Chapter 12.3, Problem 54PPSE

Solution

To find: The inverse of the matrix $\left[\begin{array}{ccc}-2& 1& -1\\ 2& 0& 4\\ 0& 2& 5\end{array}\right]$ if exist.

Inverse of the matrix $\left[\begin{array}{ccc}-2& 1& -1\\ 2& 0& 4\\ 0& 2& 5\end{array}\right]$ is $\left[\begin{array}{ccc}-4& -3.5& 2\\ -5& -5& 3\\ 2& 2& -1\end{array}\right]$ .

Given information: Matrix is $\left[\begin{array}{ccc}-2& 1& -1\\ 2& 0& 4\\ 0& 2& 5\end{array}\right]$ .

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:

.

Explanation:

Consider, the matrix as $\left[\begin{array}{ccc}-2& 1& -1\\ 2& 0& 4\\ 0& 2& 5\end{array}\right]$ .

The determinant of the matrix is,

  $\begin{array}{c}\left|\begin{array}{ccc}-2& 1& -1\\ 2& 0& 4\\ 0& 2& 5\end{array}\right|=\left(-2\right)\left(0\times 5-4\times 2\right)-1\left(2\times 5-4\times 0\right)+\left(-1\right)\left(2\times 2-0\times 0\right)\\ =\left(-2\right)\left(-8\right)-1\left(10\right)+\left(-1\right)\left(4\right)\\ =16-10-4\\ =2\end{array}$ .

Determinant of this matrix is nonzero.

This implies that inverse of the matrix exists.

Now, inverse of the matrix is,

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

  $=\frac{1}{2}\left[\begin{array}{ccc}0\times 5-4\times 2& \left(-1\right)\times 2-1\times 5& 1\times 4-\left(-1\right)\times 0\\ 4\times 0-2\times 5& \left(-2\right)\times 5-\left(-1\right)\times 0& \left(-1\right)\times 2-\left(-2\right)\times 4\\ 2\times 2-0\times 0& 1\times 0-\left(-2\right)\times 2& \left(-2\right)\times 0-1\times 2\end{array}\right]$

  $=\frac{1}{2}\left[\begin{array}{ccc}-8& -7& 4\\ -10& -10& 6\\ 4& 4& -2\end{array}\right]$

  $=\left[\begin{array}{ccc}-\frac{8}{2}& -\frac{7}{2}& \frac{4}{2}\\ -\frac{10}{2}& -\frac{10}{2}& \frac{6}{2}\\ \frac{4}{2}& \frac{4}{2}& -\frac{2}{2}\end{array}\right]$

Therefore, the inverse of the matrix is $\left[\begin{array}{ccc}-4& -3.5& 2\\ -5& -5& 3\\ 2& 2& -1\end{array}\right]$ .