Chapter 11.4, Problem 90AYU

To determine

To describe: The situation that can be represented by a matrix.

Answer to Problem 90AYU

The situation is cost of 5 pens is 26 dollars more cost of two pencils at shop A and at shop B cost of 1 pen is 6 dollars more than cost of 3 pencils. Determine the cost of pen and pencil.

Explanation of Solution

Formulate a situation as cost of 5 pens is 26 dollars more cost of two pencils at shop A and at shop B cost of 1 pen is 6 dollars more than cost of 3 pencils. Determine the cost of pen and pencil.

Consider the system of equations as,

  $\begin{array}{c}5x-2y=26\\ x+2y=6\end{array}$

Where x denote cost of pen and y denote cost of pencil.

Now, in the above system of equation we have two equations and two variables x and y.

We construct a coefficient matrix P,

  $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$

Secondly, the variable matrix X is given as,

  $X=\left[\begin{array}{c}x\\ y\end{array}\right]$

Thirdly, matrix of constants Q which are given on right hand of the equation is,

  $Q=\left[\begin{array}{c}26\\ 6\end{array}\right]$

Therefore, we can write the system as,

  $\begin{array}{c}PX=Q\\ \left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}26\\ 6\end{array}\right]\end{array}$

First, we find the inverse of the matrix, $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ .

Let the inverse of the matrix be $P^{-1}=\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]$ .

Using the relation, $P{P}^{-1}=I_2$ , we get,

  $\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, multiplying the above two matrices, we get,

  $\left[\begin{array}{ll}5a-2b\hfill & 5c-2d\hfill \\ a+2b\hfill & c+2d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, we equate the corresponding elements of the two matrices. First we write the system of equations when equated to the first column,

  $\begin{array}{c}5a-2b=1\\ a+2b=0\end{array}$

Now, we write the system of equations when equated to the second column,

  $\begin{array}{c}5c-2d=0\\ c+2d=1\end{array}$

So, we get the two augmented matrices which are as follows:

  $\left[\begin{array}{lll}5\hfill & -2\hfill & 1\hfill \\ 1\hfill & 2\hfill & 0\hfill \end{array}\right]$ and $\left[\begin{array}{lll}5\hfill & -2\hfill & 0\hfill \\ 1\hfill & 2\hfill & 1\hfill \end{array}\right]$

Combing the above two matrices such that the identity matrix is on the right hand side, we get,

  $\left[\begin{array}{llll}5\hfill & -2\hfill & 1\hfill & 0\hfill \\ 1\hfill & 2\hfill & 0\hfill & 1\hfill \end{array}\right]$

Now, we apply the Gauss-Jordan elimination to find the inverse of $\left[\begin{array}{llll}5\hfill & -2\hfill & 1\hfill & 0\hfill \\ 1\hfill & 2\hfill & 0\hfill & 1\hfill \end{array}\right]$ ,

First we have to make the entry $a_{11}$ as 1, we perform the row transformation as $R_1\iff R_2$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 5\hfill & -2\hfill & 1\hfill & 0\hfill \end{array}\right]$

Next, we have to make the entry $a_{21}$ as 0, we perform the row transformation as $R_2\to R_2-5R_1$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 0\hfill & -12\hfill & 1\hfill & -5\hfill \end{array}\right]$

Next, we have to make the entry $a_{22}$ as 1, we perform the row transformation as $R_2\to -\frac{R_2}{12}$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 0\hfill & 1\hfill & \frac{-1}{12}\hfill & \frac{5}{12}\hfill \end{array}\right]$

Next, we have to make the entry $a_{12}$ as 0, we perform the row transformation as $R_1\to R_1-2R_2$ .

  $\left[\begin{array}{llll}1\hfill & 0\hfill & \frac{1}{6}\hfill & \frac{1}{6}\hfill \\ 0\hfill & 1\hfill & \frac{-1}{12}\hfill & \frac{5}{12}\hfill \end{array}\right]$

Thus, two system can be segregated as follows,

  $\left[\begin{array}{lll}\frac{1}{6}\hfill & \frac{1}{6}\hfill & 1\hfill \\ \frac{-1}{12}\hfill & \frac{5}{12}\hfill & 0\hfill \end{array}\right]$ and $\left[\begin{array}{lll}\frac{1}{6}\hfill & \frac{1}{6}\hfill & 0\hfill \\ \frac{-1}{12}\hfill & \frac{5}{12}\hfill & 1\hfill \end{array}\right]$

Therefore, the inverse of the matrix $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ is $P^{{}^{-1}}=\left[\begin{array}{cc}\frac{1}{6}& \frac{1}{6}\\ \frac{-1}{12}& \frac{5}{12}\end{array}\right]$ .

Now, substituting the values of $P^{{}^{-1}}$ , Q, and X in the equation $X=P^{-1}Q$ ,

  $\begin{array}{c}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& \frac{1}{6}\\ \frac{-1}{12}& \frac{5}{12}\end{array}\right]\left[\begin{array}{c}26\\ 6\end{array}\right]\\ =\left[\begin{array}{c}\left(\frac{1}{6}\right)\left(26\right)+\left(\frac{1}{6}\right)\left(6\right)\\ \left(\frac{-1}{12}\right)\left(26\right)+\left(\frac{5}{12}\right)\left(6\right)\end{array}\right]\\ =\left[\begin{array}{c}5.3\\ 0.3\end{array}\right]\end{array}$

Therefore, the value of x is $5.3$ and y is $0.3$ for the system of equations.