# Determinants of higher dimension matrices

1. Sep 3, 2011

### JamesGoh

The lecturer said that a way to find the determinant of a matrix is
to do the following

det(A) = xdet(B) (1)

where A is the original matrix, B is an arbirtray matrix and x is a scalar multiplier

The lecturer also said that a simple way to find the determinant of a high dimension matrix (say a 3x3) is to put it into upper triangular form and then multiply the diagonal values (which is essentially the determinant of the triangular matrix)

Using the knowledge and the forumla given in (1), how can we determine the best value for x ?

Is it just random guess or is there a logic behind it ?

2. Sep 4, 2011

### HallsofIvy

I am afraid you have misunderstood or misremembered what your lecturer said. B cannot be an "arbitrary matrix". What is true is that det(A)= x det(B) where B is the same as A except that all the numbers in one row (or column) are divided by x.

For example, if
$$A= \begin{bmatrix}4 & 2 \\ 3 & 1\end{bmatrix}$$
then det(A)= 4(1)- 2(3)= -2.

If B is not an "arbitrary" matrix but
$$B= \begin{bmatrix}2 & 1 \\ 3 & 1\end{bmatrix}$$
the same as A except that I have divided the first row by 2,
then det(B)= 2(1)- 1(3)= -1 so det(A)= 2det(B).

That can indeed be helpful in reducing a matrix to something that is easier to work with. In fact, it is always possible to "row-reduce" a matrix to either something that has a row of all 0s (and so has determinant 0) or to the identity matrix (and so has determinant 1). You can then calculate the determinant of the original matrix by using
1) If you swap two rows, you multiply the determinant by -1.
2) If you multiply/divide a row by the number a, you multiply/divide the determinant by a.
3) If you add or subtract a multiple of one row to or from another, you do not change the determinant.

Going back to
$$A= \begin{bmatrix}4 & 2 \\ 3 & 1\end{bmatrix}$$
I can row-reduce it by
1) divide the first row by 4 and then subtract 3 times the (new) first row from the second. That gives
$$\begin{bmatrix}1 & \frac{1}{2} \\ 0 & -\frac{1}{2}\end{bmatrix}$$

2) Add the second row to the first and then multiply the second row by -1/2:
$$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$

That is now the identity matrix which obviously has determinant 1. I divided the first row by 4 and multiplied the second row by -2 so did the same to the determinant. To get back to the determinant of the original matrix, I must do the opposite: multiply by four and divide by -2: (1)(4)/(-2)= -2 as before.