Chapter 7.7, Problem 73E

a

To determine

To Find: The determinant of A

Answer to Problem 73E

Determinant of $A=6$

Explanation of Solution

Given information: Matrix $A=\left[\begin{array}{l}12\\ -22\end{array}\right]$

Concept Used: If A is a square matrix, the determinant of A is the sum of the entries in any row of A multiplied by their respective cofactors. For $2\times 2$ matrix

  $A=\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The determinant yield

  $\left|A\right|=a_1{b}_2-a_2{b}_1$

Calculation:

Consider,

  $A=\left[\begin{array}{cc}1& 2\\ -2& 2\end{array}\right]$

Determinant of $A=\left|A\right|=1\times 2+2\times 2=6$

b

To determine

To Find: inverse of A

Answer to Problem 73E

Explanation of Solution

Given information: Matrix $A=\left[\begin{array}{cc}1& 2\\ -2& 2\end{array}\right]$

Concept Used: Let V be an $n\times n$ matrix and let $l_n$ be the $n\times n$ identity matrix. If there exists a matrix $A^{-1}$ such that

  $A{A}^{-1}=I_n=A^{-1}A$

Then $A^{-1}$ is called the inverse of A

Calculation:

Begin by adjoining the identity matrix to $A$ to form the matrix

  $\left[A:I\right]=\left[\begin{array}{l}12:10\\ -22:01\end{array}\right]$

Use elementary row operations to obtain the form $\left[I:A^{-1}\right]$ as follows

  $\begin{array}{l}{R}_1\to R_1-R_2\\ \left[\begin{array}{l}30:1-1\\ -22:01\end{array}\right]\\ R_1\to \frac{1}{3}{R}_1\\ \left[\begin{array}{l}10:\frac{1}{3}\frac{-1}{3}\\ -22:01\end{array}\right]\\ R_2\to 2R_1+R_2\\ \left[\begin{array}{l}10:\frac{1}{3}\frac{-1}{3}\\ 02:\frac{2}{3}\frac{1}{3}\end{array}\right]\\ R_2\to \frac{1}{2}{R}_2\\ \left[I:A^{-1}\right]=\left[\begin{array}{l}10:\frac{1}{3}\frac{-2}{3}\\ 01\text{ : }\frac{1}{3}\frac{1}{6}\end{array}\right]\\ A^{-1}=\left[\begin{array}{l}\frac{1}{3}\frac{-1}{3}\\ \frac{1}{3}\frac{1}{6}\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_1\to R_1-2R_2\\ \left[\begin{array}{l}11:1-2\\ 2-1:01\end{array}\right]\\ R_1\to R_1-R_2\\ \left[\begin{array}{l}\text{3 0 }:1-1\\ 2-1:01\end{array}\right]\\ R_1\to \frac{1}{3}{R}_1\\ \left[\begin{array}{l}10:\frac{1}{3}\frac{-1}{3}\\ 2-1:01\end{array}\right]\\ R_2\to R_2-2R_1\\ \left[\begin{array}{l}10:\frac{1}{3}\frac{-1}{3}\\ 0-1:\frac{-2}{3}\frac{5}{3}\end{array}\right]\\ R_2\to -R_2\\ \left[I:A^{-1}\right]=\left[\begin{array}{l}10:\frac{1}{3}\frac{-1}{3}\\ 01\text{ : }\frac{2}{3}\frac{-5}{3}\end{array}\right]\\ A^{-1}=\left[\begin{array}{l}\frac{1}{3}\frac{-1}{3}\\ \frac{2}{3}\frac{-5}{3}\end{array}\right]\end{array}$

c

To determine

To Find: The determinant of $A^{-1}$

Answer to Problem 73E

The determinant of $A^{-1}=\frac{1}{6}$

Explanation of Solution

Given information: Inverse of $A:\left[\begin{array}{cc}\frac{1}{3}& \frac{-1}{3}\\ \frac{1}{3}& \frac{1}{6}\end{array}\right]$

Concept Used: If A is a square matrix, the determinant of A is the sum of the entries in any row of A multiplied by their respective cofactors. For $2\times 2$ matrix

  $A=\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The determinant yield

  $\left|A\right|=a_1{b}_2-a_2{b}_1$

Calculation:

The determinant is obtained as:

  $\left|A^{-1}\right|=\frac{1}{3}\times \frac{1}{6}+\frac{1}{9}=\frac{1}{6}$

d

To determine

To Find: Compare the determinant of $A^{-1}$ and the determinant of A

Answer to Problem 73E

Determinant of A and determinant of $A^{-1}$ are reciprocals

Explanation of Solution

Given information:

The determinant of $A^{-1}=\frac{1}{6}$

The determinant of $A=6$

Calculation:

The determinant of $A^{-1}=\frac{1}{6}$

The determinant of $A=6$

Determinant of A and determinant of $A^{-1}$ are reciprocals

  $\left|A^{-1}\right|=\frac{1}{\left|A\right|}$