91Ó°ÊÓ

(a)

The matrices and their diagonals are,

$A=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right]$

The diagonal elements are, 1

$A=\left[\begin{array}{cccc}1& \frac{8}{10}& \frac{1}{8}& \frac{10}{8}\\ \frac{10}{8}& 1& \frac{5}{32}& \frac{25}{16}\\ 8& \frac{32}{5}& 1& 10\\ \frac{8}{10}& \frac{16}{25}& \frac{1}{10}& 1\end{array}\right]$

The diagonal elements are, 1.

(b)

The values of the elements are,

$A=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right]$

The values are, $a_{12}=\frac{1}{8},a_{21}=8$

$A=\left[\begin{array}{cccc}1& \frac{8}{10}& \frac{1}{8}& \frac{10}{8}\\ \frac{10}{8}& 1& \frac{5}{32}& \frac{25}{16}\\ 8& \frac{32}{5}& 1& 10\\ \frac{8}{10}& \frac{16}{25}& \frac{1}{10}& 1\end{array}\right]$

The values are,

$a_{12}=\frac{8}{10}⇒a_{21}=\frac{10}{8}$

$a_{34}=10⇒a_{43}=\frac{1}{10}$

$a_{41}=\frac{8}{10}⇒a_{14}=\frac{10}{8}$

$a_{24}=\frac{25}{16}⇒a_{42}=\frac{16}{25}$

$a_{31}=8⇒a_{13}=\frac{1}{8}$

$a_{32}=\frac{32}{5}⇒a_{23}=\frac{5}{32}$

Therefore, the values of $a_{ij}$and $a_{ji}$are reciprocal of each other.

(c)

Consider the matrix,

$A=\left[\begin{array}{cccc}1& \frac{8}{10}& \frac{1}{8}& \frac{10}{8}\\ \frac{10}{8}& 1& \frac{5}{32}& \frac{25}{16}\\ 8& \frac{32}{5}& 1& 10\\ \frac{8}{10}& \frac{16}{25}& \frac{1}{10}& 1\end{array}\right]$

The values of the elements are,

$a_{11}=a_{12}×a_{21}⇒a_{11}=\frac{8}{10}×\frac{10}{8}=1$

$a_{22}=a_{21}×a_{12}⇒a_{22}=\frac{10}{8}×\frac{8}{10}=1$

$a_{44}=a_{43}×a_{34}⇒a_{44}=\frac{1}{10}×10=1$

$a_{24}=a_{21}×a_{14}⇒a_{24}=\frac{10}{8}×\frac{10}{8}=\frac{25}{16}$

$a_{31}=a_{34}×a_{41}⇒a_{31}=10×\frac{8}{10}=8$

$a_{32}=a_{31}×a_{12}⇒a_{32}=8×\frac{8}{10}=\frac{32}{5}$

$a_{33}=a_{31}×a_{13}⇒a_{33}=8×\frac{1}{8}=1$

The relation between $a_{ij}$, $a_{ik}$and $a_{kj}$is, $a_{ij}=a_{ik}×a_{kj}$, that is, one element is product of other two.

(d)

Now, we have

$\text{¸é²¹²Ô°ì }\left(\begin{array}{cc}1& \frac{1}{8}\\ 8& 1\end{array}\right)=1,rref\left(\begin{array}{cc}1& \frac{1}{8}\\ \text{ â¶Ä‰â¶Ä‰}8& 1\end{array}\right)=\left(\begin{array}{cc}8& 1\\ 0& 0\end{array}\right)$and$\text{¸é²¹²Ô°ì }\left(\begin{array}{cccc}1& \frac{8}{10}& \frac{1}{8}& \frac{10}{8}\\ \frac{10}{8}& 1& \frac{5}{32}& \frac{25}{16}\\ 8& \frac{32}{5}& 1& 10\\ \frac{8}{10}& \frac{16}{25}& \frac{1}{10}& 1\end{array}\right)=1,rref\left(\begin{array}{cccc}1& \frac{8}{10}& \frac{1}{8}& \frac{10}{8}\\ \frac{10}{8}& 1& \frac{5}{32}& \frac{25}{16}\\ 8& \frac{32}{5}& 1& 10\\ \frac{8}{10}& \frac{16}{25}& \frac{1}{10}& 1\end{array}\right)=\left(\begin{array}{cccc}1& \frac{4}{5}& \frac{1}{8}& \frac{5}{4}\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

The rank of the matrices is one. The reduced row-echelon form of the matrices have single row.