Cramer's Rule 3 x 3 Lesson

In this lesson, we will learn about Cramer's rule for solving a 3 x 3 linear system.

Cramer's Rule for a 3 x 3 Linear System

$$a_{1}x + b_{1}y + c_{1}z=d_{1}$$ $$a_{2}x + b_{2}y + c_{2}z=d_{2}$$ $$a_{3}x + b_{3}y + c_{3}z=d_{3}$$ First, we will set up D, this will represent the determinant of the coefficient matrix. $$D=\left| \begin{array}{ccC}a_{1}&b_{1}&c_{1}\\ a_{2}& b_{2}& c_{2}\\ a_{3}& b_{3}& c_{3}\end{array}\right|$$ Next, we will set up D_x. We replace the column from D with the x coefficients with the constants. $$D_{x}=\left| \begin{array}{ccc}d_{1}&b_{1}&c_{1}\\ d_{2}& b_{2}& c_{2}\\ d_{3}& b_{3}& c_{3}\end{array}\right|$$ Next, we will set up D_y. We replace the column from D with the y coefficients with the constants. $$D_{y}=\left| \begin{array}{ccc}a_{1}&d_{1}&c_{1}\\ a_{2}& d_{2}& c_{2}\\ a_{3}& d_{3}& c_{3}\end{array}\right|$$ Next, we will set up D_z. We replace the column from D with the z coefficients with the constants. $$D_{z}=\left| \begin{array}{ccc}a_{1}&b_{1}&d_{1}\\ a_{2}& b_{2}& d_{2}\\ a_{3}& b_{3}& d_{3}\end{array}\right|$$ As long as D is not 0, we can find our solution for the system as: $$x=\frac{D_{x}}{D}$$ $$y=\frac{D_{y}}{D}$$ $$z=\frac{D_{z}}{D}$$ Let's look at an example.
Example #1: Solve each system. $$-2x + 4y - 4z=16$$ $$-4x + 6y - 5z=18$$ $$-3x - 2y - 4z=28$$ First, let's find D, the determinant of the coefficient matrix. $$D=\left| \begin{array}{ccc}-2&4&-4\\ -4&6&-5\\-3&-2&-4\end{array}\right|=-40$$ Next, let's find D_x. We will replace the column from D with the x coefficients with the constants. $$D_{x}=\left| \begin{array}{ccc}16&4&-4\\ 18&6&-5\\28&-2&-4\end{array}\right|=0$$ Next, let's find D_y. We will replace the column from D with the y coefficients with the constants. $$D_{y}=\left| \begin{array}{ccc}-2&16&-4\\ -4&18&-5\\-3&28&-4\end{array}\right|=80$$ Next, let's find D_z. We will replace the column from D with the z coefficients with the constants. $$D_{z}=\left| \begin{array}{ccc}-2&4&16\\ -4&6&18\\-3&-2&28\end{array}\right|=240$$ Now we are ready to find the solution for our system. $$x=\frac{D_{x}}{D}=\frac{0}{-40}=0$$ $$y=\frac{D_{y}}{D}=\frac{80}{-40}=-2$$ $$z=\frac{D_{z}}{D}=\frac{240}{-40}=-6$$ Our solution for the system: $$(0, -2, -6)$$

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