Q 263

Question

Use Cramer’s Rule to Solve Systems of Equations

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

$\{\begin{cases} 3 x - z = - 3 \\ 5 y + 2 z = - 6 \\ 4 x + 3 y = - 8 \end{cases}$

Step-by-Step Solution

Verified

Answer

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

1Step 1 Given system of linear equations is,

$\{\begin{cases} 3 x - z = - 3 \\ 5 y + 2 z = - 6 \\ 4 x + 3 y = - 8 \end{cases}$

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

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

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

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

$D_{x} = |\begin{matrix} - 3 & 0 & - 1 \\ - 6 & 5 & 2 \\ - 8 & 3 & 0 \end{matrix}| = (- 3) (0 - 6) - 0 (0 + 16) - 1 (- 18 + 40) = (- 3) (- 6) - 0 (16) - 1 (22) = 18 - 22 = - 4$

To find: $x$

$x = \frac{D_{x}}{D} = \frac{- 4}{2} x = - 2$

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

$D_{y} = |\begin{matrix} 3 & - 3 & - 1 \\ 0 & - 6 & 2 \\ 4 & - 8 & 0 \end{matrix}| = 3 (0 + 16) + 3 (0 - 8) - 1 (0 + 24) = 3 (16) + 3 (- 8) - 1 (24) = 48 - 24 - 24 = 0$

To find: $y$ .

$y = \frac{D_{y}}{D} = \frac{0}{2} = 0$

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

$D_{z} = |\begin{matrix} 3 & 0 & - 3 \\ 0 & 5 & - 6 \\ 4 & 3 & - 8 \end{matrix}| = 3 (- 40 + 18) - 0 - 3 (0 - 20) = 3 (- 22) - 3 (- 20) = - 66 + 60 = - 6$