#CSIRNET #mathematics DEC 2024 | Linear Algebra | Evaluation of Linear Transformation

Hello friends in this video, we are going to discuss the problem of CSI and in December 2024 problem of linear algebra question is for a real variable X consider real vector space V a 0 plus a 1 X plus a 2 X square a 0 a 1 a 2 belongs to R. Let T V to V be the linear transformation defined by T of F equal to F plus DF by DX. Here DF by DX is a derivative of F with respect to X. Then each of the following statements are true. We have four statement first statement is T cube minus 3T square plus 3T power 2025 at X equal to X. In the second option the value may be X plus 1 the third option value may be 2025 factorial X and the fourth option value may be 2025 factorial X plus 1. Now we have given a vector space V a 0 plus a 1 X plus a 2 X square. It is a polynomial in X with the real coefficient a 0 a 1 a 2 belongs to R. Now for the given linear transformation T of F equal to F plus DF by DX but in the option we want to evaluate this transformation over X. So we are going to evaluate T over X. So here F equal to X then X plus derivative of X with respect to X is 1. So T of X equal to X plus 1. Now we need T square. So we are going to evaluate the T square. So T square of X equal to T of DX. T of X equal to X plus 1. So T of X plus 1. So here F will be X plus 1. So first term is X plus 1 and derivative of X plus 1 with respect to X. So derivative of X is 1 and derivative of 1 is 0. So therefore we have T square X equal to X plus 1 plus 1 which will be X plus 2. Now we need T cube X. So T cube X equal to T of T square X. We have value of T square X which is X plus 2. Evaluate T over X plus 2. Here F equal to X plus 2. Therefore first term is X plus 2 derivative of X plus 2 with respect to X. So derivative of X is 1 and derivative of 2 is 0. So it is 1. Therefore T cube X equal to X plus 3. Now we need T cube minus 3T square plus 3T. So first we are going to evaluate T cube minus 3T square plus 3T over X. Therefore we have T cube X minus 3T square X plus 3TX. We have value of TX, T square X and T cube X. Substitute that value. Then we have X plus 3 minus 3 times X plus 2 plus 3 times X plus 1. If we simplify then we have X plus 3 minus 3X minus 6 plus 3X plus 3. You can see the minus 3X cancel with the plus 3X. 3 plus 3 is 6. Minus 6 will be 0. Therefore we have remaining X. That means we have T cube minus 3T square plus 3T over X equal to X. Now we want T cube minus 3T square plus 3T power 2025 over X. So first we are going to calculate the square. T cube minus 3T square plus 3T power 2 of X. Therefore you can write down T cube minus 3T square plus 3T of T cube minus 3T square plus 3T X. But this has value X we already calculated. Therefore we have T cube minus 3T square plus 3T of X. But this has value X. Therefore T cube minus 3T square plus 3T power 2 of X has value X. If you repeat this process then we have T cube minus 3T square plus 3T power 2025 of X equal to X. That means X is considered as a fixed point. And therefore the option number 1 is the correct answer. Thanks for watching. For more videos like, share, subscribe and press the bell icon of Next Gen Maths.