Q. 38 - Algebra 2

Question

For Exercises 34-38, use the figure below.

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

Step-by-Step Solution

Answer

The conjecture about the transformation ${\mathbf{B}}^{\mathbf{-}\mathbf{1}}$ 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 value of the matrix ${\mathbf{B}}^{\mathbf{-}\mathbf{1}}$ is: $\left[\begin{array}{cc}\frac{\mathbf{1}}{\mathbf{2}}& \mathbf{0}\\ \mathbf{0}& \frac{\mathbf{1}}{\mathbf{2}}\\ & \end{array}\right]$.

The value after multiplying ${\mathbf{B}}^{\mathbf{-}\mathbf{1}}$ with the result of $\mathbf{BA}$ is: $\left[\begin{array}{cccc}\mathbf{0}& \mathbf{4}& \mathbf{2}& \mathbf{-}\mathbf{2}\\ \mathbf{0}& \mathbf{4}& \mathbf{6}& \mathbf{2}\\ & & & \end{array}\right]$.

The drawing to verify the conjecture is:

Step 1. Observe the given figure.

The given figure is:

Step 2. Description of step.

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 matrix $B$ which is $B^{-1}$ is given by:

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

Step 3. Write a conjecture about what transformation B - 1 describes on a coordinate plane.

It can be noticed that the matrix $B$ 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.

Step 4. Description of step.

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 $B^{-1}$ 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 of $BA$ is same as that of the vertex matrix A.

This is because the matrix B 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.

Step 5. Draw a diagram to verify the conjecture.

The diagram to verify the conjecture is: