# - Find the determinant of a 5 X 5 matrix -

1. Dec 3, 2006

### VinnyCee

--- Find the determinant of a 5 X 5 matrix ---

1. The problem statement, all variables and given/known data

Find the determinant of the matrix.

A = $$\left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 3 & 0 & 4 & 5 & 6 \\ 2 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 6 & 5 \\ 0 & 0 & 0 & 5 & 6 \end{array}\right]$$

2. Relevant equations

Laplace Expansion forumla

For an Expansion across the $$i^{th}$$ row of an n x n matrix:

det(A) = $$\sum_{j=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)$$
(for a fixed i)

For an Expansion across the $$j^{th}$$ column of an n x n matrix:

det(A) = $$\sum_{i=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)$$
(for a fixed j)

3. The attempt at a solution

So, I start by doing a Laplace Expansion across the first row and down the second column. So i = 1 and j = 2.

det(A) = $$(-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)$$

det(A) = $$(-1)\,(2)\,det\left(\left[\begin{array}{cccc} 3 & 4 & 5 & 6 \\ 2 & 2 & 3 & 4 \\ 0 & 0 & 6 & 5 \\ 0 & 0 & 5 & 6 \end{array}\right]\right)$$

I continue by doing another Laplace Expansion, this time across the first row and down the first column. So i = 1 and j = 1.

det(A) = $$(-2)\,(-1)^{1\,+\,1}\,a_{1\,1}\,det\left(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 0 & 6 & 5 \\ 0 & 5 & 6 \end{array}\right]\right)$$

det(A) = $$(-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 0 & 6 & 5 \\ 0 & 5 & 6 \end{array}\right]\right)$$

For the 3 x 3 matrix, I use Sarrus's Rule to get a determinant of 22.

det(A) = $$(-6)\,(22)\,=-132$$

However, when I plug the original matrix into my TI-92, I get det(A) = 99!

I tried a second time with i = 1 and j = 1 for the original matrix and then i = 1 and j = 2 for the 4 x 4 matrix. Here I get det(A) = -44.

Neither are right! What am I doing wrong here?

2. Dec 3, 2006

### matt grime

You haven't summed. Fix a row and you need to do the Laplace thing for every entry in that row, not just one that takes your fancy, and take the signed sum. In short, you've not used the formula properly.

It is easier to use row operations, anyway.

3. Dec 3, 2006

### VinnyCee

I found my error, it was exactly as you said. I was not doing the sum for the second non-zero term in the column!

The right equation is:

det(A) = $$(-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)\,+\,(-1)^{3\,+\,2}\,a_{3\,2}\,det\left(A_{3\,2}\right)$$

Last edited: Dec 3, 2006