a little "movie", in the link provided earlier.

For instance, the top left entry of the product matrix should be the result of multipying the top row entries of A by the left column entries of X, to give you a(1) + b(-1), which you would then set equal to the top left entry of B, giving you a - b = -5.

That is what I did after reviewing the link.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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