For Exercises 34-38, use the figure below.

38. Test your conjecture. Find${\mathbf{B}}^{-1}$ and multiply it by the result ofBA . Make a drawing to verify your conjecture.

The given figure is:

It is being given that $B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$

Therefore, the value of the determinant of matrix B is:

$\begin{array}{c}\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=\left(2\right)\left(2\right)-\left(0\right)\left(0\right)\\ =4-0\\ =4\end{array}$

The inverse of$2\times 2$matrix$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$is given by $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$and $ad-bc\ne 0$.

Therefore, the inverse of matrixB which is$B^{-1}$is given by:

$\begin{array}{c}{B}^{-1}=\frac{1}{\left(2\right)\left(2\right)-\left(0\right)\left(0\right)}\left[\begin{array}{cc}2& -\left(0\right)\\ -\left(0\right)& 2\end{array}\right]\\ =\frac{1}{4-0}\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\\ =\frac{1}{4}\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{4}& \frac{0}{4}\\ \frac{0}{4}& \frac{2}{4}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]\end{array}$

It can be noticed that the matrixB is$\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$ and the transformation was that the transformed rectangle was enlarged two times that implies all the linear measures of the transformed rectangle were two times the linear measures of the given rectangle.

As the matrix$B^{-1}$ is$\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]$ , therefore the transformation after multiplying the matrix$B^{-1}$ with vertex matrix of the given rectangle is that the transformed rectangle will be half times reduced that implies all the linear measures of the transformed rectangle will be half times the linear measures of the given rectangle.

Therefore, the conjecture about the transformation$B^{-1}$ describes on a coordinate plane is that the transformed rectangle will be half times reduced that implies all the linear measures of the transformed rectangle will be half times the linear measures of the given rectangle.

The result of BA is: $BA=\left[\begin{array}{cccc}0& 8& 4& -4\\ 0& 8& 12& 4\end{array}\right]$

The value after multiplying$B^{-1}$with the result of BA is given by:

$\begin{array}{c}{B}^{-1}BA=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]\cdot \left[\begin{array}{cccc}0& 8& 4& -4\\ 0& 8& 12& 4\end{array}\right]\\ =\left[\begin{array}{cccc}0+0& 4+0& 2+0& -2+0\\ 0+0& 0+4& 0+6& 0+2\end{array}\right]\\ =\left[\begin{array}{cccc}0& 4& 2& -2\\ 0& 4& 6& 2\end{array}\right]\end{array}$

The value after multiplying with the result of BA is: $\left[\begin{array}{cccc}0& 4& 2& -2\\ 0& 4& 6& 2\end{array}\right]$

Therefore, it can be noticed that the value after multiplying $B^{-1}$with the result ofBA is same as that of the vertex matrix A .

This is because the matrixB enlarged the given rectangle two times and matrix$B^{-1}$ has reduced the transformed rectangle half times, therefore finally the initially given rectangle is obtained.

The diagram to verify the conjecture is: