Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Form the matrix

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

Step 3. Transform equation into reduced row echelon form

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

Step 4. Find the inverse

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