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Math Help - Cramer's Rule

  1. #1
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    Cramer's Rule

    Hi, I am new to the forum. I am having trouble with solving three simultaneous equations using Cramer's Rule. I have forgotten the method.
    I know the way I used to do it was to take a column outside and label it Q P and R using the cover-up method. Not too sure were to go from there.

    Any help would be greatly appreciated.

    Here is the question:

    Use Cramer's rule to solve the following,

    3Y + Z + 2X = 10

    X - Y + Z -4 = 0

    4X + 8 = Y + 5Z
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  2. #2
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    Quote Originally Posted by Chris1981 View Post
    Hi, I am new to the forum. I am having trouble with solving three simultaneous equations using Cramer's Rule. I have forgotten the method.
    I know the way I used to do it was to take a column outside and label it Q P and R using the cover-up method. Not too sure were to go from there.

    Any help would be greatly appreciated.

    Here is the question:

    Use Cramer's rule to solve the following,

    3Y + Z + 2X = 10

    X - Y + Z -4 = 0

    4X + 8 = Y + 5Z
    First you transform the system to standard form like this

    \begin{array}{rcrcrcl}<br />
2X &+& 3Y &+& Z &=& {\color{blue}10}\\<br />
X &-& Y &+& Z &=& {\color{blue}4}\\<br />
4X &-& Y &-& 5Z &=& {\color{blue}-8}\\<br />
\end{array}

    The determinant of the coefficient matrix of this system is
    \begin{vmatrix} 2 & 3 & 1\\<br />
1 & -1 & 1\\<br />
4 & -1 & -5<br />
\end{vmatrix} = \ldots = 42<br />

    Since this is \neq 0 the system has a unique solution. To get the solution you plug in the right-hand side of the system into the corresponding column of the coefficient matrix and divide by the determinant of the coefficient matrix. For example to get the uniquely determined value of X, you replace the coefficients of X in the coefficient matrix with the right-hand side of the above system, like this

    X=\frac{\begin{vmatrix} {\color{blue}10} & 3 & 1\\<br />
{\color{blue}4} & -1 & 1\\<br />
{\color{blue}-8} & -1 & -5<br />
\end{vmatrix}}{\begin{vmatrix} 2 & 3 & 1\\<br />
1 & -1 & 1\\<br />
4 & -1 & -5<br />
\end{vmatrix}}=\ldots = 2

    Similarly for Y and Z.
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  3. #3
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    Hello, Chris1981!

    Use Cramer's rule to solve: . \begin{array}{c}3y + z + 2x \:=\: 10 \\ x - y + z -4 \:=\: 0 \\ 4x + 8 = y + 5z\end{array}
    Write the equations in standard form: . \begin{array}{ccc}2x + 3y + z &=& 10 \\ x - y + z &=& 4 \\ 4x - y - 5x &=& \text{-}8 \end{array}
    And we have: . \left[\begin{array}{ccc|c}2&3&1&10 \\ 1&\text{-}1&1& 4 \\ 4&\text{-}1&\text{-}5 & \text{-}8 \end{array}\right]



    [1] Find D, the determinant of the coefficients.

    . . D \;=\;\left|\begin{array}{ccc}2&3&1 \\ 1&\text{-}1 & 1 \\ 4 & \text{-}1 & \text{-}5 \end{array}\right| \;=\;2(5+1) - 3(\text{-}5-4) +1(\text{-}1+4) \;=\;12 + 27 + 3 \;=\;42


    [2] Find D_x. . Replace x-coefficients with the constants.

    . . D_x \;=\;\left|\begin{array}{ccc}{\color{blue}10}&3&1 \\ {\color{blue}4}&\text{-}1&1\\ {\color{blue}\text{-}8}&\text{-}1&\text{-}5\end{array}\right| \;=\; 10(5+1)-3(\text{-}20+8) + 1(\text{-}4-8) \;=\;84

    . . Divide by D\!:\;\;x \;=\;\frac{D_x}{D} \;=\;\frac{84}{42} \quad\Rightarrow\quad \boxed{x \;=\;2}



    [3] Find D_y . . Replace y-coefficients with the constants.
    . . Divide by D\!:\;\;y \;=\;\frac{D_y}{D}


    [4] Find D_z . . Replace z-coefficients with the constants.
    . . Divide by D\!:\;\;z \;=\;\frac{D_z}{D}

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  4. #4
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    Hi, thanks for the quick response. Thats helped me a lot, its coming back to me now but there was bits I'd forgotten.

    I've got re-sits for my Mech Engineering degree coming up next month, so I'm getting it all together in my head.


    thanks again!!
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