Chapter 7.2, Problem 33E

Solution

To proof: The given matrix are inverse of each other.

  $A=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$, and $B=\left[\begin{array}{cc}0.8& -0.2\\ -0.6& 0.4\end{array}\right]$

Yes, the given matrix are inverse of each other.

Given Information:

The matrix are defined as,

  $A=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$, and $B=\left[\begin{array}{cc}0.8& -0.2\\ -0.6& 0.4\end{array}\right]$

Concept Used:

Two matrix are said to be inverse of each other if and only if,

  $AB=BA=I$

Where, $I$ is identity matrix.

Calculation:

Consider the given matrix.

  $A=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$, and $B=\left[\begin{array}{cc}0.8& -0.2\\ -0.6& 0.4\end{array}\right]$

Now, calculate the matrix multiplication $AB$ .

  $\begin{array}{c}AB=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}0.8& -0.2\\ -0.6& 0.4\end{array}\right]\\ =\left[\begin{array}{cc}2\left(0.8\right)+1\left(-0.6\right)& 2\left(-0.2\right)+1\left(0.4\right)\\ 3\left(0.8\right)+4\left(-0.6\right)& 3\left(-0.2\right)+4\left(0.4\right)\end{array}\right]\\ =\left[\begin{array}{cc}1.6-0.6& -0.4+1.4\\ 2.4-2.4& -0.6+1.6\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\to I\end{array}$

Now, calculate the matrix multiplication $AB$ .

  $\begin{array}{c}AB=\left[\begin{array}{cc}0.8& -0.2\\ -0.6& 0.4\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\\ =\left[\begin{array}{cc}0.8\left(2\right)-2\left(0.3\right)& 0.8\left(1\right)-0.2\left(4\right)\\ -0.6\left(2\right)+0.4\left(3\right)& -0.6\left(1\right)+0.4\left(4\right)\end{array}\right]\\ =\left[\begin{array}{cc}1.6-0.6& 0.8-0.8\\ -1.2+1.2& -0.6+1.6\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\to I\end{array}$

As both the matrix multiplication is showing the identity matrix.

Hence, the given statement has proved.