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.

$\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]$

Short answer

The matrix $\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]$is singular.

Step-by-step solution

Step 1. Given information

The given matrix is$\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]$

Step 2. Find the inverse of the matrix.

$[A\mid I_2]=\left[\begin{array}{llll}4& -8& 1& 0\\ -1& 2& 0& 1\end{array}\right]$

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

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

$\left[\begin{array}{llll}4& -8& 1& 0\\ -1& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& -2& \frac{{\displaystyle 1}}{{\displaystyle 4}}& 0\\ -1& 2& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& -2& \frac{{\displaystyle 1}}{{\displaystyle 4}}& 0\\ 0& 0& \frac{1}{4}& 1\end{array}\right]$

Now, we can see that the reduced matrix cannot contain the identity matrix on the left of the vertical bar.

So, the given matrix is singular and its inverse does not exist.