(1 point) Let $\vec{b}_{1}=\left[\begin{array}{c}-1 \\ 3\end{array}\right]$ and $\vec{b}_{2}=\left[\begin{array}{c}3 \\ -8\end{array}\right]$. The set $B=\left\{\vec{b}_{1}, \vec{b}_{2}\right\}$ is a basis for $\mathbb{R}^{2}$. Let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be a linear transformation such that $T\left(\vec{b}_{1}\right)=2 \vec{b}_{1}+6 \vec{b}_{2}$ and $T\left(\vec{b}_{2}\right)=2 \vec{b}_{1}+2 \vec{b}_{2}$. (a) The matrix of $T$ relative to the basis $B$ is (b) The matrix of $T$ relative to the standard basis $E$ for $\mathbb{R}^{2}$ is

Step 1 ：Given the basis vectors \(\vec{b}_{1} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\) and \(\vec{b}_{2} = \begin{bmatrix} 3 \\ -8 \end{bmatrix}\) for \(\mathbb{R}^{2}\), and the linear transformation T such that \(T(\vec{b}_{1}) = 2\vec{b}_{1} + 6\vec{b}_{2}\) and \(T(\vec{b}_{2}) = 2\vec{b}_{1} + 2\vec{b}_{2}\).

Step 2 ：To find the matrix of T relative to the basis B, we express the images of the basis vectors under T as linear combinations of the basis vectors. The coefficients of these linear combinations form the entries of the matrix. For \(T(\vec{b}_{1})\), the coefficients are 2 and 6, and for \(T(\vec{b}_{2})\), the coefficients are 2 and 2. Therefore, the matrix of T relative to the basis B is \(\boxed{\begin{bmatrix} 2 & 2 \\ 6 & 2 \end{bmatrix}}\).

Step 3 ：To find the matrix of T relative to the standard basis E for \(\mathbb{R}^{2}\), we express the images of the standard basis vectors under T as linear combinations of the standard basis vectors. The coefficients of these linear combinations form the entries of the matrix. For \(T(\vec{e}_{1})\), the coefficients are 16 and -42, and for \(T(\vec{e}_{2})\), the coefficients are 4 and -10. Therefore, the matrix of T relative to the standard basis E for \(\mathbb{R}^{2}\) is \(\boxed{\begin{bmatrix} 16 & 4 \\ -42 & -10 \end{bmatrix}}\).