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 #64390 – Math – Linear Algebra

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)

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 (the sum of the products of items) 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 (the sum of the squares) are as follows:

for $1^{\text{st}}$ row, $\frac{1}{9}(2^2 + 2^2 + (-1)^2) = 1$;

for $2^{\text{nd}}$ row, $\frac{1}{9}(2^2 + (-1)^2 + 2^2) = 1$;

for $3^{\text{rd}}$ row, $\frac{1}{9}((-1)^2 + 2^2 + 2^2) = 1$.

This means that

$$A A^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = E,$$

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

Next,

$$\begin{aligned} \text{Det } A &= \frac{1}{3^3} \left(2 \cdot (-1) \cdot 2 + 2 \cdot 2 \cdot (-1) + (-1) \cdot 2 \cdot 2 - 2 \cdot 2 \cdot 2 - 2 \cdot 2 \cdot (-1) \right. \\ &\quad \left. \cdot (-1) \cdot (-1) \right) = -1 \end{aligned}$$

If $\text{Det } A = -1$, then it means that $A$ is improper.

QED.

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