Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The inverse for the matrix $\left[\begin{array}{rrr}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}\right]$ is, $\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$.

Step 1 . Given information

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

We have to find the inverse of the given matrix.

Step 2 . First find the augmented matrix A | I 3

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

Now transform it into the reduced row echelon form .

Now perform the row operations $R_3=r_3-2r_1$.

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

Perform the row operation $R_2\to \frac{r_2}{-2}$.

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

Step 3 . Perform the row operation R 1 → r 1 + r 2 .

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

Perform the row operation $R_3\to r_3+5r_2$.

$\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]\to \left[\begin{array}{ccccrr}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& -\frac{1}{2}& 2& -\frac{5}{2}& 1\end{array}\right]$

Perform the row operation $R_3\to \frac{r_3}{{\displaystyle \frac{-1}{2}}}$.

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

Step 4 . Perform the row operation R 2 → r 2 + r 3 2 .

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

Perform the row operation $R_1\to r_1-\frac{r_3}{2}$.

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

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 and $A^{-1}$ on the right of the vertical bar.

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