## Can somebody help with this matrix problem?

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

### Can somebody help with this matrix problem?

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
[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
littlebu

Posts: 8
Joined: Fri Oct 22, 2010 9:23 pm

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?

$\mbox{Find the matrix }\, A\, \mbox{such that}$

. . . . .$A\,\times\,\left[\begin{array}{rr}1&0\\-1&4\end{array}\right]\, =\, \left[\begin{array}{rr}-5&-4\\7&12\end{array}\right]$

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!

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

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?

$\mbox{Find the matrix }\, A\, \mbox{such that}$

. . . . .$A\,\times\,\left[\begin{array}{rr}1&0\\-1&4\end{array}\right]\, =\, \left[\begin{array}{rr}-5&-4\\7&12\end{array}\right]$

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.

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re:

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

$\mbox{Find the matrix }\, A\, \mbox{such that}$

. . . . .$A\,\times\,\left[\begin{array}{rr}1&0\\-1&4\end{array}\right]\, =\, \left[\begin{array}{rr}-5&-4\\7&12\end{array}\right]$

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.
littlebu

Posts: 8
Joined: Fri Oct 22, 2010 9:23 pm

### Re: Re:

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?
littlebu

Posts: 8
Joined: Fri Oct 22, 2010 9:23 pm

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 matrix multiplication.

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

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.
littlebu

Posts: 8
Joined: Fri Oct 22, 2010 9:23 pm

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

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

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.
littlebu

Posts: 8
Joined: Fri Oct 22, 2010 9:23 pm

littlebu wrote:That is what I did after reviewing the link.

But you clearly did not get the same result. And the lesson isn't helping you find your error. So it really will be necessary for you to show your work, in order for us to figure out where you're going astray.

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

Next