Chapter 3.7, Problem 24CYU

To determine

To solve: the system of equations using Cramer’s rule.

Answer to Problem 24CYU

  $\left(n,\text{ p},\text{ w}\right)=\left(5,\text{ 2},\text{ -1}\right)$

Explanation of Solution

Given:

System of equations:

  $2n+3p-4w=20$

  $4n-p+5w=13$

  $3n+2p+4w=15$

Calculation:

The given set of equations can be represented in the matrix form as,

  $\left[\begin{array}{ccc}2& 3& -4\\ 4& -1& 5\\ 3& 2& 4\end{array}\right]\left[\begin{array}{c}n\\ p\\ w\end{array}\right]=\left[\begin{array}{c}20\\ 13\\ 15\end{array}\right]$  $\left(1\right)$

The coefficient matrix, $\Delta =\left[\begin{array}{ccc}2& 3& -4\\ 4& -1& 5\\ 3& 2& 4\end{array}\right]$

Determinant of coefficient matrix is,

  $\left|\Delta \right|=\left|\begin{array}{ccc}2& 3& -4\\ 4& -1& 5\\ 3& 2& 4\end{array}\right|=2\left|\begin{array}{cc}-1& 5\\ 2& 4\end{array}\right|-4\left|\begin{array}{cc}3& -4\\ 2& 4\end{array}\right|+3\left|\begin{array}{cc}3& -4\\ -1& 5\end{array}\right|$

  $=2\left(-4-10\right)-4\left(12+8\right)+3\left(15-4\right)$

  $=2\left(-14\right)-4\left(20\right)+3\left(11\right)$

  $=-28-80+33$

  $=-75$

The n-matrix ${\Delta }_n$ can be obtained by replacing the first column of $\Delta$ by the column of numbers in the right hand side of the matrix equation $\left(1\right)$ ,

  ${\Delta }_n=\left[\begin{array}{ccc}2& 3& -4\\ 4& -1& 5\\ 3& 2& 4\end{array}\right]$

Determinant of ${\Delta }_n$ is,

  $\left|{\Delta }_n\right|=\left|\begin{array}{ccc}20& 3& -4\\ 13& -1& 5\\ 15& 2& 4\end{array}\right|=20\left|\begin{array}{cc}-1& 5\\ 2& 4\end{array}\right|-13\left|\begin{array}{cc}3& -4\\ 2& 4\end{array}\right|+15\left|\begin{array}{cc}3& -4\\ -1& 5\end{array}\right|$

  $=20\left(-4-10\right)-13\left(12+8\right)+15\left(15-4\right)$

  $=20\left(-14\right)-13\left(20\right)+15\left(11\right)$

  $=-280-260+165$

  $=-375$

The p-matrix ${\Delta }_p$ can be obtained by replacing the second column of $\Delta$ by the column of numbers in the right hand side of the matrix equation $\left(1\right)$ ,

  ${\Delta }_p=\left[\begin{array}{ccc}2& 20& -4\\ 4& 13& 5\\ 3& 15& 4\end{array}\right]$

Determinant of ${\Delta }_p$ is,

  $\left|{\Delta }_b\right|=\left|\begin{array}{ccc}2& 20& -4\\ 4& 13& 5\\ 3& 15& 4\end{array}\right|=2\left|\begin{array}{cc}13& 5\\ 15& 4\end{array}\right|-4\left|\begin{array}{cc}20& -4\\ 15& 4\end{array}\right|+3\left|\begin{array}{cc}20& -4\\ 13& 5\end{array}\right|$

  $=2\left(52-75\right)-4\left(80+60\right)+3\left(100+52\right)$

  $=2\left(-23\right)-4\left(140\right)+3\left(152\right)$

  $=-46-560+456$

  $=-150$

The w-matrix ${\Delta }_w$ can be obtained by replacing the third column of $\Delta$ by the column of numbers in the right hand side of the matrix equation $\left(1\right)$ ,

  ${\Delta }_w=\left[\begin{array}{ccc}2& 3& 20\\ 4& -1& 13\\ 3& 2& 15\end{array}\right]$

Determinant of ${\Delta }_w$ is,

  $\left|{\Delta }_w\right|=\left|\begin{array}{ccc}2& 3& 20\\ 4& -1& 13\\ 3& 2& 15\end{array}\right|=2\left|\begin{array}{cc}-1& 13\\ 2& 15\end{array}\right|-4\left|\begin{array}{cc}3& 20\\ 2& 15\end{array}\right|+3\left|\begin{array}{cc}3& 20\\ -1& 13\end{array}\right|$

  $=2\left(-15-26\right)-4\left(45-40\right)+3\left(39+20\right)$

  $=2\left(-41\right)-4\left(5\right)+3\left(59\right)$

  $=-82-20+177$

  $=75$

  $n=\frac{\left|{\Delta }_n\right|}{\left|\Delta \right|}=\frac{-375}{-75}=5$

  $b=\frac{\left|{\Delta }_b\right|}{\left|\Delta \right|}=\frac{-150}{-75}=2$

  $c=\frac{\left|{\Delta }_c\right|}{\left|\Delta \right|}=\frac{75}{-75}=-1$

Conclusion:

The solution for the given set of equations is, $\left(n,\text{ p},\text{ w}\right)=\left(5,\text{ 2},\text{ -1}\right)$