- Find the determinant of a 5 X 5 matrix -

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The discussion focuses on finding the determinant of a 5x5 matrix using Laplace expansion. The initial calculations led to incorrect determinant values due to improper application of the expansion formula, specifically neglecting to sum contributions from all relevant terms in the chosen row or column. A participant identified that they needed to account for all non-zero entries in the row during the calculation. The correct approach involves summing the determinants of the resulting submatrices for each entry in the row. Ultimately, the conversation emphasizes the importance of following the determinant calculation formula accurately to avoid errors.
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--- Find the determinant of a 5 X 5 matrix ---

Homework Statement



Find the determinant of the matrix.

A = \left[\begin{array}{ccccc}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\<br /> 3 &amp; 0 &amp; 4 &amp; 5 &amp; 6 \\<br /> 2 &amp; 1 &amp; 2 &amp; 3 &amp; 4 \\<br /> 0 &amp; 0 &amp; 0 &amp; 6 &amp; 5 \\<br /> 0 &amp; 0 &amp; 0 &amp; 5 &amp; 6<br /> \end{array}\right]



Homework Equations



Laplace Expansion formula

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 &amp; 4 &amp; 5 &amp; 6 \\<br /> 2 &amp; 2 &amp; 3 &amp; 4 \\<br /> 0 &amp; 0 &amp; 6 &amp; 5 \\<br /> 0 &amp; 0 &amp; 5 &amp; 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 &amp; 3 &amp; 4 \\<br /> 0 &amp; 6 &amp; 5 \\<br /> 0 &amp; 5 &amp; 6 <br /> \end{array}\right]\right)

det(A) = (-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc}<br /> 2 &amp; 3 &amp; 4 \\<br /> 0 &amp; 6 &amp; 5 \\<br /> 0 &amp; 5 &amp; 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?
 
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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.
 
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)
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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