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I have a matrix

$$ A = \begin{pmatrix} 1 & 2 & 3 & 4\\ -1 & 1 & 2 & 3\\ 1 & -1 & 1 & 2\\ -1 & 1 & -1 & 1\\ \end{pmatrix} $$

I should be able to add multiples of any row to another row and not effect the determinant.

By adding $1$ times the first row to the second row, $-1$ times the first to the third, and $1$ times the first to the fourth, I should still have the same determinant.

This results in the matrix $$ B = \begin{pmatrix} 1 & 2 & 3 & 4\\ 0 & 3 & 5 & 7\\ 0 & -3 & -2 & -2\\ 0 & 3 & 2 & 5\\ \end{pmatrix} $$

$\det(A) = 27,\quad \det(B) = 87$

What am I missing, I am expecting $\det(B) = 27$ since I am only adding multiples of the first row to the others.

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By my calculation $\det(B)=27$, just as it should (assuming that your calculation of $\det(A)$ is correct). Concretely, $$\begin{align*}\det(B)&=(3)(-2)(5)+(5)(-2)(3)+(7)(-3)(2)-(7)(-2)(3)\\&\qquad-(5)‌​(-3)(5)-(3)(-2)(2)\\&=-30-30+75+12=87-60=27\;.\end{align*}$$ –  Brian M. Scott Apr 5 '12 at 2:50
    
How did you calculate those determinants to be 27 and 87? It's a rather unwieldy computation for 4×4 matrices (except by the trick of doing some row/column operations first to make it more tractable. The difference could easily be due to a sign error somewhere -- especially since two of the terms in the complete expansion of $\det B$ have absolute value equal to half of the error. –  Henning Makholm Apr 5 '12 at 2:50
    
sorry to waste your time on this, thanks for pointing that out though. –  brandon Apr 5 '12 at 2:59

1 Answer 1

up vote 2 down vote accepted

Apparently you made a mistake in computing ${\rm det}(B).$ I tried Maple, and ${\rm det}(B) = 27$ as well.

Here is Wolfram|Alpha for both ${\rm det}(A)$ and ${\rm det}(B)$.

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You're right. I thought I was going crazy. I put a - sign in front of entry 4,2 and the determinant got thrown off a bit. Thanks for the help. I typed it over a good 5 times and made that mistake every time –  brandon Apr 5 '12 at 2:58

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