Chapter 4.5, Problem 24CYU

To determine

To find: The solution of equations using Cramer’s rule.

Answer to Problem 24CYU

The solution of the system is $\left(\frac{-375}{-127},\frac{-150}{-127},\frac{135}{-127}\right)$.

Explanation of Solution

Given,

The set of equation is given as,

  $\begin{array}{l}2n+3p-4w=20\\ 4n-p+5w=13\\ 3n+2p+4w=15\end{array}$

Calculation:

Calculate the determinant of n .

  $\begin{array}{c}\left|n\right|=\left|\begin{array}{ccc}20& 3& -4\\ 13& -1& 5\\ 15& 2& 4\end{array}\right|\\ =20\left(-4-10\right)-3\left(52-75\right)-4\left(26+15\right)\\ =-375\end{array}$

Calculate the determinant of p .

  $\begin{array}{c}\left|p\right|=\left|\begin{array}{ccc}2& 20& -4\\ 4& 13& 5\\ 3& 15& 4\end{array}\right|\\ =2\left(52-75\right)-20\left(16-15\right)-4\left(60-39\right)\\ =-150\end{array}$

Calculate the determinant of w .

  $\begin{array}{c}\left|w\right|=\left|\begin{array}{ccc}2& 3& 20\\ 4& -1& 13\\ 3& 2& 15\end{array}\right|\\ =2\left(-15-26\right)-3\left(60-39\right)+20\left(8+3\right)\\ =135\end{array}$

Calculate the determinant C .

  $\begin{array}{c}\left|C\right|=\left|\begin{array}{ccc}2& 3& -4\\ 4& -1& 5\\ 3& 2& 4\end{array}\right|\\ =2\left(-4-10\right)-3\left(16-15\right)-4\left(8+3\right)\\ =-127\end{array}$

Calculate the value of x, yandz.

  $\begin{array}{c}x=\frac{-375}{-127}\\ y=\frac{-150}{-127}\\ z=\frac{135}{-127}\end{array}$

Thus, the solution of the system is $\left(\frac{-375}{-127},\frac{-150}{-127},\frac{135}{-127}\right)$ .