1.A2+4A−5IA^2+4A-5IA2+4A−5I where A=[124−3]A = \begin{bmatrix}1&2\\4&-3\end{bmatrix}A=[142−3] equalsa.4[2120]4\begin{bmatrix}2&1\\2&0\end{bmatrix}4[2210]b.4[0−122]4\begin{bmatrix}0&-1\\2&2\end{bmatrix}4[02−12]c.32[2120]32\begin{bmatrix}2&1\\2&0\end{bmatrix}32[2210]d.32[1110]32\begin{bmatrix}1&1\\1&0\end{bmatrix}32[1110]Login to continueOnly logged in users canattempt or see the solution.
