To determine whether the linear transformation represented by each matrix is one-to-one or onto, we need to assess the following:

1. One-to-one: A transformation is one-to-one (injective) if the kernel of the matrix only contains the zero vector, i.e., the matrix has full column rank. This means there are no non-zero vectors that map to the zero vector.
2. Onto: A transformation is onto (surjective) if the image of the matrix spans the entire codomain. This is the case if the matrix has full row rank.

Here’s the breakdown for each matrix:

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Matrix A

1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ - **One-to-one**: Yes. The matrix is an identity matrix, so it has full rank (rank = 3). Therefore, the transformation is one-to-one. - **Onto**: Yes. The matrix has full row rank (rank = 3), so the transformation is onto. --- ### Matrix B $$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix}$$ - **One-to-one**: Yes. The matrix has full column rank (rank = 2), and the transformation is one-to-one. - **Onto**: No. The matrix only has 2 rows, so it cannot span 3-dimensional space. It is not onto. --- ### Matrix C $$C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$ - **One-to-one**: Yes. The matrix has full column rank (rank = 2), so the transformation is one-to-one. - **Onto**: No. There are only two rows, so the matrix cannot map to 3-dimensional space. It is not onto. --- ### Matrix D $$D = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ - **One-to-one**: No. This matrix has rank 0, so the transformation is not injective. - **Onto**: No. The matrix does not span any non-zero space (rank = 0), so it is not onto. --- ### Matrix E $$E = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$ - **One-to-one**: Yes. The matrix has full column rank (rank = 3), so it is one-to-one. - **Onto**: Yes. The matrix has full row rank (rank = 3), so it is onto. --- ### Matrix F $$F = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$$ - **One-to-one**: Yes. The matrix has full column rank (rank = 3), so the transformation is one-to-one. - **Onto**: Yes. The matrix has full row rank (rank = 3), so it is onto. --- ### Matrix G $$G = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 2 & 0 \\ 4 & 4 & 4 \end{bmatrix}$$ - **One-to-one**: No. The matrix does not have full column rank, as the second and third columns are linearly dependent. Therefore, it is not one-to-one. - **Onto**: No. The matrix has rank 2 (not 3), so it is not onto. --- ### Summary: - **A**: One-to-one, Onto - **B**: One-to-one, Not Onto - **C**: One-to-one, Not Onto - **D**: Not One-to-one, Not Onto - **E**: One-to-one, Onto - **F**: One-to-one, Onto - **G**: Not One-to-one, Not Onto Would you like any further explanations or clarifications on these results? --- Here are some related questions for you to explore: 1. How do the concepts of rank relate to the properties of one-to-one and onto transformations? 2. Can a matrix have full column rank but not full row rank? What does this imply about the transformation? 3. What happens when the rank of a matrix is less than both the number of rows and columns? 4. What is the significance of the determinant in checking the injectiveness of a matrix? 5. How can you use row reduction to check if a matrix is one-to-one or onto? Tip: A simple way to check if a matrix is one-to-one is to look at whether the columns are linearly independent. If the columns are linearly independent, the matrix will be injective.