Chapter 8.3, Problem 23E

To determine

Tofind:the inverse of the matrix if it’s exists.

Answer to Problem 23E

  ${\text{A}}^{\text{-1}}=\left|\begin{array}{cccc}\text{-}\frac{1}{8}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \frac{1}{4}& \text{0}\\ \text{0}& \text{0}& \text{0}& -\frac{1}{5}\end{array}\right|$

Explanation of Solution

Given:

  ${\text{A}}^{\text{-1}}=\left[\begin{array}{cccc}\text{-}\frac{1}{8}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \frac{1}{4}& \text{0}\\ \text{0}& \text{0}& \text{0}& -\frac{1}{5}\end{array}\right]$

Concept used:

Steps to solve the inverse of matrix by

Step1: Form the augment matrix $\left[\text{A|I}\right]$ .

Step2:Perform row operation on $\left[\text{A|I}\right]$ to get the matrix of the form $\left[\text{I|B}\right]$ .

Calculation:

Here $\text{A=}\left[\begin{array}{ccc}\text{1}& \text{1}& \text{1}\\ \text{3}& \text{5}& \text{4}\\ \text{3}& \text{6}& \text{5}\end{array}\right]$

To find the inverse of A, let’s form the augment matrix $\left[\text{A|I}\right]$ as:

  $\text{A=}\left[\begin{array}{cccc}\text{-8}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \text{4}& \text{0}\\ \text{0}& \text{0}& \text{0}& \text{-5}\end{array}|\begin{array}{cccc}\text{1}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \text{1}& \text{0}\\ \text{0}& \text{0}& \text{0}& \text{1}\end{array}\right]$

Now perform row operation on $\left[\text{A|I}\right]$ to get the matrix of the form $\left[\text{I|B}\right]$ as:

  $\text{-}\left(\frac{1}{8}\right){\text{R}}_{\text{1}}\to \text{A=}\left[\begin{array}{cccc}1& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& 1& \text{0}\\ \text{0}& \text{0}& \text{0}& 1\end{array}|\begin{array}{cccc}\text{-}\frac{1}{8}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \frac{1}{4}& \text{0}\\ \text{0}& \text{0}& \text{0}& -\frac{1}{5}\end{array}\right]$ = $\left[\text{I|B}\right]$

Hence the matrix B is the last transformation is the desired inverse matrix.

  ${\text{A}}^{\text{-1}}=\left[\begin{array}{cccc}\text{-}\frac{1}{8}& \text{0}& \text{0}& \text{0}\\ \text{0}& \text{1}& \text{0}& \text{0}\\ \text{0}& \text{0}& \frac{1}{4}& \text{0}\\ \text{0}& \text{0}& \text{0}& -\frac{1}{5}\end{array}\right]$ .