91Ó°ÊÓ

Matrix transformation can be of two types rotation and reflection only for square matrices. When the determinant value of the matrix is1 then it is termedrotationand if the value is-1then it isreflection.

The matrix representing rotation about the x-axis is $x=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)$.

The matrix representing rotation about z-axis is $y=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)$.

In the first case, the rotation is first $90°$about the x-axis and then $90°$rotation about the z-axis. Hence the resultant matrix is obtained by the formula $M=ZX.$.

role="math" localid="1659001373511" $$\begin{gathered} M=ZX \\ =\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right) \end{gathered}$$

In the second case, the rotation is first about the z-axis and then rotation about axis. Hence the resultant matrix is obtained by the formula .

$$\begin{gathered} N=XZ \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right) \\ =\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right) \end{gathered}$$

Calculate role="math" localid="1659001937602" $detManddetN.$

role="math" localid="1659001921671" $$\begin{gathered} detM=0\left[\left(0×0\right)-\left(0×1\right)\right]-0\left[\left(1×0\right)-\left(0×0\right)\right]+1\left[\left(1×1\right)-\left(0×0\right)\right] \\ =0-0+1 \\ =1 \\ detN=0\left[\left(0×0\right)-\left(0×1\right)\right]-\left(-1\right)\left[\left(0×0\right)-\left(1×\left(-1\right)\right)\right]+0\left[\left(0×0\right)-\left(1×0\right)\right] \\ =0+1\left[-1\right]+0 \\ =1 \end{gathered}$$

$detManddetN$are both 1 equal hence both are rotations.

To determine the rotation axes for both, use the formula$Lr=r$, where role="math" localid="1659002229628" $r=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$, to find the axes of rotation.

Calculate for M .

role="math" localid="1659002765114" $$\begin{gathered} Mr=r \\ \backslash \left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \end{gathered}$$

Hence the equations obtained are $z=x,x=y, and y=z$. It is seen that the vector$\left(1,1,-1\right)$remains unchanged after the transformation hence the axis of rotation is .$i+j+k$

Further check for the angle of rotation.

$$\begin{gathered} M^2=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right) \\ {M}^1=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ =I \end{gathered}$$

Since$M^3=I$hence the angle of rotation is $120°$.

Calculate for N.

role="math" localid="1659002921980" $$\begin{gathered} Nr=r \\ \left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \end{gathered}$$

Therefore, the equations obtained are$-y=x,-z=y, and x=z$. It is seen that the vector$(1,-1,1)$remains unchanged after the transformation hence the axis of rotation is $i-j+k.$.

Further, check for the angle of rotation.

$$\begin{gathered} N^2=\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right)\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 1\\ -1& 0& 0\\ 1& -1& 0\end{array}\right) \\ {N}^3=\left(\begin{array}{ccc}0& 0& 1\\ -1& 0& 0\\ 1& -1& 0\end{array}\right)\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right) \\ =I \end{gathered}$$

Since,$N^3=I$ hence, the angle of rotation is$120°$ .