Chapter 7.7, Problem 75E

a.

To determine

To Find: The determinant of A

Answer to Problem 75E

Determinant of $A=2$

Explanation of Solution

Given information: Matrix is $A=\left[\begin{array}{ccc}1& -3& -2\\ -1& 3& 1\\ 0& 2& -2\end{array}\right]$

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

Calculation:

Consider,

  $A=\left[\begin{array}{ccc}1& -3& -2\\ -1& 3& 1\\ 0& 2& -2\end{array}\right]$

Determinant of A $=\left|A\right|=1\times \left(-6-2\right)+3\times \left(2-0\right)-2\times \left(-2-0\right)=2$

b

To determine

To Find: inverse of $A=\left[\begin{array}{ccc}1& -3& -2\\ -1& 3& 1\\ 0& 2& -2\end{array}\right]$

Answer to Problem 75E

Explanation of Solution

Given information: Matrix is $A=\left[\begin{array}{ccc}1& -3& -2\\ -1& 3& 1\\ 0& 2& -2\end{array}\right]$

Concept Used: Let V be an $n\times n$ matrix and let $I_n$ 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

  $\left[A:I\right]=\left[\begin{array}{ccc}1& -3& -2\\ -1& 3& 1\\ 0& 2& -2\end{array} \begin{array}{c}:\\ :\\ :\end{array} \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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

  $\begin{array}{l}{R}_2\to R_1+R_2\\ \left[\begin{array}{ccc}1& -3& -2\\ 0& 0& -1\\ 0& 2& -2\end{array} \begin{array}{c}:\\ :\\ :\end{array} \begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right]\\ R_1\to R_1-R_2\\ \left[\begin{array}{ccc}1& -5& 0\\ 0& 0& -1\\ 0& 2& -2\end{array} \begin{array}{c}:\\ :\\ :\end{array} \begin{array}{ccc}1& 0& -1\\ 1& 1& 0\\ 0& 0& 1\end{array}\right]\\ R_3\to \frac{1}{2}{R}_3\\ \left[\begin{array}{ccc}1& -5& 0\\ 0& 0& -1\\ 0& 1& -1\end{array} \begin{array}{c}:\\ :\\ :\end{array} \begin{array}{ccc}1& 0& -1\\ 1& 1& 0\\ 0& 0& \frac{1}{2}\end{array}\right]\end{array}$

Similarly,

Hence,

  $\left[I:A^{-1}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array} \begin{array}{c}:\\ :\\ :\end{array} \begin{array}{ccc}-4& -5& \frac{3}{2}\\ -1& -1& \frac{1}{2}\\ -1& -1& 0\end{array}\right]$

The inverse is obtained as:

  $A^{-1}=\left[\begin{array}{ccc}-4& -5& \frac{3}{2}\\ -1& -1& \frac{1}{2}\\ -1& -1& 0\end{array}\right]$

c.

To determine

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

Answer to Problem 75E

The determinant of $A^{-1}=\frac{1}{2}$

Explanation of Solution

Given information: Inverse of A: $\left[\begin{array}{ccc}-4& -5& \frac{3}{2}\\ -1& -1& \frac{1}{2}\\ -1& -1& 0\end{array}\right]$

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

Calculation:

The determinant is obtained as:

  $\left|A^{-1}\right|=-4\times \left(0+\frac{1}{2}\right)+5\times \frac{1}{2}+0=\frac{1}{2}$

d

To determine

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

Answer to Problem 75E

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

Explanation of Solution

Given information:

The determinant of $\left|A^{-1}\right|=\frac{1}{2}$

The determinant of $\left|A\right|=2$

Calculation:

The determinant of $\left|A^{-1}\right|=\frac{1}{2}$

The determinant of $\left|A\right|=2$

Determinant of $\left|A\right|$ and determinant of $\left|A^{-1}\right|$ are reciprocals

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