Question 1165858
To determine whether multiplication by a matrix $A$ is a **one-to-one transformation**, we need to analyze the properties of the linear transformation $T(\mathbf{x}) = A\mathbf{x}$.

A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is **one-to-one (injective)** if and only if **$T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution** ($\mathbf{x} = \mathbf{0}$). Equivalently, this means the **null space (kernel)** of $A$ contains only the zero vector, or that the matrix $A$ has **linearly independent columns**.

## 1. Analyze the Matrix and Transformation

The given matrix $A$ is a $3 \times 2$ matrix (3 rows, 2 columns):
$$A = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix}$$

The transformation $T$ maps vectors from $\mathbb{R}^2$ to $\mathbb{R}^3$, i.e., $T: \mathbb{R}^2 \to \mathbb{R}^3$.
$$T \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$

## 2. Check for One-to-One Property

We check the condition by solving the homogeneous equation $A\mathbf{x} = \mathbf{0}$:

$$\begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

This corresponds to the system of equations:
1. $x_1 - x_2 = 0$
2. $2x_1 + 0x_2 = 0$
3. $3x_1 - 4x_2 = 0$

From equation (2):
$$2x_1 = 0 \implies x_1 = 0$$

Substitute $x_1 = 0$ into equation (1):
$$0 - x_2 = 0 \implies x_2 = 0$$

Substitute $x_1 = 0$ and $x_2 = 0$ into equation (3) to verify consistency:
$$3(0) - 4(0) = 0 \implies 0 = 0$$

Since the only solution to $A\mathbf{x} = \mathbf{0}$ is the **trivial solution** ($\mathbf{x} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$), the columns of $A$ are **linearly independent**.

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## Conclusion

Yes, multiplication by matrix $A$ **is a one-to-one transformation**.