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Thread: 3.2.9 find 4x4 matrix det by row replacement

  1. #1
    Super Member bigwave's Avatar
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    Angry 3.2.9 find 4x4 matrix det by row replacement

    $\textsf{ compute the determinant by cofactor expansions}$
    $$\left[\begin{array}{rrrr}
    6& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{a}{\sim}
    %
    \left[\begin{array}{rrrr}
    0& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{b}{\sim}
    %
    \left[\begin{array}{rrrr}
    1& \,7& \,2& \,0\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{c}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ -15& 1& 0& -21\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]$$

    $$\stackrel{d}{\sim}\left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{e}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{f}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\
    \frac{23}{2}& \frac{55}{2}& 8& 8
    \end{array}\right]\stackrel{g}{\sim}
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\
    \frac{75}{16}& \frac{-5}{16}& 0&0\\
    1& 7& 2& 0\\
    \frac{23}{2}& \frac{55}{2}& 8& 8
    \end{array}\right]$$
    $\textbf{Solution:}\\$
    $\textit{a. $-3\times R_3$ add to $R_1$ }$
    $\textit{b. $1 \times R_2$ add to $R_1$}$
    $\textit{c. exchange $R_3$ and $R_1$}$
    $\textit{d. $-2\times R_4$ add to $R_2$}$
    $\textit{e. $3\times R_4$ add to $R_2$ }$
    $\textit{f. $7/2\times R_3$ add to $R_4$}$
    $\textit{g.$-3/8\times R_4$ add to $R_2$}$
    $(2)(-5/16)(2)(8)=-10$
    ok the bk answer is 10 so not sure where the ??? is
    Last edited by bigwave; Apr 3rd 2018 at 03:31 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: 3.2.9 find 4x4 matrix det by row replacement

    I thought this was to be done by co-factor expansion?

    ie.

    $\displaystyle \left | \begin{matrix} 6 & 0 & 0 & 5 \\ 1 & 7 & 2 & -5 \\ 2 & 0 & 0 & 0 \\ 8 & 3 & 1 & 8 \end{matrix} \right | = 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix} \right | - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right | $

    etc.

    And yes, the determinant is 10.

    By the way... Step c) multiplies your determinant by -1. Just switching rows does this.

    -Dan
    Last edited by topsquark; Apr 3rd 2018 at 04:35 PM.
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  3. #3
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    Re: 3.2.9 find 4x4 matrix det by row replacement

    Quote Originally Posted by bigwave View Post
    $\textsf{ compute the determinant by cofactor expansions}$
    $$\left[\begin{array}{rrrr}
    6& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{a}{\sim}
    %
    \left[\begin{array}{rrrr}
    0& \,0& \,0& \,5\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{b}{\sim}
    %
    \left[\begin{array}{rrrr}
    1& \,7& \,2& \,0\\ 1& 7& 2&-5\\ 2& 0& 0& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{c}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ -15& 1& 0& -21\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]$$

    $$\stackrel{d}{\sim}\left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{e}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\8& 3& 1& 8
    \end{array}\right]\stackrel{f}{\sim}
    %
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\ 9& 10& 3&3\\ 1& 7& 2& 0\\
    \frac{23}{2}& \frac{55}{2}& 8& 8
    \end{array}\right]\stackrel{g}{\sim}
    \left[\begin{array}{rrrr}
    2& \,0& \,0& \,0\\
    \frac{75}{16}& \frac{-5}{16}& 0&0\\
    1& 7& 2& 0\\
    \frac{23}{2}& \frac{55}{2}& 8& 8
    \end{array}\right]$$
    $\textbf{Solution:}\\$
    $\textit{a. $-3\times R_3$ add to $R_1$ }$
    $\textit{b. $1 \times R_2$ add to $R_1$}$
    $\textit{c. exchange $R_3$ and $R_1$}$
    $\textit{d. $-2\times R_4$ add to $R_2$}$
    $\textit{e. $3\times R_4$ add to $R_2$ }$
    $\textit{f. $7/2\times R_3$ add to $R_4$}$
    $\textit{g.$-3/8\times R_4$ add to $R_2$}$
    $(2)(-5/16)(2)(8)=-10$
    ok the bk answer is 10 so not sure where the ??? is
    Interchanging two rows causes a sign change in the determinant.
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  4. #4
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    Re: 3.2.9 find 4x4 matrix det by row replacement

    Quote Originally Posted by JaguarXJS View Post
    Interchanging two rows causes a sign change in the determinant.
    Oh come one! Just evaluate the minor factors

    $\displaystyle 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix}\right| =0 $. This results from the all zero row.

    Then
    $\displaystyle - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right|=-5(-2)[(7)(1)-(2)(3)]=10 $
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    Forum Admin topsquark's Avatar
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    Re: 3.2.9 find 4x4 matrix det by row replacement

    Quote Originally Posted by Plato View Post
    Oh come one! Just evaluate the minor factors

    $\displaystyle 6 \left | \begin{matrix} 7 & 2 & -5 \\ 0 & 0 & 0 \\ 3 & 1 & 8 \end{matrix}\right| =0 $. This results from the all zero row.

    Then
    $\displaystyle - 5 \left | \begin{matrix} 1 & 7 & 2 \\ 2 & 0 & 0 \\ 8 & 3 & 1 \end{matrix} \right|=-5(-2)[(7)(1)-(2)(3)]=10 $
    Agreed. I simply pointed that out to show him how to correct his work with the row reduction. And, of course, it's also as easy to expand along row 3...

    -Dan
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