1.If A=(1525−2515)A = \begin{pmatrix} \tfrac{1}{\sqrt{5}} & \tfrac{2}{\sqrt{5}} \\ -\tfrac{2}{\sqrt{5}} & \tfrac{1}{\sqrt{5}} \end{pmatrix}A=(51−525251), B=(10i1)B = \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix}B=(1i01), i=−1i = \sqrt{-1}i=−1, and Q=ATBAQ = A^T B AQ=ATBA, then the inverse of the matrix AQ2021ATA Q^{2021} A^TAQ2021AT is equal to:a.(1−202101)\begin{pmatrix} 1 & -2021 \\ 0 & 1 \end{pmatrix}(10−20211)b.(10−2021i1)\begin{pmatrix} 1 & 0 \\ -2021 i & 1 \end{pmatrix}(1−2021i01)c.(15−2021202115)\begin{pmatrix} \tfrac{1}{\sqrt{5}} & -2021 \\ 2021 & \tfrac{1}{\sqrt{5}} \end{pmatrix}(512021−202151)d.(102021i1)\begin{pmatrix} 1 & 0 \\ 2021 i & 1 \end{pmatrix}(12021i01)Login to continueOnly logged in users canattempt or see the solution.
