Question

What is the quickest way to solve for X_3(s)/F(s) in a 3x3 matrix can someone help me?

Answer 1

\[\left(\begin{matrix} X_1(s) \\X_2(s) \\ X_3(s) \end{matrix}\right) = \left[\begin{matrix}a & b & c\\ d & e & f\ \\ g & h & i \end{matrix}\right]\left(\begin{matrix}0 \\ F(s) \\ 0 \end{matrix}\right)\]

Answer 2

could I do (g*0+hF(s) + i*0 )/determinant?

Answer 3

@e.mccormick your field. right?

Answer 4

@allsmiles I really don't understand your question, can you please post the original problem?

Answer 5

I wrote the matrix out wrong sorry. But my question is to just solve for X_3/F(s) give this matrix equation

Answer 6

I will rewrite the matrix

Answer 7

\[\left(\begin{matrix}X_1 \\ X_2 \\ X_3 \end{matrix}\right) = \left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right]^{-1}\left(\begin{matrix}0 \\ F(s) \\ 0\end{matrix}\right) \]

Answer 8

F(s) being a function. I want to solve for X_3 / F(s)

Answer 9

You have an inverse the second time, not the first...

Answer 10

Yeah I wrote my matrix out wrong I apologize

Answer 11

kk. So this was original;ly Ax=b, and you did A^{-1}Ax=A^{-1}b to get it to that point?

Answer 12

yes

Answer 13

It would depend on it being an invertable matrix for that to work... but if you do that, then it is just a matrix multiplication at that point.

Answer 14

ah sorry I don't quite understand my book got this

Answer 15

\[X_3 = \left[\begin{matrix}a & b & 0 \\ d & e & F(s) \\ g & h & 0\end{matrix}\right]/\left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right]\]

Answer 16

how did they get to this point? I can solve it that way

Answer 17

Is the second part of that [] or | |. THey have different meanings.

Answer 18

oh they are both | | sorry, does that stand for finding the determinant

Answer 19

Yes.

Answer 20

yeah both matrix are | |. so it's | | / | |

Answer 21

but I have no idea how they got that expression

Answer 22

Through doing the assorted operations. Seen this: \(A^{-1}=\frac{1}{|A|}adj(A)\)

Answer 23

but then why is the expression equal to just X_3?

Answer 24

Beacuse once you find the adjoint, it is multiplied by that vector.

Answer 25

I am quite new to matrices, does adjoint just mean inverse?

Answer 26

No. the inverse of the determinant times the adjoint is the inverse. The adjoint is the transpose of the cofactors and the cofactors is the minors with signs applied.

Answer 27

Oh I see okay so the top matrix of the | | / | | I posted above is finding the adjoint of A and the bottom is the determinant of A?

Answer 28

I'm still kind of lost on how they replaced the right side with 0 F(s) 0

Answer 29

sorry for so many questions..

Answer 30

Well, they said it was that, right? Did it actually start with this:
\[\left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right]\left(\begin{matrix}X_1 \\ X_2 \\ X_3 \end{matrix}\right)=\left(\begin{matrix}0 \\ F(s) \\ 0\end{matrix}\right)\]

Answer 31

Yeah that's how it started

Answer 32

Now, do you know what that right hand side would look like if you multiplied it?

Answer 33

Sorry... left hand side....

Answer 34

woudln't you get 3 equations..

Answer 35

lol oops I meant aX_1 + bX_2+cX_3 = 0

Answer 36

so if I want to solve

Answer 37

for the X_2 I would replace the second column by 0 F(s) 0 instead

Answer 38

Yah, looks like the email worked... lets see if I can post again finally.

Answer 39

okay should I still go about finding the inverse?

Answer 40

According to Craimer's Rule, if you have any matrix A, and vectors x and b, if Ax=b, you can form new matrices by replacing columns with the b. For this, lets take a 3x3 because that is what you have. \(A_1\) is A but with column 1 replaced by b. \(A_2\) has column 2 replaced, and so on. The the corresponding x values are the determinant of the new matrix divided by the determinant of the original, or precisely that:

\[\left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right]\left(\begin{matrix}X_1 \\ X_2 \\ X_3 \end{matrix}\right)=\left(\begin{matrix}0 \\ F(s) \\ 0\end{matrix}\right)\implies
\\
X_3 = \frac{\left|\begin{matrix}a & b & 0 \\ d & e & F(s) \\ g & h & 0\end{matrix}\right|}{\left|\begin{matrix}a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right|}\]With the between step being just making \(A_3\) to put on the top.

Answer 41

No point to finding the inverse for that.... I mean, it could be done, but it is a ton of math for no reason.

Answer 42

Did you need it solved past that point? If so, do cofactor expansion on the third column.

Answer 43

Oh snap thanks so much I get it!!!

Answer 44

but what rule did you know to solve for X_3?

Answer 45

like dividing it by the determinant?

Answer 46

Oh, I could use any of them. That is Craimer's Rule.

Answer 47

Oh that is so cool does this work with any size of matrices?

Answer 48

Thanks so much for your help and writing all that out I really appreciate it!!

Answer 49

By the same rule:
\[X_2 = \frac{\left|\begin{matrix}a & 0 & c \\ d & F(s) & f \\ g & 0 & i\end{matrix}\right|}{\left|\begin{matrix}a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right|}\\

X_1 = \frac{\left|\begin{matrix}0 & b & c \\ F(s) & e & f \\ 0 & h & i\end{matrix}\right|}{\left|\begin{matrix}a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right|}\]
Craimer's rule is valid for any size matrix. If you actually take those determinants, you can solve it a little bit more. Don't know if there would be any reason to with this.

Answer 50

See how which column I replaces is in line with what column it would be if I multiplied out the original equation? All the \(X_1\) values are replaced with the solution when solving for \(X_1\).

Answer 51

Yes I see how it works now, wish I could give you more stars. You have no idea how much this will help me, thanks again!

Answer 52

And do you know what cofactor expansion is? Have they gone over that?