I assume that we handle this according to the pic i attched? (I only took the $\displaystyle a_{11}$ part as the example)
And could someone please tell me how we use matrices to solve linear equations?
Okay i found this site: Cramer's Rule
But you need to know how to find the determinants of a 3x3 matrix... Which I am confused about.
If someone could only explain this part to me:
Here's how to find the determinant of a 3x3 matrix. but this method (called the cofactor expansion method) can be generalized to general nxn matrices.
i'll use a,b,c,d... instead of $\displaystyle a_{11}, a_{12}, \cdots$ because those will get annoying to type after a while. hope it doesn't mess you up
Let's say we are given a 3x3 matrix $\displaystyle \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right)$
you begin by imagining a checker board pattern of pluses and minuses, with a + sign starting at the top left corner, that is, think of
$\displaystyle \left( \begin{array}{ccc} + & - & + \\ - & + & - \\ + & - & + \end{array} \right)$
now that accounts for the signs in front of the terms that you have. this is why you had $\displaystyle + a_{11} ... - a_{12} ... + a_{13}$ in your example.
now, in using the cofactor expansion method, you can expand along ANY row OR column you wish. your example showed an expansion along the first row, but it may be convenient to expand along the second column, for instance. for example, that's what i'd do if i noticed that e = h = 0, and all other terms were non-zero, because then, expanding along the second column would require doing the determinant of only 1 2x2 matrix.
so the determinant in such a case would be: $\displaystyle \mbox{det} \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) = -b \left| \begin{array}{cc} d & f \\ g & i \end{array} \right|$ ..........when we write a matrix with straight braces, it means the determinant of that matrix
now we will see how they got such a formula.
when you choose a column or row to expand along, begin with the first element in that row or column, and affix the sign in front of it as directed by the checker board pattern. upon doing so, you should eliminate every term in that element's row and column. (so if we are considering the element b, we would affix a minus sign in front of it and eliminate the elements a,c,e and h, since they fall in the same row and column of b). this will result in a new matrix that is one dimension lower, in this case, a 2x2 matrix (the matrix with row 1: d,f and row 2:g,i). find the determinant of that matrix and multiply by the original term. repeat this process for each term in that row or column you decided to expand along.
for example, using my matrix, let's call the matrix A. finding the determinant by expanding along the third row would be:
$\displaystyle \mbox{det}A = g \left| \begin{array}{cc} b & c \\ e & f \end{array} \right| - h \left| \begin{array}{cc} a & c \\ d & f \end{array} \right| + i \left| \begin{array}{cc} a & b \\ d & e \end{array} \right|$
expanding along the second column would be:
$\displaystyle \mbox{det}A = -b \left| \begin{array}{cc} d & f \\ g & i \end{array} \right| + e \left| \begin{array}{cc} a & c \\ g & i \end{array} \right| - h \left| \begin{array}{cc} a & c \\ d & f \end{array} \right|$
Here's an exercise: what would the formula be for expanding along the second row?
did you have any problems with Cramer's rule itself?
Thanks a lot Jhevon.
For expanding the second row, :
$\displaystyle \mbox{det}A = -d \left| \begin{array}{cc} b & c \\ h & i \end{array} \right| + e \left| \begin{array}{cc} a & c \\ g & i \end{array} \right| - f \left| \begin{array}{cc} a & b \\ g & h \end{array} \right| $
And no, Cramer's rule seems easy enough
The reason I suggested it is that I'm not comfortable with inverting 3x3 (or larger) matrices. I admit it's probably simply that I haven't done it enough. Generally in Quantum Mechanics (where I initially learned how to do many of the tricks I know with matrices) you don't have to invert them, so I have never worked with it much.
So I like Cramers because I don't have to rely on my calculator to finish the problem.
-Dan