Q 261 - 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.

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

Step-by-Step Solution

Answer

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

Step 1 Given system of linear equation is,

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

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

Step 2 Find the determinants D by using the coefficients of the variable.

$$\begin{gathered} D=\left|\begin{array}{ccc}2& 5& 0\\ 0& 3& -1\\ 4& 0& 3\end{array}\right| \\ =2\left(9-0\right)-5\left(0+4\right)+0 \\ =18-20 \\ =-2 \end{gathered}$$

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

$$\begin{gathered} D_x=\left|\begin{array}{ccc}4& 5& 0\\ 3& 3& -1\\ -3& 0& 3\end{array}\right| \\ =4\left(9-0\right)-5\left(9-3\right)+0 \\ =4\left(9\right)-5\left(6\right) \\ =36-30 \\ =6 \end{gathered}$$

To find: $x$,

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{6}{-2} \\ x=-3 \end{gathered}$$

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

$$\begin{gathered} D_y=\left|\begin{array}{ccc}2& 4& 0\\ 0& 3& -1\\ 4& -3& 3\end{array}\right| \\ =2\left(9-3\right)-4\left(0+4\right)+0 \\ =2\left(6\right)-4\left(4\right) \\ =12-16 \\ =-4 \end{gathered}$$

To find: $y$,

$$\begin{gathered} y=\frac{D_y}{D} \\ =\frac{-4}{-2} \\ =2 \end{gathered}$$

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

$$\begin{gathered} D_z=\left|\begin{array}{ccc}2& 5& 4\\ 0& 3& 3\\ 4& 0& -3\end{array}\right| \\ =2\left(-9-0\right)-5\left(0-12\right)+4\left(0-12\right) \\ =2\left(-9\right)-5\left(-12\right)+4\left(-12\right) \\ =-18+60-48 \\ =-6 \end{gathered}$$

To find: $z$,

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

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

Step 6 Write the solution as an ordered triad.

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

Substitute $x=-3,y=2,z=3$ into the equation $2x+5y=4$.

 $$\begin{gathered} 2\left(-3\right)+5\left(2\right)=4 \\ -6+10=4 \\ 4=4 \end{gathered}$$

which is true.

Step 7 Substitute x = - 3 , y = 2 , z = 3 into the equation 3 y - z = 3 .

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

which is true.