Let's analyze the transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by:

[ T(x_1, x_2, x_3) = \left( 6 \sqrt[3]{x_1 x_2 x_3}, 6x_1 - 4x_2 - x_3, 8 \sqrt[3]{x_3^2 x_1} \right) ]

We need to determine which of the following statements is true: A. $T$ is a linear transformation.
B. $T$ does not preserve the zero vector.
C. $T$ does not preserve addition.
D. $T$ does not preserve scalar multiplication.

Step 1: Linear Transformation Check

For $T$ to be a linear transformation, it must satisfy two properties:

1. Preserve addition:
   $T(\mathbf{v} + \mathbf{w}) = T(\mathbf{v}) + T(\mathbf{w})$ for any vectors $\mathbf{v}, \mathbf{w} \in \mathbb{R}^3$.

2. Preserve scalar multiplication:
   $T(c \mathbf{v}) = c T(\mathbf{v})$ for any scalar $c \in \mathbb{R}$ and vector $\mathbf{v} \in \mathbb{R}^3$.

Let's check these properties for the given transformation.

Step 2: Preservation of Addition

We need to check if $T$ preserves vector addition, i.e., if for any vectors $\mathbf{v} = (v_1, v_2, v_3)$ and $\mathbf{w} = (w_1, w_2, w_3)$, the following holds: $T(v_1 + w_1, v_2 + w_2, v_3 + w_3) = T(v_1, v_2, v_3) + T(w_1, w_2, w_3)$

However, from the first and third components of the transformation: $T(v_1 + w_1, v_2 + w_2, v_3 + w_3) = \left( 6 \sqrt[3]{(v_1+w_1)(v_2+w_2)(v_3+w_3)}, \dots \right)$ This clearly does not equal: $T(v_1, v_2, v_3) + T(w_1, w_2, w_3)$ because cube roots and products do not distribute over sums in this way. Hence, T does not preserve addition.

Step 3: Preservation of Scalar Multiplication

We need to check if $T$ preserves scalar multiplication, i.e., if for any scalar $c$ and vector $\mathbf{v} = (v_1, v_2, v_3)$, the following holds: $T(c v_1, c v_2, c v_3) = c T(v_1, v_2, v_3)$

For the first component of $T$: [ T(c v_1, c v_2, c v_3) = \left( 6 \sqrt[3]{(c v_1)(c v_2)(c v_3)}, \dots \right) = \left( 6 \sqrt[3]{c^3 v_1 v_2 v_3}, \dots \right) = \left( 6 c \sqrt[3]{v_1 v_2 v_3}, \dots \right) ] This is not equal to $c T(v_1, v_2, v_3)$, as the second component does not multiply the scalar linearly, and the cube root affects the scalar $c$ in the first and third components. Hence, T does not preserve scalar multiplication.

Step 4: Zero Vector Check

For $T$ to preserve the zero vector, we need to verify if $T(0, 0, 0) = (0, 0, 0)$.

The transformation of the zero vector is: [ T(0, 0, 0) = \left( 6 \sqrt[3]{0 \cdot 0 \cdot 0}, 6 \cdot 0 - 4 \cdot 0 - 0, 8 \sqrt[3]{0^2 \cdot 0} \right) = (0, 0, 0) ] Thus, T preserves the zero vector.

Conclusion:

• A. $T$ is not a linear transformation.
• B. $T$ preserves the zero vector.
• C. $T$ does not preserve addition.
• D. $T$ does not preserve scalar multiplication.

The correct answers are C and D.

Would you like more details on any of these concepts?

Here are 5 related questions to explore:

1. What are the properties of a linear transformation?
2. How does a non-linear transformation behave geometrically?
3. How can we test if a transformation preserves scalar multiplication in different dimensions?
4. What are some examples of linear and non-linear transformations in physics?
5. Why does preserving the zero vector matter in transformations?

Tip: Always test both addition and scalar multiplication to verify if a transformation is linear.