what is the solution of the matrix equation
{ 5, -2 2,1]x= [4 -4]

A) [2 -4]
B) [-10 -24]
C) [-10 24]
D) [10 24]

Question

what is the solution of the matrix equation 
{ 5, -2  2,1]x= [4  -4]

A) [2  -4]
B) [-10  -24]
C) [-10  24]
D) [10    24]

hero · Answer

I assume this is your matrix equation

e.mccormick · Answer

Or perhaps:

$\begin{bmatrix}
5 & -2\ 
2 & 1
\end{bmatrix}x=\begin{bmatrix}
4\ 
-4
\end{bmatrix}$

hero · Answer

It is in the form $Ax = B$

Find the inverse of matrix A then take the inverse of both sides to isolate x:

$x = A^{-1}B$

hero · Answer

Then multiply the inverse of A with B to get your final result.

anonymous · Answer

wow sorry i messed up that 1 is a -1 and the positive 4 is a 2 and yes that is my matrix

hero · Answer

$$\begin{bmatrix}
5 & -2\ 
2 & -1
\end{bmatrix}x=\begin{bmatrix}
2\ 
-4
\end{bmatrix}$$

anonymous · Answer

yes perfect now how do you the next step sorry I'm not very good with matrixes

hero · Answer

$$A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \ -c &  a\end{bmatrix}$$

anonymous · Answer

so would the answer be c [-10]
                                        [  24]

e.mccormick · Answer

What Hero is using is a principal of matrices that is based off of the multiplicative inverse.

$5\cdot \dfrac{1}{5}=1$

That is true for pretty much any number and also works in matrices.  

So what he is saying is that your original form is:

$Ax=b$

You need to cancel the $A$ on the left, but that will make a change in the right.  To do so, you need the  multiplicative inverse:

$Ax=b\implies $

$A^{-1}Ax=A^{-1}b\implies $

$1x=A^{-1}b\implies $

$x=A^{-1}b$

So the first step in this is to find $A^{-1}$.

anonymous · Answer

you guys are like genius compared to me in this so is a -4?

e.mccormick · Answer

We have just done this before.

Then he showed you a form of this:

If $A=\begin{bmatrix}
a & b\ 
c & d
\end{bmatrix}$ then:

$A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}
d & -b\ 
-c & a
\end{bmatrix}$

That help?

anonymous · Answer

damn I'm sorry so its 1/ (-1)(5) - (-2)(2)? @e.mccormick

e.mccormick · Answer

For that part, yes.  Then you multiply through the modified matrix (adjoint matrix) with that.

anonymous · Answer

man i feel so dumb could you show me how to do it

e.mccormick · Answer

Well, what did $\dfrac{1}{(-1)(5) - (-2)(2)}$ get you?

anonymous · Answer

okay so you -5 - (-4)= 9

e.mccormick · Answer

Ummm...

When you distribute the $-(-4)$, what do you get?

anonymous · Answer

-1 sorry

e.mccormick · Answer

OK.  Good!

So we are at:

$A^{-1}=-1\begin{bmatrix}
d & -b\ 
-c & a
\end{bmatrix}$

So now, the a, b, c, and d parts.  You take these from the original matrix:

$A=\begin{bmatrix}
a & b\ 
c & d
\end{bmatrix}$

So, what are they?

anonymous · Answer

[-5  2]
[ -2 1 ]  ?

e.mccormick · Answer

OK, you did the sign swap, but look at the position of a and d.

If you learned about a cofactor matrix, this is all about that.

anonymous · Answer

okay so (-1  2)
             (-2  5)

e.mccormick · Answer

Just a little FYI in case my terms are confusing you:

Depending on where you are with linear algebra, or just algebra, they may throw all these terms at you or not.  If you are learning this as part of algebra, and not a linear class, al they usually talk about are:

matrix
transpose matrix
cofactor matrix
inverse matrix
determinant

Then they just give you a few set formulas to remember.

If you are taking linear algebra, they then say:

adjoint/adjunct/adjugate matrix = transpose of cofactor matrix
eigen values
eigen vectors
column vectors
row vectors
pivots/free variables

So if I do not know where you are, I might use some terms you are unfamiliar with.

anonymous · Answer

thats okay I've seen these before matrix cofactor matrix inverse matrix

e.mccormick · Answer

OK.  So, here is where you are at:

$A^{-1}=-1\begin{bmatrix}
d & -b\ 
-c & a
\end{bmatrix}=-1\begin{bmatrix}
-1 & 2\ 
-2 & 5
\end{bmatrix}
$

e.mccormick · Answer

So, multiply through the -1 (swap the signs).

anonymous · Answer

[1 - 2]
[2  -5] right

e.mccormick · Answer

Yes.  So, do you understand what you need to do with that now?  How to plug it in?

anonymous · Answer

no really, I'm really sorry man i know its probably a pain for you to do this

anonymous · Answer

not*

e.mccormick · Answer

It is just a scroll up at this point.  OK, we are at:

$A^{-1}=\begin{bmatrix}
1 & -2\ 
2 & -5
\end{bmatrix}
$

Now, remember, this was just to find what is needed for:

$x=A^{-1}b$

From the original, $b=\begin{bmatrix}2\ -4 \end{bmatrix}$ and you just found $A^{-1}$ so:

$x=\begin{bmatrix}1 & -2\ 2 & -5\end{bmatrix}\begin{bmatrix}2\ -4 \end{bmatrix}$

Can you do that now?


Oh, and how I Hero and I did the whole formatted things is with $\LaTeX$ `$x=\begin{bmatrix}1 & -2\ 2 & -5\end{bmatrix}\begin{bmatrix}2\ -4 \end{bmatrix}$`

anonymous · Answer

okay i see how you did the format thing and would it be the answer D

e.mccormick · Answer

Yep. That would be it in the end. Here is a reference to that whole inverse thing we used. It should be similar to what you see in an algebra book: http://www.mathwords.com/i/inverse_of_a_matrix.htm If you can get this process down, and it really is just a process, you should do fine on any test.

anonymous · Answer

yesss! Thanks a lot man i really appreciate it, your a legend

e.mccormick · Answer

Hero is a legend.  I just helped out a little.

anonymous · Answer

haha both you guys are, but anyways i really appreciate it, thanks

e.mccormick · Answer

np.  Have fun!

anonymous · Answer

hah il try too