Chapter 8, Problem 6CT

To determine

Find the inverse of the given matrix.

Answer to Problem 6CT

The inverse of the matrix $A$ is $\left[\begin{array}{ccc}\frac{5}{18}& -\frac{4}{9}& \frac{1}{3}\\ -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrix is

  $\left[\begin{array}{ccc}-2& 4& -6\\ 2& 1& 0\\ 4& -2& 5\end{array}\right]$ .

Calculation:

Let $A$ bean $n\times n$ and let $I_n$ be the $n\times n$ identitymatrix. 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 $A$ . The symbol $A^{-1}$ is read “ $A$ inverse.”

Let $A=\left[\begin{array}{ccc}-2& 4& -6\\ 2& 1& 0\\ 4& -2& 5\end{array}\right]$

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

  $\left[A:I\right]=\left[\begin{array}{ccccccc}-2& 4& -6& :& 1& 0& 0\\ 2& 1& 0& :& 0& 1& 0\\ 4& -2& 5& :& 0& 0& 1\end{array}\right]$

Use Gauss-Jordan elimination.

  $\begin{array}{l}{R}_1\leftrightarrow \frac{R_1}{2}\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 2& 1& 0& :& 0& 1& 0\\ 4& -2& 5& :& 0& 0& 1\end{array}\right]\\ R_2\to R_2-2R_1\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& -3& 6& :& -1& 1& 0\\ 4& -2& 5& :& 0& 0& 1\end{array}\right]\\ R_3\to R_3-4R_1\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& -3& 6& :& -1& 1& 0\\ 0& -10& 17& :& -2& 0& 1\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_2\to \frac{R_2}{-3}\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& 1& -2& :& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& -10& 17& :& -2& 0& 1\end{array}\right]\\ R_3\to R_3+10R_2\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& 1& -2& :& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& 0& -3& :& \frac{4}{3}& -\frac{10}{3}& 1\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_3\to \frac{R_3}{-3}\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& 1& -2& :& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& 0& 1& :& -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]\\ R_2\to R_2+2R_3\\ \left[\begin{array}{ccccccc}1& 2& -3& :& \frac{1}{2}& 0& 0\\ 0& 1& 0& :& -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ 0& 0& 1& :& -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]\end{array}$

  $\begin{array}{l}{R}_1\to R_1+3R_3\\ \left[\begin{array}{ccccccc}1& 2& 0& :& -\frac{5}{6}& \frac{10}{3}& -1\\ 0& 1& 0& :& -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ 0& 0& 1& :& -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]\\ R_1\to R_1-2R_2\\ \left[\begin{array}{ccccccc}1& 0& 0& :& \frac{5}{18}& -\frac{4}{9}& \frac{1}{3}\\ 0& 1& 0& :& -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ 0& 0& 1& :& -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]\end{array}$

Now the inverse matrix is

  $\left[\begin{array}{ccc}\frac{5}{18}& -\frac{4}{9}& \frac{1}{3}\\ -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]$

Hence the inverse of the matrix $A$ is $\left[\begin{array}{ccc}\frac{5}{18}& -\frac{4}{9}& \frac{1}{3}\\ -\frac{5}{9}& \frac{17}{9}& -\frac{2}{3}\\ -\frac{4}{9}& \frac{10}{9}& -\frac{1}{3}\end{array}\right]$ .