Q. 39 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $31-40$, each matrix is nonsingular. Find the inverse of each matrix.

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

Step-by-Step Solution

Answer

The inverse of the matrix $\left[\begin{array}{ccc}1& 1& 1\\ 3& 2& -1\\ 3& 1& 2\end{array}\right]$ is, $\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]$

Step 1 . Given information

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

We have to find the inverse of the given matrix.

Step 2 . First find A | I 3

$\left[A\mid I_3\right]=\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right]$

In order to transform the matrix $\left[A|I_3\right]$ into reduced row echelon form first perform the row operations $R_2=r_2-3r_1$ and then $R_3=r_3-3r_1$.

$$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right] \\ \to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \end{gathered}$$

Step 3 . Now, perform the row operation R 2 = - 1 . r 2 followed by R 1 = r 1 - r 2

$$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \\ \to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \end{gathered}$$

Perform $R_3=r_3+2r_2$ and then $R_3=\frac{1}{7}{r}_3$.

$$\begin{gathered} \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 7& 3& -2& 1\end{array}\right] \\ \to \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \end{gathered}$$

Step 4 . Now perform the row operations R 1 = r 1 + 3 r 3 followed by R 2 = r 2 - 4 r 3 .

$$\begin{gathered} \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \\ \to \left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 0& \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \end{gathered}$$

We can see that the reduced row echelon form of $\left[A|I_3\right]$ contains the identity matrix $I_3$ on the left of the vertical bar.

The $3$ by $3$ matrix on the right side of the vertical bar is the inverse of $A$.

$A^{-1}=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]$.