Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 31– 40, each matrix is non singular. Find the inverse of each matrix.
$\left[\begin{array}{cc}-4& 1\\ 6& -2\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix $\left[\begin{array}{cc}-4& 1\\ 6& -2\end{array}\right]$ is $\left[\begin{array}{cc}-1& -\frac{1}{2}\\ -3& -2\end{array}\right]$

Step 1. Given Information

The matrix $\left[\begin{array}{cc}-4& 1\\ 6& -2\end{array}\right]$ is non singular.

Step 2. Form the matrix

$[A\overline{)I_2]=\left[\begin{array}{cc}-4& 1\\ 6& -2\end{array}\right]\overline{)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}}$

Step 3. Transform equation into reduced row echelon form

$$\begin{gathered} R_1\to \frac{R_1+R_2}{2}=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 6& -2\end{array}\right]\overline{)\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ 0& 1\end{array}\right]} \\ {R}_2\to R_2-6R_1=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& 1\end{array}\right]\overline{)\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -3& -2\end{array}\right]} \\ {R}_1\to R_1+\frac{1}{2}{R}_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\overline{)\left[\begin{array}{cc}-1& -\frac{1}{2}\\ -3& -2\end{array}\right]} \end{gathered}$$

Step 4. Find the inverse

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