Matrices question Watch

David_Skiller
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Hi tsr,

I am given matrices :

A = [2,3,3]
[3,-1,-2]
[5,1, -1]

B = [10]
[1]
[6]

C = [3,6,-3]
[-7,-17,13]
[8,13,-11]



and asked to evaluate :

AC
CA
CB


which I have done.

The question then asks me to hence, or otherwise, solve

2x+3x+3z = 10
3x-y-2z = 1
5x+y-z = 6

I know I can solve by writing Ax = B


and solving for x.

But is there a quicker way using the information i've worked out previously, and thus using the "hence" because although i can solve my multiplying by the inverse it is quite long
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David_Skiller
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Slowbro93
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(Original post by David_Skiller)
Hi tsr,

I am given matrices :

A = [2,3,3]
[3,-1,-2]
[5,1, -1]

B = [10]
[1]
[6]

C = [3,6,-3]
[-7,-17,13]
[8,13,-11]



and asked to evaluate :

AC
CA
CB


which I have done.

The question then asks me to hence, or otherwise, solve

2x+3x+3z = 10
3x-y-2z = 1
5x+y-z = 6

I know I can solve by writing Ax = B


and solving for x.

But is there a quicker way using the information i've worked out previously, and thus using the "hence" because although i can solve my multiplying by the inverse it is quite long
Firstly, what did you get when you multiplied AC and CA Does this remind you of something in particualr?
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David_Skiller
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(Original post by cpdavis)
Firstly, what did you get when you multiplied AC and CA Does this remind you of something in particualr?
I got 900,090,009 for CA (where , means new row)

and I got 9 -30 0, 0 29 0, 0 10 9 for AC
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Slowbro93
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(Original post by David_Skiller)
I got 900,090,009 for CA (where , means new row)

and I got 9 -30 0, 0 29 0, 0 10 9 for AC
You should have the same answer for AC and CA. If you divided through by a particular constant, what relation do AC and CA have?

On a side note, I would recommend learning LaTeX to make it easier for communicating your question :yep:
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Khallil
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(Original post by David_Skiller)
...
It's been a while since I've done any matrices, but here goes:

You should've found that:

\mathbf{A}\mathbf{C} = 9\mathbf{I}

\mathbf{C}\mathbf{A} = 9\mathbf{I}

When the product of two matrices yields the identity matrix, you should know that the matrices are inverses of one another, i.e.:

(*) \ \frac{1}{9} \mathbf{C} = \mathbf{A^{-1}}

For the hence part of your question, note that they're telling you to solve the equality:

\mathbf{A} \begin{pmatrix} x \\ y \\ z \end{pmatrix}= \mathbf{B}

Using the fact (*) that we derived earlier, can you see the link between the RHS and \mathbf{C}\mathbf{B} \ ?
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