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

by VinnyCee
Tags: determinant, matrix
 P: 492 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?
 P: 492 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) = $$(-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)\,+\,(-1)^{3\,+\,2}\,a_{3\,2}\,det\left(A_{3\,2}\right)$$