Chapter 7.4, Problem 96E

To determine

To find: The number of each denomination by solving the system of equations using matrices.

Answer to Problem 96E

Number of $1 bill $=15$ .

Number of $5 bill $=8$ .

Number of $10 bill $=2$ .

Number of $20 bill $=1$ .

Explanation of Solution

Given information: A server has a total of $95 in denominations of $1, $5, $10 and $20. Total number of paper bills is 26. Number of $5 bill is 4 times greater than $10 bill and number of $1 bill is 1 less than twice number of $5 bill.

Calculation:

Let,

  $a=$ Number of $1 bill.

  $b=$ Number of $5 bill.

  $c=$ Number of $10 bill.

  $d=$ Number of $20 bill.

Using the given information we can make the system of equations as,

  $\begin{array}{l}a+5b+10c+20d=95\\ b-4c=0\\ a-2b=-1\\ a+b+c+d=26\end{array}$

Now the augmented matrix can be written as,

  $\left(\begin{array}{ccccc}1& 5& 10& 20& 95\\ 0& 1& -4& 0& 0\\ 1& -2& 0& 0& -1\\ 1& 1& 1& 1& 26\end{array}\right)$

Using row transformations,

  $\begin{array}{l}{R}_3\to R_3-R_1\\ R_4\to R_4-R_1\\ \left(\begin{array}{ccccc}1& 5& 10& 20& 95\\ 0& 1& -4& 0& 0\\ 0& -7& -10& -20& -96\\ 0& -4& -9& -19& -69\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to R_3+7R_2\\ R_4\to R_4+4R_2\\ \left(\begin{array}{ccccc}1& 5& 10& 20& 95\\ 0& 1& -4& 0& 0\\ 0& 0& -38& -20& -96\\ 0& 0& -25& -19& -69\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1-5R_2\\ \left(\begin{array}{ccccc}1& 0& 30& 20& 95\\ 0& 1& -4& 0& 0\\ 0& 0& -38& -20& -96\\ 0& 0& -25& -19& -69\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to -\frac{1}{38}{R}_3\\ \left(\begin{array}{ccccc}1& 0& 30& 20& 95\\ 0& 1& -4& 0& 0\\ 0& 0& 1& \frac{10}{19}& \frac{48}{19}\\ 0& 0& -25& -19& -69\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_4\to R_4+25R_3\\ \left(\begin{array}{ccccc}1& 0& 30& 20& 95\\ 0& 1& -4& 0& 0\\ 0& 0& 1& \frac{10}{19}& \frac{48}{19}\\ 0& 0& 0& -\frac{111}{19}& -\frac{111}{19}\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_4\to -\frac{19}{111}{R}_4\\ \left(\begin{array}{ccccc}1& 0& 30& 20& 95\\ 0& 1& -4& 0& 0\\ 0& 0& 1& \frac{10}{19}& \frac{48}{19}\\ 0& 0& 0& 1& 1\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_2\to R_2+4R_3\\ R_1\to R_1-30R_3\\ \left(\begin{array}{ccccc}1& 0& 0& \frac{80}{19}& \frac{365}{19}\\ 0& 1& 0& \frac{40}{19}& \frac{192}{19}\\ 0& 0& 1& \frac{10}{19}& \frac{48}{19}\\ 0& 0& 0& 1& 1\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to R_3-\frac{10}{19}{R}_4\\ R_2\to R_2-\frac{40}{19}{R}_4\\ R_1\to R_1-\frac{80}{19}{R}_4\\ \left(\begin{array}{ccccc}1& 0& 0& 0& 15\\ 0& 1& 0& 0& 8\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& 1\end{array}\right)\end{array}$

The matrix is now in reduced row echelon form.

Converting back to the system of linear equations,

  $a=15,b=8,c=2,d=1$ .

Therefore,

Number of $1 bill $=15$ .

Number of $5 bill $=8$ .

Number of $10 bill $=2$ .

Number of $20 bill $=1$ .