Q. 12 - Precalculus Enhanced with Graphing Utilities

Question

find the inverse of each nonsingular matrix.

$\left[\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right]$

Step-by-Step Solution

Answer

find the inverse of each nonsingular matrix.

$\left[\begin{array}{ccc}3& 3& -4\\ -2& -2& 3\\ -4& -5& 7\end{array}\right]$

Step 1. Given

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

Step 2. Calculation of Deterrminant

$$\begin{gathered} \left|\begin{array}{ccc}1& -1& 1\\ 2& 5& -1\\ 2& 3& 0\end{array}\right| \\ =1\left(5\times 0-(-1)\times 3\right)-\left(-1\right)\left(2\times 0-\left(-1\right)\times 2\right)+\left(2\times 3-5\times 2\right) \\ =1\left(3\right)+1\left(2\right)+1(-4) \\ =3-2-4 \\ =-3 \\ thus matrix is non-\sin gular \end{gathered}$$

Step 3. Calculation for Augmented matrix

$$\begin{gathered} \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 2& 5& -1& 0& 1& 0\\ 2& 3& 0& 0& 0& 1\end{array}\right] \\ Applying row transformation \\ \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 2& 5& -1& 0& 1& 0\\ 2& 3& 0& 0& 0& 1\end{array}\right]\stackrel{R_2-2R_1}{\to \left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 2& 5& -1& 0& 1& 0\\ 0& 5& -2& -2& 0& 1\end{array}\right]} \\ \stackrel{\frac{R_2}{7}}{\to }\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& 1& -\frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& 0\\ 0& 5& -2& -2& 0& 1\end{array}\right] \\ \stackrel{R_1+R_2}{\to }\left[\begin{array}{cccccc}1& 0& \frac{4}{7}& \frac{5}{7}& \frac{1}{7}& 0\\ 0& 1& -\frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& 0\\ 0& 5& -2& -2& 0& 1\end{array}\right] \\ \stackrel{R_3-5R_2}{\to }\left[\begin{array}{cccccc}1& 0& \frac{4}{7}& \frac{5}{7}& \frac{1}{7}& 0\\ 0& 1& -\frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& 0\\ 0& 0& \frac{1}{7}& -\frac{4}{7}& -\frac{5}{7}& 1\end{array}\right] \\ \stackrel{7R_3}{\to }\left[\begin{array}{cccccc}1& 0& \frac{4}{7}& \frac{5}{7}& \frac{1}{7}& 0\\ 0& 1& -\frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& 0\\ 0& 0& 1& -4& -5& 7\end{array}\right] \\ \stackrel{R_1-\frac{4}{7}{R}_3}{\to }\left[\begin{array}{cccccc}1-\frac{4}{7}\left(0\right)& 0-\frac{4}{7}\left(0\right)& \frac{4}{7}-\frac{4}{7}\left(1\right)& \frac{5}{7}-\frac{4}{7}\left(-4\right)& \frac{1}{7}-\frac{4}{7}\left(-5\right)& 0-\frac{4}{7}\left(7\right)\\ 0& 1& -\frac{3}{7}& -\frac{2}{7}& \frac{1}{7}& 0\\ 0& 0& 1& -4& -5& 7\end{array}\right] \\ \stackrel{R_2+\frac{3}{7}{R}_3}{\to }\left[\begin{array}{cccccc}1& 0& 0& 3& 3& -4\\ 0& 1& 0& -2& -2& 0\\ 0& 0& 1& -4& -5& 7\end{array}\right] \\ Therefore the inverse is \\ \left[\begin{array}{ccc}3& 3& -4\\ -2& -2& 0\\ -4& -5& 7\end{array}\right] \end{gathered}$$