Question

Q. Show that the transformation between the coordinates X1, X2, X3 and X1’ , X2’ , X3’ defined by
X1’=1/3 (2x1+2x2-x3)
X2’=1/3 (2x1-x2+2x3)
X3’=1/3(-x1+2x2+2x3)
Is orthogonal and left-handed(improper)

Answer 1

Answer on Question #64381 – Math – Linear Algebra

Question

Show that the transformation between the coordinates $x_1, x_2, x_3$ and $x_1', x_2', x_3'$ defined by

$$x_1' = \frac{1}{3}(2x_1 + 2x_2 - x_3),$$ $$x_2' = \frac{1}{3}(2x_1 - x_2 + 2x_3),$$ $$x_3' = \frac{1}{3}(-x_1 + 2x_2 + 2x_3)$$

is orthogonal and left-handed (improper).

Solution

For the matrix

$$A = \frac{1}{3} \begin{pmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ -1 & 2 & 2 \end{pmatrix},$$

the dot products (scalar products) of the different rows will be the sum of the products of items, they are as follows:

for $1^{\text{st}}$ and $2^{\text{nd}}$ rows,

$$\frac{1}{3}(2 \cdot 2 + 2 \cdot (-1) + (-1) \cdot 2) = 0;$$

for $1^{\text{st}}$ and $3^{\text{rd}}$ rows,

$$\frac{1}{3}(2 \cdot (-1) + 2 \cdot 2 + (-1) \cdot 2) = 0;$$

for $2^{\text{nd}}$ and $3^{\text{rd}}$ rows,

$$\frac{1}{3}(2 \cdot (-1) + (-1) \cdot 2 + 2 \cdot 2) = 0.$$

The scalar products of each row by itself will be the sum of the squares, they are as follows:

for $1^{\mathrm{st}}$ and $2^{\mathrm{nd}}$ rows,

$$\frac {1}{9} (2 ^ {2} + 2 ^ {2} + (- 1) ^ {2}) = 1;$$

for $1^{\mathrm{st}}$ and $3^{\mathrm{rd}}$ rows,

$$\frac {1}{9} (2 ^ {2} + (- 1) ^ {2} + 2 ^ {2}) = 1;$$

for $2^{\mathrm{nd}}$ and $3^{\mathrm{rd}}$ rows,

$$\frac {1}{9} ((- 1) ^ {2} + 2 ^ {2} + 2 ^ {2}) = 1.$$

This means that

$$A A ^ {T} = \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = E,$$

i.e., $A$ is an orthogonal matrix.

Next,

$$\begin{array}{l} \det A = \left| \begin{array}{c c c} \frac {2}{3} & \frac {2}{3} & - \frac {1}{3} \\ \frac {2}{3} & - \frac {1}{3} & \frac {2}{3} \\ - \frac {1}{3} & \frac {2}{3} & \frac {2}{3} \end{array} \right| = \\ = \frac {1}{3 ^ {3}} \left(2 \cdot (- 1) \cdot 2 + 2 \cdot 2 \cdot (- 1) + (- 1) \cdot 2 \cdot 2 - 2 \cdot 2 \cdot 2 - 2 \cdot 2 \cdot (- 1) \cdot \right. \\ \left. \cdot (- 1) \cdot (- 1)\right) = - 1 \\ \end{array}$$

If $\det A = -1$, then it means that $A$ is improper.

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