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Math Help - 4 ?'s Help Me Please!!!

  1. #1
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    4 ?'s Help Me Please!!!

    questions on attachment!!
    Attached Files Attached Files
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Lane
    questions on attachment!!
    1. Find the 3 cofactors of the second row of:

    <br />
A=\left [ \begin{array}{ccc}<br />
{-7}& {-5}&{6}\\<br />
{9}& {4}&{-3}\\<br />
{8}&{-1}&{2}<br />
\end{array} \right]<br />

    The cofactor C_{i,j} is (-1)^{i+j} times the determinant of the matrix obtained from A by deleting the ith row
    and the jth column of A.

    The question is asking for C_{2,1},\ C_{2,2} and C_{2,3}.

    <br />
C_{2,1}=(-1)^{2+1}\ <br />
\left | \begin{array}{cc}<br />
{-5}& {6}\\<br />
{-1}& {2}<br />
\end{array} \right|=-1[(-5)\times 2-(-1)\times 6]=4<br />

    <br />
C_{2,2}=(-1)^{2+2}\ <br />
\left | \begin{array}{cc}<br />
{-7}& {6}\\<br />
{8}& {2}<br />
\end{array} \right|=\ ?

    <br />
C_{2,3}=\ ?,

    where ? denotes something left to the reader to complete.

    RonL
    Last edited by CaptainBlack; July 24th 2006 at 07:51 PM.
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  3. #3
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    Hello, Lane1

    You could help us by indicating where your difficulty lies.
    Otherwise, we're not helping you ... we're just doing your homework.


    1. Find the three cofactors of the second row of the matrix.

    . . . . \begin{pmatrix}\text{-}7 & \text{-}5 & 6 \\ 9 & 4 & \text{-}3 \\ 8 & \text{-}1 & 2\end{pmatrix}

    Cofactor of 9:\;\;\begin{vmatrix}\text{-}5 & 6\\\text{-}1&2\end{vmatrix}

    Cofactor of 4:\;\;\begin{vmatrix}\text{-}7&6\\8&2\end{vmatrix}

    Cofactor of \text{-}3:\;\;\begin{vmatrix}\text{-}7&\text{-}5\\8&\text{-}1\end{vmatrix}


    2. Find the determinant of: \begin{vmatrix}4&3&2\\3&2&4\\1&3&1\end{vmatrix}
    . . . 4\begin{vmatrix}2&4\\3&1\end{vmatrix} - 3\begin{vmatrix}5&4\\1&1\end{vmatrix} + 2\begin{vmatrix}5&2\\1&3\end{vmatrix} \;= \;4(2-12) - 3(5-4) + 2(15-2)

    . . . = \;4(-10) - 3(1) + 2(13) \;= \;-40 - 3 + 26 \;= \; \boxed{-17}


    3. Use Cramer's Rule to solve (if possible) the system of equations.

    . . . . \begin{array}{cc}4x + 4y \:= \:1\\x + 5y \:=\:\text{-}4\end{array}
    D\:=\:\begin{vmatrix}4&4\\1&5\end{vmatrix} \:= \:20 - 4 \:= \:16

    D_x\:=\:\begin{vmatrix}1&4\\-4&5\end{vmatrix}\:=\:5+16\:= \:21\quad\Rightarrow\quad \boxed{x \,= \,\frac{21}{16}}

    D_y\:=\:\begin{vmatrix}4&1\\1&\text{-}4\end{vmatrix}\:=\:\text{-}16 - 1\:=\:-17\quad\Rightarrow\quad \boxed{y \,= \,-\frac{17}{16}}


    the vertices of a triangle are given below. .Use a determinant
    and the vertices of the triangle to find the area of the triangle.
    . . . . . (-2.5,\;3)\;\;(3,\;7.5)\;\;(5,\;5)

    The area of a triangle with vertices (x_1,y_1),\;(x_2,y_2),\;(x_3,y_3) is given by:

    . . . A \;= \;\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}

    So we have: . A \;= \;\frac{1}{2}\begin{vmatrix}\text{-}2.5 & 3 & 1 \\ 3 & 7.5 & 1 \\ 5 & 5 & 1\end{vmatrix}

    I'll let you crank it out . . .

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  4. #4
    Grand Panjandrum
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    2. Find the determinant of:

    <br />
A=\left [ \begin{array}{ccc}<br />
{4}& {3}&{2}\\<br />
{5}& {2}&{4}\\<br />
{1}&{3}&{1}<br />
\end{array} \right]<br />

    Let's do this by expanding about the first row:

    <br />
|A|=A_{1,1}\ C_{1,1}+A_{1,2}\ C_{1,2}+A_{1,3}\ C_{1,3}= <br />
4\ \left | \begin{array}{cc}<br />
{2}& {4}\\<br />
{3}& {1}<br />
\end{array} \right|- <br />
3\ \left | \begin{array}{cc}<br />
{5}& {4}\\<br />
{1}& {1}<br />
\end{array} \right|+<br />
2\ \left | \begin{array}{cc}<br />
{5}& {2}\\<br />
{1}& {3}<br />
\end{array} \right|<br />
,

    where C_{i,j} is the i,jth cofactor od A

    RonL
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