Chapter 2.5, Problem 52E

To determine

To find: scores of Jessi for the two tests.

Answer to Problem 52E

Score of test 1 = 86

Score of test 2 = 93

Explanation of Solution

Given information:

Let x and y be the scores of test 1 and test 2 respectively.

The equations depicting the data for points of two test are as follows:

  $\begin{array}{l}x+y=179\\ y=x+7\to x-y=-7\end{array}$

Calculation:

Set up the matrix for the first two equations as follows:

  $\begin{array}{l}\begin{array}{ccc}A& X& B\end{array}\\ \left[\begin{array}{cc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}1\\ -1\end{array}\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}179\\ -7\end{array}\right]\end{array}$

For $A=\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]$ , the inverse formula is $$

A−1=1ad−bc[d−c−b  a] . Put the values in the above formula and evaluate the inverse for the given matrix as follows:

  $\begin{array}{c}{A}^{-1}=\frac{1}{-1-1}\left[\begin{array}{cc}\begin{array}{l}-1\\ -1\end{array}& \begin{array}{l}-1\\ \text{ 1}\end{array}\end{array}\right]\\ =\left[\begin{array}{cc}\begin{array}{l}0.5\\ 0.5\end{array}& \begin{array}{l}0.5\\ -0.5\end{array}\end{array}\right]\end{array}$

Solve $A^{-1}B=X$ as follows:

  $\begin{array}{c}\left[\begin{array}{cc}\begin{array}{l}0.5\\ 0.5\end{array}& \begin{array}{l}0.5\\ -0.5\end{array}\end{array}\right]\cdot \left[\begin{array}{l}179\\ -7\end{array}\right]=\left[\begin{array}{l}x\\ y\end{array}\right]\\ \left[\begin{array}{l}86\\ 93\end{array}\right]=\left[\begin{array}{l}x\\ y\end{array}\right]\end{array}$

Thus, the value of x and y is 86 and 93 respectively.