- #1

VinnyCee

- 489

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

## Homework Statement

Find the determinant of the matrix.

A = [tex]\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][/tex]

## Homework Equations

__Laplace Expansion forumla__

For an Expansion across the [tex]i^{th}[/tex] row of an n x n matrix:

det(A) = [tex]\sum_{j=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)[/tex]

(for a fixed i)

For an Expansion across the [tex]j^{th}[/tex] column of an n x n matrix:

det(A) = [tex]\sum_{i=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)[/tex]

(for a fixed j)

## 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) = [tex](-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)[/tex]

det(A) = [tex](-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)[/tex]

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) = [tex](-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)[/tex]

det(A) = [tex](-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc}

2 & 3 & 4 \\

0 & 6 & 5 \\

0 & 5 & 6

\end{array}\right]\right)[/tex]

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

det(A) = [tex](-6)\,(22)\,=-132[/tex]

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?