Solving System of Equations Using Cramer's Rule - Problem #42 | Jan 14th, 2013

[Jameson]

Consider the three equtions:

[math]x-3y+3z=-4[/math]
[math]2x+3y-z=15[/math]
[math]4x-3y-z=19[/math]

Using Cramer's Rule, solve the system. Show your work, especially how you calculate the determinants.
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) soroban
3) Sudharaka

Solution (from soroban): [sp] Solve by Cramer's Rule: .[/color]$\begin{array}{ccc}x - 3y + 3 &=& \text{-}4 \\ 2x + 3y - z &=& 15 \\ 4x - 3y - z &=& 19 \end{array}$$D \:=\:\begin{vmatrix}1&\text{-}3&3 \\ 2&3&\text{-}1 \\ 4&\text{-}3&\text{-}1 \end{vmatrix} \;=\;1\begin{vmatrix}3&\text{-}1\\\text{-}3&\text{-}1\end{vmatrix} - (\text{-}3)\begin{vmatrix}2&\text{-}1 \\ 4&\text{-}1\end{vmatrix} + 3\begin{vmatrix}2&3\\4&\text{-}3\end{vmatrix} $

. . [/color]$=\;1(\text{-}3-3) + 3(\text{-}2+4) + 3(\text{-}6-12) \;=\;1(\text{-}6) + 3(2) + 3(\text{-}18)$

. . [/color]$=\;\text{-}6 + 6 - 54 \quad\Rightarrow\quad \boxed{D \:=\:\text{-}54} $$D_x \;=\;\begin{vmatrix}\text{-}4&\text{-}3&3 \\ 15 & 3&\text{-}1 \\ 19&\text{-}3&\text{1}1\end{vmatrix} \;=\;\text{-}4\begin{vmatrix}3&\text{-}1\\\text{-}3&\text{-}1\end{vmatrix} - (\text{-}3)\begin{vmatrix}15&\text{-}1\\19&\text{-}1\end{vmatrix} + 3\begin{vmatrix}15&3\\19&\text{-}3\end{vmatrix}$

. . . [/color]$=\;\text{-}4(\text{-}3-3) + 34(\text{-}15 + 19) + 3(\text{-}45-57)$

. . . [/color]$=\; \text{-}4(\text{-}6) + 3(4) + 3(\text{-}102) \;=\; 24 + 12 - 306 \;=\;\text{-}270$

$x \;=\;\dfrac{D_x}{D} \;=\;\dfrac{\text{-}270}{\text{-}54} \quad\Rightarrow\quad \boxed{x \:=\:5}$$D_y \;=\;\begin{vmatrix} 1&\text{-}4&3 \\ 2 &15&\text{-}1 \\ 4&19&\text{-}1\end{vmatrix} \;=\;1\begin{vmatrix}15&\text{-}1\\19&\text{-}1\end{vmatrix} - (\text{-}4)\begin{vmatrix}2&\text{-}1\\4&\text{-}1\end{vmatrix} + 3\begin{vmatrix}2&15\\4&19\end{vmatrix}$

. . . [/color]$=\;1(\text{-}15+19) + 4(\text{-}2 + 4) + 3(38-60)$

. . . [/color]$=\; 1(4) + 4(2) + 3(\text{-}22) \;=\; 4 + 8 - 66 \;=\;\text{-}54$

$y \;=\;\dfrac{D_y}{D} \;=\;\dfrac{\text{-}54}{\text{-}54} \quad\Rightarrow\quad \boxed{y \:=\:1}$ $D_z \;=\;\begin{vmatrix}1&\text{-}3&\text{-}4 \\ 2&3&15 \\ 4&\text{-}3&19\end{vmatrix} \;=\; 1\begin{vmatrix}3&15\\\text{-}3&19
\end{vmatrix} - (\text{-}3)\begin{vmatrix}2&15\\4&19\end{vmatrix} - 4\begin{vmatrix}2&3\\4&\text{-}3\end{vmatrix}$

. . . [/color]$=\;1(57+45) + 3(38-60) - 4(\text{-}6-12)$

. . . [/color]$=\; 1(102) + 3(\text{-}22) - 4(\text{-}18) \;=\; 102 - 66 + 72 \;=\;108$

$z \;=\;\dfrac{D_z}{D} \;=\;\dfrac{108}{\text{-}54} \quad\Rightarrow\quad \boxed{z \:=\:\text{-}2}$
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