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Thread: Proof that det(A) = -det(B) if B is A after row interchange

  1. #1
    Junior Member
    Jul 2008

    Proof that det(A) = -det(B) if B is A after row interchange

    Hello everyone,

    I am having trouble understanding my instructor's proof due to two questions that I haven't been able to figure out.

    Please note that for my elementary linear algebra course, the instructor has said that permutations (and thus any proofs involving this concept) will not be covered in the course.

    I have pasted the proof and my questions below. Thank you very much!


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  2. #2
    MHF Contributor
    Mar 2010
    $\displaystyle A=\begin{bmatrix}
    a_{11} & a_{12}\\
    a_{21} & a_{22}
    \end{bmatrix}\rightarrow det(A)=a_{11}a_{22}-a_{21}a_{12}$

    $\displaystyle det(E_1A)=\begin{bmatrix}
    a_{21} & a_{22}\\
    a_{11} & a_{12}

    In regards to adding determinants, it is possible to associate a real number, scalar, with each non-singular square matrix; therefore, since the determinant is just a constant, adding determinants yields:$\displaystyle det(A)+det(B)=k+c \ \mbox{where} \ k,c\in\mathbb{Z}$

    $\displaystyle det(A)+det(B)\neq det(A+B)$
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  3. #3
    Senior Member
    Feb 2010
    Determinants are linear in any row (column) of the matrix.

    $\displaystyle \det \left( \begin{array}{ccc}
    & \vdots & \\
    \textrm{---} & r+s & \textrm{---} \\
    & \vdots & \end{array} \right) = \det \left( \begin{array}{ccc}
    & \vdots & \\
    \textrm{---} & r & \textrm{---} \\
    & \vdots & \end{array} \right) + \det \left( \begin{array}{ccc}
    & \vdots & \\
    \textrm{---} & s & \textrm{---} \\
    & \vdots & \end{array} \right)$

    The line starting 0=det(A)+det(B) follows because the matrix after the second equal sign contains linearly dependant rows. Then you just apply linearity in the rows until you get det(A)+det(B) out.
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