91Ó°ÊÓ

Let's find the images of the basis elements \(x^2, x\) and \(1\) under L. We have: i) \(L(x^2) = (x^2)'+ x^2(0) = 2x\) ii) \(L(x)= (x)' + (x)(0) = 1\) iii) \(L(1) = (1)' + (1)(0)= 0\)

Now, let's express the images of the basis elements found in Step 1 in terms of the basis \(\{2, 1-x\}\): i) The image of \(x^2\) under L is \(2x\). To express it in terms of \(\{2, 1-x\}\), we need to find the coefficients \(a\) and \(b\) such that \(2x = a(2) + b(1-x)\). From this equation, we have: \(a=0\) and \(b=-2\). So, \(2x = 0(2) + (-2)(1-x)\) ii) The image of \(x\) under L is \(1\). To express it in terms of \(\{2, 1-x\}\), we need to find the coefficients \(a\) and \(b\) such that \(1 = a(2) + b(1-x)\). From this equation, we have: \(a=\frac{1}{2}\) and \(b=-\frac{1}{2}\). So, \(1 = \frac{1}{2}(2) + ( -\frac{1}{2})(1-x)\) iii) The image of \(1\) under L is \(0\). Since \(0\) is a trivial linear combination, we have: \(0 = 0(2) + 0(1-x)\)

Now, we will construct the matrix representation of L using the coefficients found in Step 2. The matrix A representing L with respect to the given bases is: \[ A = \begin{bmatrix} 0 & 1/2 & 0 \\ -2 & -1/2 & 0 \\ \end{bmatrix} \]

Now, we are ready to find the transformed polynomials for the given input polynomials in terms of the basis \(\{2, 1-x\}\): (a) \(L(x^{2}+2x-3)\): Reach the final answer using the matrix representation on \(p(x)= x^{2}+2x−3\) A[p(x)] = \[ \begin{bmatrix}0 & 1/2 & 0 \\ -2 & -1/2 & 0 \\\end{bmatrix} \begin{bmatrix}1 \\ 2 \\ -3 \\\end{bmatrix} = \begin{bmatrix}\frac{1}{2} \\ -\frac{7}{2} \\ \end{bmatrix}\] Therefore, \(L(x^{2}+2 x-3)= \frac{1}{2}(2) + \frac{-7}{2}(1-x)\) (b) \(L(x^{2}+1)\): Reach the final answer using the matrix representation on \(p(x)= x^{2}+1\) A[p(x)] = \[ \begin{bmatrix}0 & 1/2 & 0 \\ -2 & -1/2 & 0 \\\end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 1 \\\end{bmatrix} = \begin{bmatrix}\frac{1}{2} \\ -\frac{3}{2} \\ \end{bmatrix}\] Therefore, \(L(x^{2}+1)= \frac{1}{2}(2) + \frac{-3}{2}(1-x)\) (c) \(L(3x)\): Reach the final answer using the matrix representation on \(p(x)= 3x\) A[p(x)] = \[ \begin{bmatrix}0 & 1/2 & 0 \\ -2 & -1/2 & 0 \\\end{bmatrix} \begin{bmatrix}0 \\ 3 \\ 0 \\\end{bmatrix} = \begin{bmatrix}\frac{3}{2} \\ -\frac{3}{2} \\ \end{bmatrix}\] Therefore, \(L(3x)= \frac{3}{2}(2) + \frac{-3}{2}(1-x)\) (d) \(L(4x^{2} + 2x)\): Reach the final answer using the matrix representation on \(p(x)= 4x^{2}+2x\) A[p(x)] = \[ \begin{bmatrix}0 & 1/2 & 0 \\ -2 & -1/2 & 0 \\\end{bmatrix} \begin{bmatrix}4 \\ 2 \\ 0 \\\end{bmatrix} = \begin{bmatrix}1 \\ -8 \\ \end{bmatrix}\] Therefore, \(L(4x^{2}+ 2x)= 1(2) + (-8)(1-x)\)