Question -112.5 points HoltLinAlg2 3.2.016 Expand the given matrix expression and combine as many terms as possible. Assume that all matrices are n × n. (A DAA) +-12.5 points HoltLinAlg2 3.2.019 The given matrix equation is false in general. Explain why the equation is false. Assume that all matrices are nxn. (A+B)2 = A2 + 2AB + B2 O (A B)2 ABBA B2. This is only -112.5 points HoltLinAlg2 3.2.016 Expand the given matrix expression and combine as many terms as possible. Assume that all matrices are n × n. (A DAA) +-12.5 points HoltLinAlg2 3.2.019 The given matrix equation is false in general. Explain why the equation is false. Assume that all matrices are nxn. (A+B)2 = A2 + 2AB + B2 O (A B)2 ABBA B2. This is only equal to A22ABB2 when AB BA, which in general is not true. O (A+ B)2 #A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB-BA, which in general is not true. (AB)2 -A2 BABA82 -A282. This is only equal to A22ABB2 when A-0nn or 8- Onni which in general is not true. O (A B)2 - A2 + AB AB B2-A2 + B2. This is only equal to A2 2AB B2 when A or , which in general is not true. Give a 2 × 2 example showing the equation is false. 1 2 3 4

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. ...