~~DISCUSSION|3×3 Number Discussion~~
~~TOC~~
Lab46 Tutorials
Solving 3 x 3 Number Systems
An Exploration of Methods
Eq1→ <latex>3x - 3y + 2z = 11</latex>
Eq2→ <latex>3x + y - 3z = 11</latex>
Eq3→ <latex>15x - y - 2z = 0</latex>
Use the addition method to remove one of the variables from the equation. To do this, choose two of the equations and multiply each by any given multiple, such that each equation contains opposing (complimentary) coefficients for at least one variable. This step may, in some situations, cause two variables to drop out if you're lucky. In this case, multiplying <latex>3x + y - 3z = 11</latex> by <latex>-1</latex> will result in <latex>-3x - y + 3z</latex>. Then, taking the sum of <latex>3x - 3y + 2z = 11</latex> and <latex>-3x - y + 3z</latex>, results in <latex>-4y + 5z = 0</latex>, an equation with one less variable.
Eq1 + (-1)Eq2
<latex>
\begin{array}{3}
-1(3x + y - 3z = 11)
\underline{+ 3x - 3y + 2z = 11}
-4y + 5z = 0
\end{array}
</latex>
At first, it may seem that you're only multiplying one equation by -3 and leaving the other alone, but it's important to note that leaving an equation alone is the same as multiplying it by 1.
Use the addition method to remove the same variable from a different set of equations. To do this, choose two different equations and multiply each by a given multiple, such that each equation contains opposing coefficients for at least one variable. In this case, adding the value of <latex>3x - 3y + 2z = 11</latex> multiplied by <latex>-5</latex> to <latex>15x - y - 2z = 0</latex> produces a result of <latex>14y - 12z = -55</latex>.
(-5)Eq1 + Eq3
<latex>
\begin{array}{3}
-5(3x - 3y + 2z = 11)
\underline{+ 15x - y - 2z = 0}
14y - 12z = -55
.\end{array}
</latex>
Two new equations have been introduced. Use these new equations to find the missing variable.
Eq4 → <latex>-4y + 5z = 0 </latex>
Eq5 → <latex>14y - 12z = -55</latex>
Create 3 2 x 2 martices as follows:
<latex>
y_1=-4 , y_2=14
</latex>
<latex>
z_1=5 , z_2=-12
</latex>
<latex>
C_1=0 , C_2=-55
</latex>
<latex>
D
= \left|
\begin{array}{ccc}
y_1 … z_1
\vdots \ddots \vdots
y_2 … z_2 \end{array}
\right|
</latex>
<latex>
D_y
= \left|
\begin{array}{ccc}
C_1 … z_1
\vdots \ddots \vdots
C_2 … z_2\end{array}
\right| \to y = \frac{D_y}{D} \to y = \frac{(0)(-12)-(-55)(5)}{(-4)(-12)-(14)(5)} \to y = \frac{25}{2}
</latex>
<latex>
D_z
= \left|
\begin{array}{ccc}
y_1 … C_1
\vdots \ddots \vdots
y_2 … C_2\end{array}
\right| \to z = \frac{D_z}{D} \to y = \frac{(-4)(-55)-(14)(0)}{(-4)(-12)-(14)(5)} \to z = -10
</latex>
Now that <latex> y </latex> and <latex> z </latex> have been determined, place these values into one of the initial equations to find the value of <latex> x </latex>.
<latex>
y = \frac{25}{2}
z = -10
</latex>
<latex>
3x - 3y + 2z = 11 \to 3x - 3(\frac{25}{2}) + 2(-10) = 11 \to x = \frac{3(\frac{25}{2}) + 2(-10) - 11}{3}
x = \frac{-13}{6}
</latex>
General Form
<latex>a_1x + b_1y + c_1z = d_1</latex>
<latex>a_2x + b_2y + c_2z = d_2</latex>
<latex>a_3x + b_3y + c_3z = d_3</latex>
Equations given in general form can be solved using determinants with the following formulas:
<latex>
x= \frac{d_1b_2c_2 + d_3b_1c_2 + d_2b_3c_1 - d_3b_2c_1 - d_1b_3c_2 - d_2b_1c_3}{a_1b_2c_3 + a_3b_1c_2 + a_2b_3c_1 - a_3b_2c_1 - a_1b_2c_3 - a_2b_1c_3}
y= \frac{a_1d_2c_3 + a_3d_1c_2 + a_2d_3c_1 - a_3d_2c_1 - a_1d_3c_2 - a_2d_1c_3}{a_1b_2c_3 + a_3b_1c_2 + a_2b_3c_1 - a_3b_2c_1 - a_1b_2c_3 - a_2b_1c_3}
z= \frac{a_1b_2d_3 + a_3b_1d_2 + a_2b_3d_1 - a_3b_2d_1 - a_1b_3d_2 - a_2b_1d_3}{a_1b_2c_3 + a_3b_1c_2 + a_2b_3c_1 - a_3b_2c_1 - a_1b_2c_3 - a_2b_1c_3}
</latex>