(1 point) If $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is a linear transformation such that \[ T\left(\left[\begin{array}{l} 1 \\ 2 \end{array}\right]\right)=\left[\begin{array}{c} 5 \\ 11 \\ 7 \end{array}\right], \text { and } T\left(\left[\begin{array}{c} 2 \\ -3 \end{array}\right]\right)=\left[\begin{array}{c} 17 \\ -6 \\ -7 \end{array}\right] \] then the standard matrix of $T$ is

Step 1 ：The standard matrix of a linear transformation T is the matrix A such that T(x) = Ax for all x in the domain of T. In this case, we are given two vectors in the domain of T and their corresponding images under T. We can use these to find the standard matrix of T.

Step 2 ：The standard matrix of T is a 3x2 matrix, because T maps from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{3}\). The columns of the standard matrix are the images of the standard basis vectors under T. We can find these by expressing the standard basis vectors as linear combinations of the given vectors, and then applying T to these linear combinations.

Step 3 ：The standard basis vectors in \(\mathbb{R}^{2}\) are [1, 0] and [0, 1]. We can express these as linear combinations of the given vectors as follows: [1, 0] = a*[1, 2] + b*[2, -3] and [0, 1] = c*[1, 2] + d*[2, -3].

Step 4 ：We can solve these systems of equations to find a, b, c, and d. Then we can apply T to these linear combinations to find the columns of the standard matrix.

Step 5 ：By solving the equations, we get the solutions as {a: 3/7, b: 2/7} for the first system and {c: 2/7, d: -1/7} for the second system.

Step 6 ：Applying T to these linear combinations, we get the columns of the standard matrix as [7, 3, 1] and [-1, 4, 3].

Step 7 ：Thus, the standard matrix of the linear transformation T is \(\boxed{\begin{bmatrix} 7 & -1 \\ 3 & 4 \\ 1 & 3 \end{bmatrix}}\).