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Math Help - Help ME Please!!!

  1. #1
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    Help ME Please!!!

    questions on attachment!
    Attached Thumbnails Attached Thumbnails Help ME Please!!!-gash.jpg  
    Last edited by CaptainBlack; July 23rd 2006 at 09:15 PM.
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  2. #2
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    problem nr. 4 only

    Quote Originally Posted by Lane
    questions on attachment!
    Hello, Lane,

    use Cramer's rule:

    \left|\begin{array}{cc}5&9\\7&-10\end{array}\right|=5\cdot (-10)-9\cdot7=-113. So it's answer C.

    Greetings

    EB
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  3. #3
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    problem nr. 5 only

    Quote Originally Posted by Lane
    questions on attachment!
    Hello, Lane,


    it's me again.
    Use the same method as in prob. nr. 4.:

    \left|\begin{array}{cc}x-9&2\\3&x-4\end{array}\right|=0

    (x-9)(x-4)-6=0\ \Longleftrightarrow\ x^2-13x+30=0. Solving this quadratic equation you should get x= 3 or x = 10.

    Greetings

    EB
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  4. #4
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    problem nr. 2 only

    Quote Originally Posted by Lane
    questions on attachment!
    Hello, Lane

    (I assume that there is a typo in the first row)

    \begin{array}{r}x+3y+z=-12\\2x+7y-2z=-25\\x-y-6z=15\end{array}

    Calculate: R2-2*R1 and R3-R1. You'll get:

    \begin{array}{r}x+3y+z=-12\\y-4z=-1\\-4y-7z=27\end{array}

    Now calculate R3+4*R2:

    \begin{array}{r}x+3y+z=-12\\y-4z=-1\\-23z=23\end{array}

    So you get z = -1. Now re-substitute this value into equation 2 and afterward plug in the values for y and z into the first equation. You should get: [4, -5, -1]

    Greetings

    EB
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  5. #5
    Grand Panjandrum
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    Problem #3 only

    The augmented matrix is equivalent to the system of equations:

    <br />
\begin{array} {ccccc}<br />
1x&+0y&-1z&=&-5\\<br />
0x&+1y&-3z&=&4\\<br />
0x&+0y&+0z&=&0<br />
\end{array}<br />

    Now check the candidate solutions to see which satisfy these equations.

    The first equation is satisfied by those candidate solutions with x=t-5,\ z=t, and the second by those candidates with y=3t+4,\ z=t, so it looks like (C) is the required answer.

    RonL
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  6. #6
    Grand Panjandrum
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    Problem #1 only

    Let:

    <br />
M=\left[<br />
\begin{array}{ccc}<br />
3&10&-3\\<br />
1&5&-9\\<br />
-4&-7&1<br />
\end{array} \right]<br />

    A: Swap rows 1 and 2 gives:

    <br />
M'=\left[<br />
\begin{array}{ccc}<br />
1&5&-9\\<br />
3&10&-3\\<br />
-4&-7&1<br />
\end{array} \right]<br />

    B: Replace row 2 by row 2 plus -3 times row 1

    <br />
M''=\left[<br />
\begin{array}{ccc}<br />
1&5&-9\\<br />
3-3&10-15&-3+27\\<br />
-4&-7&1<br />
\end{array} \right]= <br />
\left[<br />
\begin{array}{ccc}<br />
1&5&-9\\<br />
0&-5&25\\<br />
-4&-7&1<br />
\end{array} \right]<br />

    I am sure you can now do part C yourself.

    RonL
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