- #1
VinnyCee
- 489
- 0
--- Find the determinant of a 5 X 5 matrix ---
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]
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)
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?
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?