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Thread: Determinant question

  1. #1
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    Determinant question

    If you use row-reduction to make a n x n matrix A into a triangular matrix then you can take the product of the diagonal to achieve det(A)

    and if you multiply a row/column by a non-zero scalar $k$ the effect is $k\det(A)$

    also $\det(kA) = k^n\det(A)$

    thus therein lies my question.

    if I use row reduction then multiply some row by $k$ , call the new triangular matrix B then it becomes $k\det(B)$

    so what is the effect on $\det(A)$ because $k\det(B) \neq \det(A)$

    is it $k\det(B) = \det(kA) = k^n\det(A)$?
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  2. #2
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    Re: Determinant question

    Hey Jonroberts74.

    A scalar multiplication of a row is not the same as a scalar multiplication of an entire matrix: this is why one is k^n and the other is just k.

    If you are doing a row-operation then it will be multiplying one row by a constant but if you multiply the entire matrix by a scalar then that is not an elementary row operation in general (unless it's rank 1).

    Basically you should look at each matrix decomposition and just use the fact that det(AB) = det(A)det(B) if A and B are well defined (i.e. square).
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  3. #3
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    Re: Determinant question

    ah okay I got it now.

    If I made A into a triangular matrix, call it B, and for whatever reason I multiply a row/col in B by a scalar $k$. then $\det(B)=k\det(A) \Rightarrow \frac{\det(B)}{k}=\det(A)$

    thanks
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