Finding the determinant of a 3x3 matrix.

Mozart

Hello, I just finished doing a question in which I had to find the values of x y and z from 3 linear equations using Cramer's rule. I used row augmentation for the x and y but then for z I couldn't see any way of using row augmentation.

So I looked in my notes and saw an example using expansion along row 2 to find detA and expansion along column 1 as well to find detA. (Note I was dealing with 3x3 determinants)

My question is, when I run into 3x3 determinants and row augmentation isn't an option do I always expand along row 2 or expand along column 1. Is this some sort of rule because I first attempted to expand along row 1 and it yielded the wrong answer... When I did an expansion along row 2 I got the right answer.

Thanks.

cristo

You must have made a mistake in your expansion. When calculating a determinant, you can expand by any row or column that you like; it's normally best to pick the one with the most zeros in it though!

HallsofIvy

What do you mean by "row augmentation"? And why would you be able to do it for x and y but not for z?

radou

Are you referring to Laplace's rule? It would be great if you would write down the exact problem.

HallsofIvy

Kramers rule says that if [itex]a_1x+ b_1y+ c_1z= d_1[/itex], [itex]a_2x+ b_2y+ c_2z= d_2[/itex] and [itex]a_3x+ b_3y+ c_3z= d_3[/itex] then
[tex]x= \frac{\left|\begin{array}{ccccc}d_1 && b_1 && c_1 \\ d_2 && b_2 && c_2 \\ d_3 && b_3 && c_3\end{array}\right|}{\left|\begin{array}{ccccc}a_1 && b_1 && c_1 \\ a_2 && b_2 && c_2 \\ a_3 && b_3 && c_3\end{array}\right|}[/tex]
Notice that in the numerator, the first column has been replaced by the column from the right side of the equations. Replacing the second column instead gives y, replacing the third column gives z.

I am still not sure what is meant by "row augmentation". One method that I learned for 3 by 3 determinants (and only 3 by 3) was to repeat the first two columns, calculate the three diagonals from upper left to lower right, add them, then from upper right to lower left, add them, and finally subtract the second number from the first. Of course, using rows would work as well as columns. Perhaps that is what the OP meant but I don't see why z would give any particular problem.

Mozart

Thanks for the replies. My teacher uses this method and calls it row augmentation. Pretty much if he has a 3x3 matrix he'll make either any row or column a value followed by zeros by adding some other row to it or some column. It's hard to explain but say row 1 is 1 2 3 he'll add say row 2 (lets say its 1 -2 -3) He'll show it like R2+R1= and then write the new determinant he then crosses out the row containing the two zeros and the column that joins up with the non zero value and then is able to reduce it to a 2x2 determinant. So to calculate the resulting determinant it looks like
2(-1)^2 l....l Sorry if this is not making sense but thats what I meant by row augmentation. There is probably some other name for it.

radou

If that's what you meant by row augmentation, then of course it makes sense, since it is known what happenes with the value of the determinant when you add a row/column multiplied by a non zero scalar to another row/column. After a few steps, you can obtain a determinant with a row/column with only one non zero value, and that very row/column makes a good choice for the application of the Laplace's rule, which leaves you with a determinant of a lower order, in your case, order 2.

mybsaccownt

oops, it's already been mentioned

oh well, i'll leave it up anyway cause I spent a few minutes on typing it up :-P

there is a trick that's really nice for 3x3 matrices, but it only works for 3x3 matrices

what you do is set up...(I would call it an augmented matrix as well)

a11 a12 a13|a11 a12
a21 a22 a23|a21 a22
a31 a32 a33|a31 a32

and you take the sum of the product of the diagonals to the right minus the sum of the product of the diagonals to the left

so: (a11*a22*a33+a12*a23*a31+a13*a21*a32) - (a13*a22*a31+a11*a23*a32+a12*a21*a33)

it is a nice method though for 3x3 matrices, but make sure you know co-factor expansion anyway, it's good stuff

radou

That's the Sarus rule, unless I'm mistaken. Nice trick.

HallsofIvy

Row "augmentation"? I would call that row "reduction" since the point is to reduce a row to something simple, like 1 0 0. If you can "reduce" a matrix to "upper triangular" (only 0s below the main diagonal) with row operations, then the determinant is the product of the numbers on the main diagonal (taking into account that if you multiply an entire row by a number the determinant is multiplied by that number and if you swap two rows, the determinant is multiplied by -1).