3 Consider the map \( F: \mathbb{R}^{3} \rightarrow \mathbb{R}^{4} \) given by
\[
F(x, y, z)=(x-z, y+z, 4 x+3 y-2 z, z) .
\]
a. [1 point \( ] \) Prove that \( F \) is a linear map.
b. [1 point \( ] \) Determine the matrix associated with the linear map \( F \).
Answer
Determine matrix: F(x, y, z)=(x-z, y+z, 4x+3y-2z, z) can be represented as a matrix multiplication with A as the matrix: \( \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 4 & 3 & -2 \\ 0 & 0 & 1 \end{pmatrix} \). Therefore, the matrix associated with the linear map F is \( A=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 4 & 3 & -2 \\ 0 & 0 & 1 \end{pmatrix} \).
Step 1 ：Prove linearity: Let \( u=(x_1, y_1, z_1) \) and \( v=(x_2, y_2, z_2) \) in \( \mathbb{R}^{3} \) and scalar \( \alpha \). Then, \( F(u+v)=F(x_1+x_2, y_1+y_2, z_1+z_2)=(x_1+x_2-z_1-z_2, y_1+y_2+z_1+z_2, 4(x_1+x_2)+3(y_1+y_2)-2(z_1+z_2), z_1+z_2) \) and \( F(\alpha u)=F(\alpha x_1, \alpha y_1, \alpha z_1)=(\alpha x_1-\alpha z_1, \alpha y_1+\alpha z_1, 4\alpha x_1+3\alpha y_1-2\alpha z_1, \alpha z_1) \).
Step 2 ：Apply properties: \( F(u)+F(v)=(x_1-z_1, y_1+z_1, 4x_1+3y_1-2z_1, z_1)+(x_2-z_2, y_2+z_2, 4x_2+3y_2-2z_2, z_2)=(x_1+z_1-x_2-z_2, y_1+y_2+z_1+z_2, 4(x_1+x_2)+3(y_1+y_2)-2(z_1+z_2), z_1+z_2) \) and \( \alpha F(u)=(\alpha(x_1-z_1), \alpha(y_1+z_1), \alpha(4x_1+3y_1-2z_1), \alpha z_1) \). Since \( F(u+v)=F(u)+F(v) \) and \( F(\alpha u)=\alpha F(u) \), F is a linear map.
Step 3 ：Determine matrix: F(x, y, z)=(x-z, y+z, 4x+3y-2z, z) can be represented as a matrix multiplication with A as the matrix: \( \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 4 & 3 & -2 \\ 0 & 0 & 1 \end{pmatrix} \). Therefore, the matrix associated with the linear map F is \( A=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 4 & 3 & -2 \\ 0 & 0 & 1 \end{pmatrix} \).
