VinnyCee
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--- Find the determinant of a 5 X 5 matrix ---
Find the determinant of the matrix.
A = \left[\begin{array}{ccccc}<br /> 1 & 2 & 3 & 4 & 5 \\<br /> 3 & 0 & 4 & 5 & 6 \\<br /> 2 & 1 & 2 & 3 & 4 \\<br /> 0 & 0 & 0 & 6 & 5 \\<br /> 0 & 0 & 0 & 5 & 6<br /> \end{array}\right]
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)
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}<br /> 3 & 4 & 5 & 6 \\<br /> 2 & 2 & 3 & 4 \\<br /> 0 & 0 & 6 & 5 \\<br /> 0 & 0 & 5 & 6<br /> \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}<br /> 2 & 3 & 4 \\<br /> 0 & 6 & 5 \\<br /> 0 & 5 & 6 <br /> \end{array}\right]\right)
det(A) = (-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc}<br /> 2 & 3 & 4 \\<br /> 0 & 6 & 5 \\<br /> 0 & 5 & 6 <br /> \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?
Homework Statement
Find the determinant of the matrix.
A = \left[\begin{array}{ccccc}<br /> 1 & 2 & 3 & 4 & 5 \\<br /> 3 & 0 & 4 & 5 & 6 \\<br /> 2 & 1 & 2 & 3 & 4 \\<br /> 0 & 0 & 0 & 6 & 5 \\<br /> 0 & 0 & 0 & 5 & 6<br /> \end{array}\right]
Homework 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)
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}<br /> 3 & 4 & 5 & 6 \\<br /> 2 & 2 & 3 & 4 \\<br /> 0 & 0 & 6 & 5 \\<br /> 0 & 0 & 5 & 6<br /> \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}<br /> 2 & 3 & 4 \\<br /> 0 & 6 & 5 \\<br /> 0 & 5 & 6 <br /> \end{array}\right]\right)
det(A) = (-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc}<br /> 2 & 3 & 4 \\<br /> 0 & 6 & 5 \\<br /> 0 & 5 & 6 <br /> \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?