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 Post subject: det A
PostPosted: Tue, 11 Nov 2008 01:56:58 UTC 
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i tried so hard to find a way in order to solve these two questions and i think that i can't.
if someone can help me because it's really important for me??????
please??
Image

use the value of det A to compute:
Image

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PostPosted: Tue, 11 Nov 2008 02:04:09 UTC 
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This looks weird. Determinant is supposed to be defined for square matrix, i.e. nxn matrix. This question involves 4x5 matrix, I'm not sure whether the method of cofactor expansion is still valid here.


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PostPosted: Tue, 11 Nov 2008 02:43:44 UTC 
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i am so sorry the true matrix is this

Image

Image

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PostPosted: Tue, 11 Nov 2008 05:42:09 UTC 
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We can use cofactor expansion with respect to any column or any row. So usually we choose the column or row with the most number of 0 to simplify the calculation.

For this question, I expand at second row since there are 4 zeros there. Then, I expand the submatrix by picking the row with most number of zero too but need to take care of the sign.

\det(A)=1\left|\begin{array}{cccc}4&8&8&5\\6&8&8&7\\0&8&3&0\\0&2&0&0\end{array}\right|=2\left|\begin{array}{ccc}4&8&5\\6&8&7\\0&3&0\end{array}\right|=2(-3)\left|\begin{array}{cc}4&5\\6&7\end{array}\right|=12.

Part (b) require some properties of determinant, namely switching two rows/columns changes the sign of determinant and multiple of rows/columns can be added without changing the determinant value. Let me denote the question as
\det(B)-2\det(C).
Let E_1 denote the elementary operation of switching R1(row 1) and R3 of B. E_2 be (R2-R4) and E_3 be (R5-R2). Then,
\det(B)=\det(E_3E_2E_1A)=\det(E_3)\det(E_2)\det(E_1)\det(A)=1(1)(-1)(12).
Hope you can proceed to find det(C). For more information on the properties of determinant, you can try these websites:
http://en.wikipedia.org/wiki/Determinant
http://mathworld.wolfram.com/Determinant.html


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PostPosted: Fri, 14 Nov 2008 18:48:04 UTC 
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thank you very much for your help!!!!!!!!!!!!!

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