Question

Use Cramer’s Rule to Solve Systems of Equations.

In the following exercises, solve each system of equations using Cramer’s Rule.

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

Short answer

The system is consistent and dependent and it has infinitely many solutions.

Step-by-step solution

Given system of equations are,

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

The objective is , we need to solve the system of equations by using the Cramer's rule.

Step 2 Find the determinant D by using the coefficients of the variables.

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 1& -3\\ 0& 1& -1\\ -1& 2& 0\end{array}\right| \\ =1\left(0+2\right)-1\left(0-1\right)-3\left(0+1\right) \\ =1\left(2\right)-1\left(-1\right)-3\left(1\right) \\ =2+1-3 \\ =0 \end{gathered}$$

Here we cannot use the Cramer's rule to solve this system. But by looking at the determinants $D_x,D_y,D_z$, we can determine whether the given system is inconsistent or dependent.

Step 3 Evaluate the determinant Dx by using the constants to replace the coefficients of x.

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-1& 1& -3\\ 0& 1& -1\\ 1& 2& 0\end{array}\right| \\ =-1\left(0+2\right)-1\left(0+1\right)-3\left(0-1\right) \\ =-1\left(2\right)-1\left(1\right)-3\left(-1\right) \\ =-2-1+3 \\ =0 \end{gathered}$$

Step 4 Evaluate the determinant Dyby using the constants to replace the coefficients of y.

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& -1& -3\\ 0& 0& -1\\ -1& 1& 0\end{array}\right| \\ =1\left(0+1\right)+1\left(0-1\right)-3\left(0-0\right) \\ =1\left(1\right)+1\left(-1\right)-3\left(0\right) \\ =1-1-0 \\ =0 \end{gathered}$$

Step 5 Evaluate the determinant Dzby using the constants to replace the coefficients of z .

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 1& -1\\ 0& 1& 0\\ -1& 2& 1\end{array}\right| \\ =1\left(1-0\right)-1\left(0-0\right)-1\left(0+1\right) \\ =1\left(1\right)-1\left(0\right)-1\left(1\right) \\ =1-0-1 \\ =0 \end{gathered}$$

Here all the determinants are equal to zero.

So, the system is consistent and dependent and it has infinitely many solutions.