Q. 33 - Algebra 2

Question

The matrix used to rotate a figure $\mathbf{270}\mathbf{°}$ counter clockwise about the origin is $\left[\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\\ & \end{array}\right]$. Compare this matrix with the matrix used to rotate a figure $\mathbf{90}\mathbf{°}$ counter clockwise about the origin.

1. Are they inverses?
2. Does your answer make sense based on the geometry? Use a drawing to support your answer.

Step-by-Step Solution

Answer

1. Yes, they are inverses.
2. Yes, our answer makes sense based on the geometry. The drawing to support our answer is:

a. Step 1. Description of step.

The matrix A which is used to rotate a figure $270°$ counter clockwise about the origin is given by:

$A=\left[\begin{array}{cc}0& 1\\ -1& 0\\ & \end{array}\right]$

The matrix $B$ which is used to rotate a figure $90°$ counter clockwise about the origin is given by:

$B=\left[\begin{array}{cc}0& -1\\ 1& 0\\ & \end{array}\right]$

Step 2. Description of step.

The two matrices A and B are said to be inverses if $AB=BA=I$, where $I$ is an identity matrix of the same order as that of A and B.

Now, find out the product of the matrices A and B.

Therefore, it can be obtained that:

$\begin{array}{c}AB=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\\ =\left[\begin{array}{cc}0+1& 0+0\\ 0+0& 1+0\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

Therefore, it can be noticed that product of the matrices A and B is equal to the identity matrix.

Therefore, the matrices A and B are inverses of each other.

Step 3. Description of step.

Therefore, yes, they are inverses.

b. Step 1. Description of step.

Yes, the matrices which are used to rotate a figure $270°$ counter clockwise about the origin and  $90°$ counter clockwise about the origin are inverse of each other.

Consider the triangle $\Delta ABC$ having coordinates as $A\left(1,2\right)$, $B\left(2,0\right)$ and $C\left(3,1\right)$.

Therefore, the vertex matrix $C$ of the given triangle $\Delta ABC$ is given by:

$C=\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ & & \end{array}\right]$

If the matrices which are used rotate a figure $270°$ counter clockwise about the origin and  $90°$ counter clockwise about the origin would have been inverse of each other, then after rotating consecutively the figure $270°$ counter clockwise about the origin and  $90°$ counter clockwise about the origin, the same figure would have been obtained.

Step 2. Description of step.

Rotate the triangle $\Delta ABC$ $270°$ counter clockwise about the origin by multiplying the matrix A on the left with the vertex matrix $C$.

Therefore, it can be obtained that the vertex matrix of the triangle $\Delta A\text{'}B\text{'}C\text{'}$ obtained after rotating the triangle $\Delta ABC$ $270°$ counter clockwise about the origin is given by:

$\begin{array}{c}AC=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\end{array}\right]\\ =\left[\begin{array}{ccc}0+2& 0+0& 0+1\\ -1+0& -2+0& -3+0\end{array}\right]\\ =\left[\begin{array}{ccc}2& 0& 1\\ -1& -2& -3\end{array}\right]\end{array}$

Therefore, the vertex matrix of the triangle $\Delta A\text{'}B\text{'}C\text{'}$ obtained after rotating the triangle $\Delta ABC$ $270°$ counter clockwise about the origin is: $\left[\begin{array}{ccc}2& 0& 1\\ -1& -2& -3\\ & & \end{array}\right]$.

 

Step 3. Description of step.

Rotate the triangle $\Delta A\text{'}B\text{'}C\text{'}$ $90°$ counter clockwise about the origin by multiplying the matrix B on the left with the vertex matrix $AC$.

Therefore, it can be obtained that the vertex matrix of the triangle $\Delta A\text{'}\text{'}B\text{'}\text{'}C\text{'}\text{'}$ obtained after rotating the triangle $\Delta A\text{'}B\text{'}C\text{'}$ $90°$ counter clockwise about the origin is given by:

$\begin{array}{c}BAC=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{ccc}2& 0& 1\\ -1& -2& -3\end{array}\right]\\ =\left[\begin{array}{ccc}0+1& 0+2& 0+3\\ 2+0& 0+0& 1+0\end{array}\right]\\ =\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\end{array}\right]\end{array}$

Therefore, the vertex matrix of the triangle $\Delta A\text{'}\text{'}B\text{'}\text{'}C\text{'}\text{'}$ obtained after rotating the triangle $\Delta A\text{'}B\text{'}C\text{'}$ $90°$ counter clockwise about the origin is: $\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ & & \end{array}\right]$.

Therefore, it can be noticed that the matrix $\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ & & \end{array}\right]$ is not equal to the matrix $C=\left[\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ & & \end{array}\right]$.

Therefore, same figure will be obtained.

Therefore, they are inverse of each other.

Step 4. Description of step.

The graph showing the vertices of the triangles $\Delta ABC$, $\Delta A\text{'}B\text{'}C\text{'}$ and $\Delta A\text{'}\text{'}B\text{'}\text{'}C\text{'}\text{'}$ is:

Therefore, here also it can be noticed that the triangles $\Delta ABC$ is same as that of $\Delta A\text{'}\text{'}B\text{'}\text{'}C\text{'}\text{'}$.

Therefore, the matrices which are used to rotate a figure $270°$ counter clockwise about the origin and  $90°$ counter clockwise about the origin are inverse of each other.

Therefore, yes, our answer makes sense based on the geometry.