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1) Given expression is = (A+I)(A2+A)                                         = A(A2+A)+I(A2+A)                                         = A3+A2+A2+A                                         = A3+2A2+A 2) The given matrix equation (A+B)2 = A2+2AB+B2 is false in general because : (A+B)2 = A2+AB+BA+B2. This is only equal to A2+2AB+B2 when AB = BA, which in general is not true. Here, A = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] and B = \left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right] Now, AB = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right] = \left[\begin{array}{ll}13 & 16 \\ 29 & 36\end{array}\right] and BA = \left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] = \left[\begin{array}{ll}15 & 22 \\ 23 & 34\end{array}\right] Since \left[\begin{array}{ll}13 & 16 \\ 29 & 36\end{array}\right] 7 \left[\begin{array}{ll}15 & 22 \\ 23 & 34\end{array}\right], then AB 7 BA. 3) Given expression is = (A+B)(A2-AB+B2)                                         = A(A2-AB+B2)+B(A2-AB+B2)                                         = A3-A2B+AB2+BA2-BAB+B3 The given equation is false because A2B 7 BA2 and AB2 7 BAB. Here, A = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] and B = \left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right] Now, A2B = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right] = \left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}13 & 16 \\ 29 & 36\end{array}\right] = \left[\begin{array}{cc}71 & 88 \\ 155 & 192\end{array}\right] And, BA2 = \left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] = \left[\begin{array}{ll}15 & 22 \\ 23 & 34\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] = \left[\begin{array}{cc}81 & 118 \\ 125 & 182\end{array}\right] And, AB2 = AB*B = \left[\begin{array}{ll}13 & 16 \\ 29 & 36\end{array}\right]\left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right] = \left[\begin{array}{ll}119 & 148 \\ 267 & 332\end{array}\right] And, BAB = \left[\begin{array}{ll}3 & 4 \\ 5 & 6\end{array}\right]\left[\begin{array}{ll}13 & 16 \\ 29 & 36\end{array}\right] = \left[\begin{array}{ll}155 & 192 \\ 239 & 296\end{array}\right] Therefore, A2B 7 BA2 and AB2 7 BAB. ...