Question

Use Cramer’s Rule to Solve Systems of Equations

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

$\left\{\begin{array}{l}3x-z=-3\\ 5y+2z=-6\\ 4x+3y=-8\end{array}\right.$

Short answer

The solution for the system of linear equation is,$\left(-2,0,-3\right)$.

Step-by-step solution

Given system of linear equations is,

$\left\{\begin{array}{l}3x-z=-3\\ 5y+2z=-6\\ 4x+3y=-8\end{array}\right.$

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

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

$$\begin{gathered} D=\left|\begin{array}{ccc}3& 0& -1\\ 0& 5& 2\\ 4& 3& 0\end{array}\right| \\ =3\left(0-6\right)-0\left(0-8\right)-1\left(0-20\right) \\ =3\left(-6\right)-0-1\left(-20\right) \\ =-18+20 \\ =2 \end{gathered}$$

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

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-3& 0& -1\\ -6& 5& 2\\ -8& 3& 0\end{array}\right| \\ =\left(-3\right)\left(0-6\right)-0\left(0+16\right)-1\left(-18+40\right) \\ =\left(-3\right)\left(-6\right)-0\left(16\right)-1\left(22\right) \\ =18-22 \\ =-4 \end{gathered}$$

To find: $x$

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{-4}{2} \\ x=-2 \end{gathered}$$

Step 4 Find the determinant Dy by using the constants to replace the coefficients of y.

$$\begin{gathered} D_y=\left|\begin{array}{ccc}3& -3& -1\\ 0& -6& 2\\ 4& -8& 0\end{array}\right| \\ =3\left(0+16\right)+3\left(0-8\right)-1\left(0+24\right) \\ =3\left(16\right)+3\left(-8\right)-1\left(24\right) \\ =48-24-24 \\ =0 \end{gathered}$$

To find: $y$.

$$\begin{gathered} y=\frac{D_y}{D} \\ =\frac{0}{2} \\ =0 \end{gathered}$$

Step 5 Find the determinant Dz by using the constants to replace the coefficients of z.

$$\begin{gathered} D_z=\left|\begin{array}{ccc}3& 0& -3\\ 0& 5& -6\\ 4& 3& -8\end{array}\right| \\ =3\left(-40+18\right)-0-3\left(0-20\right) \\ =3\left(-22\right)-3\left(-20\right) \\ =-66+60 \\ =-6 \end{gathered}$$

To find: $z$.

$$\begin{gathered} z=\frac{D_z}{D} \\ =\frac{-6}{2} \\ =-3 \end{gathered}$$

So, the solution for the system of linear equation is,$\left(-2,0,-3\right)$.

Step 6 write the solution as an ordered triad.

The ordered triad is, $\left(x,y,z\right)=\left(-2,0,-3\right)$.

Substitute $x=-2,y=0,z=-3$into the equation $3x-z=-3$.

$$\begin{gathered} 3\left(-2\right)-\left(-3\right)=-3 \\ -6+3=-3 \\ -3=-3 \end{gathered}$$

which is true.

Step 7 substitute x=-2,y=0,z=-3 into the equation 5y+2z=-6.

$$\begin{gathered} 5\left(0\right)+2\left(-3\right)=-6 \\ 0-6=-6 \\ -6=-6 \end{gathered}$$

which is true.