11.8 Solving systems with cramer's rule  (Page 3/11)

Augment the matrix with the first two columns and then follow the formula. Thus,

Use Cramer’s Rule.

Then,

The solution is $\left(1,3,\mathrm{-2}\right).$

We begin by finding the determinants $D,D_x,\text{and }{D}_y.$

We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables.

1. Multiply equation (1) by $\mathrm{-2.}$
2. Add the result to equation $\left(2\right).$

We obtain the equation $0=\mathrm{-8},$ which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See [link] .

Let’s find the determinant first. Set up a matrix augmented by the first two columns.

Then,

As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out.

1. Multiply equation (1) by $\mathrm{-2}$ and add the result to equation (3):
   $$\frac{\begin{array}{r}\hfill -2x+4y-6x=0\\ \hfill 2x-4y+6z=0\end{array}}{0=0}$$
2. Obtaining an answer of $0=0,$ a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See [link] .