Q. 17 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 17–19, find the inverse, if there is one, of each matrix. If there is not an inverse, say that the matrix is singular.

$\left[\begin{array}{cc}4& 6\\ 1& 3\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the given matrix is $\left[\begin{array}{cc}\frac{1}{2}& -1\\ -\frac{1}{6}& \frac{2}{3}\end{array}\right]$

Step 1. Given information

The given matrix is $\left[\begin{array}{cc}4& 6\\ 1& 3\end{array}\right]$

Step 2. Find the inverse of the matrix.

$[A\mid I_2]=\left[\begin{array}{llll}4& 6& 1& 0\\ 1& 3& 0& 1\end{array}\right]$

Transform the matrix $[A\mid I_2]$ into reduced row echelon form.

Perform the operations $R_1=\frac{1}{4}{R}_1$ and then $R_2=R_2-R_1$,

$\left[\begin{array}{llll}4& 6& 1& 0\\ 1& 3& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& \frac{{\displaystyle 3}}{{\displaystyle 2}}& \frac{{\displaystyle 1}}{{\displaystyle 4}}& 0\\ 1& 3& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& \frac{{\displaystyle 3}}{{\displaystyle 2}}& \frac{{\displaystyle 1}}{{\displaystyle 4}}& 0\\ 0& \frac{{\displaystyle 3}}{{\displaystyle 2}}& -\frac{{\displaystyle 1}}{{\displaystyle 4}}& 1\end{array}\right]$

Now, perform the operations $R_1=R_1-R_2$ and then $R_2=\frac{2}{3}{R}_2$,

$\left[\begin{array}{cccc}1& \frac{{\displaystyle 3}}{{\displaystyle 2}}& \frac{{\displaystyle 1}}{{\displaystyle 4}}& 0\\ 0& \frac{{\displaystyle 3}}{{\displaystyle 2}}& -\frac{{\displaystyle 1}}{{\displaystyle 4}}& 1\end{array}\right]\to \left[\begin{array}{cccc}1& 0& \frac{{\displaystyle 1}}{{\displaystyle 2}}& -1\\ 0& \frac{{\displaystyle 3}}{{\displaystyle 2}}& -\frac{{\displaystyle 1}}{{\displaystyle 4}}& 1\end{array}\right]\to \left[\begin{array}{ccrr}1& 0& \frac{{\displaystyle 1}}{{\displaystyle 2}}& -1\\ 0& 1& -\frac{{\displaystyle 1}}{{\displaystyle 6}}& \frac{{\displaystyle 2}}{{\displaystyle 3}}\end{array}\right]$

Now, we can see that the identity matrix is on the left of the vertical bar.

Then the matrix on the right of the vertical bar is the inverse of $A$.

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}\frac{1}{2}& -1\\ -\frac{1}{6}& \frac{2}{3}\end{array}\right]$