Question 1165858: Let matrix3x2 A = 1, -1 , 2, 0, 3, -4. Determine whether multiplication by A is a one-to-one transformation.
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! 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**.
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