To show that the transformation $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ defined by

$T(x, y, z) = \left(2x - 4y + 3z + b, 6x + cxyz\right)$

is linear if and only if $b = 0$ and $c = 0$, we must first recall the definition of a linear transformation. A transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is linear if it satisfies two properties for all vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $\alpha \in \mathbb{R}$:

1. Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$.
2. Homogeneity: $T(\alpha \mathbf{u}) = \alpha T(\mathbf{u})$.

Let's check if $T$ satisfies these properties.

Additivity

Consider two arbitrary vectors $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in $\mathbb{R}^3$. We compute $T((x_1, y_1, z_1) + (x_2, y_2, z_2))$ and compare it with $T(x_1, y_1, z_1) + T(x_2, y_2, z_2)$.

$T((x_1, y_1, z_1) + (x_2, y_2, z_2)) = T(x_1 + x_2, y_1 + y_2, z_1 + z_2)$ $= \left( 2(x_1 + x_2) - 4(y_1 + y_2) + 3(z_1 + z_2) + b, 6(x_1 + x_2) + c(x_1 + x_2)(y_1 + y_2)(z_1 + z_2) \right)$

Expanding the terms:

$= \left( (2x_1 - 4y_1 + 3z_1) + (2x_2 - 4y_2 + 3z_2) + b, 6x_1 + 6x_2 + c(x_1y_1z_1 + x_1y_2z_2 + x_2y_1z_2 + x_2y_2z_1) \right)$

On the other hand:

$T(x_1, y_1, z_1) + T(x_2, y_2, z_2) = \left( 2x_1 - 4y_1 + 3z_1 + b, 6x_1 + c x_1 y_1 z_1 \right) + \left( 2x_2 - 4y_2 + 3z_2 + b, 6x_2 + c x_2 y_2 z_2 \right)$

$= \left( (2x_1 - 4y_1 + 3z_1) + (2x_2 - 4y_2 + 3z_2) + 2b, 6x_1 + 6x_2 + c(x_1y_1z_1 + x_2y_2z_2) \right)$

For additivity to hold, the results must be identical:

1. The first component will match if $b = 0$.
2. The second component will match if $c(x_1y_2z_2 + x_2y_1z_1 + x_2y_2z_1 + x_1y_1z_2) = 0$ for all $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. This requires $c = 0$.

Homogeneity

Consider a scalar $\alpha$ and a vector $(x, y, z)$. We compute $T(\alpha(x, y, z))$ and compare it with $\alpha T(x, y, z)$.

$T(\alpha x, \alpha y, \alpha z) = \left( 2\alpha x - 4\alpha y + 3\alpha z + b, 6\alpha x + c(\alpha x)(\alpha y)(\alpha z) \right)$

This simplifies to:

$= \left( \alpha(2x - 4y + 3z) + b, 6\alpha x + c\alpha^3 xyz \right)$

On the other hand:

$\alpha T(x, y, z) = \alpha \left( 2x - 4y + 3z + b, 6x + cxyz \right)$

$= \left( \alpha(2x - 4y + 3z + b), 6\alpha x + \alpha cxyz \right)$

For homogeneity to hold, the results must be identical:

1. The first component will match if $b = 0$.
2. The second component will match if $c(\alpha^3 - \alpha)xyz = 0$ for all $\alpha$ and $(x, y, z)$. This requires $c = 0$.

Conclusion

The transformation $T$ is linear if and only if both $b = 0$ and $c = 0$.

Would you like more details on this topic or any other questions? Here are some related questions to consider:

1. What is a linear transformation in general, and what are some examples?
2. How do we determine if a function is linear?
3. What are the properties of vector spaces related to linear transformations?
4. Can we discuss the concept of linearity in the context of matrices?
5. How does the rank of a transformation relate to its linearity?

Tip: Always test both additivity and homogeneity when checking for linearity.