Chapter 9.5, Problem 90E

To determine

To find: The determinant of each matrix of the encircles.

Answer to Problem 90E

  $A_n=0.5n^2+0.5n$

Explanation of Solution

Given:The given figure of encircled groups of numbers to form $2\times 2$ matrix.

  

There are total 5 encircled.

First circle as represents in matrix form and to find determinant.

  $\left|A_1\right|=\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=2-1=1$

Second encircled,

  $\left|A_2\right|=\left|\begin{array}{cc}2& 1\\ 3& 3\end{array}\right|=6-3=3$

Third encircled,

  $\left|A_3\right|=\left|\begin{array}{cc}3& 1\\ 6& 4\end{array}\right|=12-6=6$

Fourth encircled,

  $\left|A_4\right|=\left|\begin{array}{cc}4& 1\\ 10& 5\end{array}\right|=20-10=10$

Fifth encircled,

  $\left|A_5\right|=\left|\begin{array}{cc}5& 1\\ 15& 6\end{array}\right|=30-15=15$

Now make series of five determinant value, $1,3,6,10,15$

  $\begin{array}{cccccccccc}\text{Series}& 1& & 3& & 6& & 10& & 15\\ \text{First Difference}& & 2& & 3& & 4& & 5& \\ \text{Second Difference}& & & 1& & 1& & 1& & \end{array}$

Second difference is constant. So, the patter in the form of quadratic.

Let pattern be $A_n=a{n}^2+bn+c$

Put $A_1=1,A_2=3,A_3=6$

  $\begin{array}{c}a+b+c=1\\ 4a+2b+c=3\\ 9a+3b+c=6\end{array}$

Solve above three equation and find out a, b and c

So, $a=\frac{1}{2},b=\frac{1}{3},c=0$

Therefore, the pattern of the matrix is $A_n=0.5n^2+0.5n$