Chapter 8.3, Problem 77E

a.

To determine

To write a $2\times 2$ matrix and a $3\times 3$ matrix in the form of $A=\left[\begin{array}{cccccc}{a}_{11}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& a_{22}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& a_{33}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& a_{nn}\end{array}\right]$ and find inverse of each.

Answer to Problem 77E

  $B^{-|}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]and{C}^{-|}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$

Explanation of Solution

Given information:

Given $A=\left[\begin{array}{cccccc}{a}_{11}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& a_{22}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& a_{33}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& a_{nn}\end{array}\right]$ .

Let $B=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\text{and}C=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ be two matrices of orders $2\times 2$ and $3\times 3$ respectively in the form of matrix $A$ .

  $\begin{array}{l}{B}^{-|}=\frac{adjB}{\left|B\right|}\\ whereadjB=\left[\begin{array}{cc}b& 0\\ 0& a\end{array}\right]and\left|B\right|=ab\\ \Rightarrow B^{-|}=\frac{1}{ab}\left[\begin{array}{cc}b& 0\\ 0& a\end{array}\right]\\ \therefore B^{-|}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]\\ C^{-|}=\frac{adjC}{\left|C\right|}\\ whereadjC=\left[\begin{array}{ccc}bc& 0& 0\\ 0& ac& 0\\ 0& 0& ab\end{array}\right],\left|C\right|=abc\\ \Rightarrow C^{-|}=\frac{1}{abc}\left[\begin{array}{ccc}bc& 0& 0\\ 0& ac& 0\\ 0& 0& ab\end{array}\right]\\ \therefore C^{-|}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]\end{array}$

Therefore, $B^{-|}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]and{C}^{-|}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$

b.

To determine

To make a conjecture of about the inverse of $A=\left[\begin{array}{cccccc}{a}_{11}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& a_{22}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& a_{33}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& a_{nn}\end{array}\right]$ from the above results.

Answer to Problem 77E

  $A^{-|}=\left[\begin{array}{cccccc}\frac{1}{a_{11}}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& \frac{1}{a_{22}}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& \frac{1}{a_{33}}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& \frac{1}{a_{nn}}\end{array}\right]$ .

Explanation of Solution

Given information:

Given $A=\left[\begin{array}{cccccc}{a}_{11}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& a_{22}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& a_{33}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& a_{nn}\end{array}\right]$ , inverses of $B=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\text{and}C=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ are $B^{-|}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]and{C}^{-|}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$ respectively.

Since inverses of $B=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\text{and}C=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ are $B^{-|}=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& \frac{1}{b}\end{array}\right]and{C}^{-|}=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$ respectively.

From above result we observe that the inverse of $A$ is again in the form of matrix $A$ , whose diagonal elements are corresponding inverse elements of $A$ .

Thus, $A^{-|}=\left[\begin{array}{cccccc}\frac{1}{a_{11}}& 0& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& \frac{1}{a_{22}}& 0& \mathrm{..}& \mathrm{..}& 0\\ 0& 0& \frac{1}{a_{33}}& \mathrm{..}& \mathrm{..}& 0\\ :& :& :& :& :& :\\ :& :& :& :& :& :\\ 0& 0& 0& \mathrm{..}& \mathrm{..}& \frac{1}{a_{nn}}\end{array}\right]$