Q 262 - Intermediate Algebra

Question

Use Cramer’s Rule to Solve Systems of Equations 

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

$\left\{\begin{array}{l}2y + 3z = -1 \\ 5x + 3y = -6 \\ 7x + z = 1\end{array}\right.$ 

Step-by-Step Solution

Answer

The solution for the system of linear equation is,$\left(0,-2,1\right)$.

Step 1 Given system of linear equation is,

$\left\{\begin{array}{l}2y + 3z = -1 \\ 5x + 3y = -6 \\ 7x + z = 1\end{array}\right.$

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

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

$$\begin{gathered} D=\left|\begin{array}{ccc}0& 2& 3\\ 5& 3& 0\\ 7& 0& 1\end{array}\right| \\ =0-2\left(5-0\right)+3\left(0-21\right) \\ =0-2\left(5\right)+3\left(-21\right) \\ =-10-63 \\ =-73 \end{gathered}$$

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

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-1& 2& 3\\ -6& 3& 0\\ 1& 0& 1\end{array}\right| \\ =\left(-1\right)\left(3-0\right)-2\left(-6-0\right)+3\left(0-3\right) \\ =-3+12-9 \\ =0 \end{gathered}$$

To find: $x$

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{0}{-73} \\ =0 \end{gathered}$$

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

$$\begin{gathered} D_y=\left|\begin{array}{ccc}0& -1& 3\\ 5& -6& 0\\ 7& 1& 1\end{array}\right| \\ =0-\left(-1\right)\left(5-0\right)+3\left(5+42\right) \\ =5+3\left(47\right) \\ =146 \end{gathered}$$

To find: $y$

$$\begin{gathered} y=\frac{D_y}{D} \\ =\frac{146}{-73} \\ y=-2 \end{gathered}$$

Step 5 Find the determinants D z by using the constants to replace the coefficients of z .

$$\begin{gathered} D_z=\left|\begin{array}{ccc}0& 2& -1\\ 5& 3& -6\\ 7& 0& 1\end{array}\right| \\ =0-2\left(5+42\right)-1\left(0-21\right) \\ =-2\left(47\right)-1\left(-21\right) \\ =-94+21 \\ =-73 \end{gathered}$$

To find: $z$

$$\begin{gathered} z=\frac{D_z}{D} \\ =\frac{-73}{-73} \\ z=1 \end{gathered}$$

The solution for the system of linear equation is,$\left(0,-2,1\right)$.

Step 6 write the solution as an ordered triad.

The ordered triad is, $\left(x,y,z\right)=\left(0,-2,1\right)$.

Substitute $x=0,y=-2,z=1$ into the equation $2y+3z=-1$.

 $$\begin{gathered} 2\left(-2\right)+3\left(1\right)=-1 \\ -4+3=-1 \\ -1=-1 \end{gathered}$$

which is true.

Step 7 substitute x = 0 , y = - 2 , z = 1 into the equation 7 x + z = 1 .

$$\begin{gathered} 7\left(0\right)+1=1 \\ 0+1=1 \\ 1=1 \end{gathered}$$

which is true.