Chapter 12.3, Problem 56PPSE

Solution

To find: The inverse of the matrix $\left[\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right]$ if exist.

Inverse of the matrix $\left[\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right]$ is $\left[\begin{array}{ccc}0& \frac{1}{7}& \frac{2}{7}\\ \frac{1}{4}& \frac{3}{14}& -\frac{1}{14}\\ \frac{1}{2}& 0& 0\end{array}\right]$ .

Given information: Matrix is $\left[\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right]$ .

Formula used: The determinant of the matrix $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ is,

  $\begin{array}{c}\left|A\right|=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ =a_{11}\left(a_{22}{a}_{33}-a_{23}{a}_{32}\right)-a_{12}\left(a_{21}{a}_{33}-a_{23}{a}_{31}\right)+a_{13}\left(a_{21}{a}_{32}-a_{22}{a}_{31}\right)\end{array}$ .

If determinant is nonzero, then inverse of the matrix exists.

Inverse of the matrix is .

Explanation:

Consider, the matrix as $\left[\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right]$ .

The determinant of the matrix is,

  $\begin{array}{c}\left|\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right|=0\left(4\times 1-\left(-2\right)\times \left(-2\right)\right)-0\left(1\times 1-\left(-2\right)\times 3\right)+2\left(1\times \left(-2\right)-4\times 3\right)\\ =0-0+2\times \left(-14\right)\\ =-28\end{array}$ .

Determinant of this matrix is nonzero.

This implies that inverse of the matrix exists.

Now, inverse of the matrix is,

  ${\left[\begin{array}{ccc}0& 0& 2\\ 1& 4& -2\\ 3& -2& 1\end{array}\right]}^{-1}=\frac{1}{\left(-28\right)}\left[\begin{array}{ccc}\left|\begin{array}{cc}4& -2\\ -2& 1\end{array}\right|& \left|\begin{array}{cc}2& 0\\ 1& -2\end{array}\right|& \left|\begin{array}{cc}0& 2\\ 4& -2\end{array}\right|\\ \left|\begin{array}{cc}-2& 1\\ 1& 3\end{array}\right|& \left|\begin{array}{cc}0& 2\\ 3& 1\end{array}\right|& \left|\begin{array}{cc}2& 0\\ -2& 1\end{array}\right|\\ \left|\begin{array}{cc}1& 4\\ 3& -2\end{array}\right|& \left|\begin{array}{cc}0& 0\\ -2& 3\end{array}\right|& \left|\begin{array}{cc}0& 0\\ 1& 4\end{array}\right|\end{array}\right]$

  $=\frac{1}{-28}\left[\begin{array}{ccc}0& -4& -8\\ -7& -6& 2\\ -14& 0& 0\end{array}\right]$

  $=\left[\begin{array}{ccc}\frac{0}{-28}& \frac{-4}{-28}& \frac{-8}{-28}\\ \frac{-7}{-28}& \frac{-6}{-28}& \frac{2}{-28}\\ \frac{-14}{-28}& \frac{0}{-28}& \frac{0}{-28}\end{array}\right]$

  $=\left[\begin{array}{ccc}0& \frac{1}{7}& \frac{2}{7}\\ \frac{1}{4}& \frac{3}{14}& -\frac{1}{14}\\ \frac{1}{2}& 0& 0\end{array}\right]$

Therefore, the inverse of the matrix is $\left[\begin{array}{ccc}0& \frac{1}{7}& \frac{2}{7}\\ \frac{1}{4}& \frac{3}{14}& -\frac{1}{14}\\ \frac{1}{2}& 0& 0\end{array}\right]$ .