Question

In Problems 31– 40, each matrix is non singular. Find the inverse of each matrix.

$\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

Short answer

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

Step-by-step solution

Step 1. Given Information

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

Step 2. Form the matrix

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

Step 3. Transform equation into reduced row echelon form

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

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

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

Step 4. Find the inverse

The inverse of Ais $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$