QuestionsQuestion 1If A=[aij]3×3A=[a_{ij}]_{3\times 3}A=[aij]3×3, then show that:(i)I3A=AI_3A=AI3A=A(ii)AI3=AAI_3=AAI3=ASolutionTheoryQuestion 2If A=[0−12321−104]A=\begin{bmatrix}0&-1&2\\3&2&1\\-1&0&4\end{bmatrix}A=03−1−120214, B=[21−1124−121]B=\begin{bmatrix}2&1&-1\\1&2&4\\-1&2&1\end{bmatrix}B=21−1122−141 and C=[10−2−15034−1]C=\begin{bmatrix}1&0&-2\\-1&5&0\\3&4&-1\end{bmatrix}C=1−13054−20−1, then find:(i)A−BA-BA−B(ii)B−CB-CB−C(iii)(A−B)−C(A-B)-C(A−B)−C(iv)A−(B−C)A-(B-C)A−(B−C)SolutionTheoryQuestion 3If A=[−i2i1−i]A=\begin{bmatrix}-i&2i\\1&-i\end{bmatrix}A=[−i12i−i], B=[−i12i1]B=\begin{bmatrix}-i&1\\2i&1\end{bmatrix}B=[−i2i11] and C=[2i2i−ii]C=\begin{bmatrix}2i&2i\\-i&i\end{bmatrix}C=[2i−i2ii], then show that:(i)(AB)C=A(BC)(AB)C=A(BC)(AB)C=A(BC)(ii)A(B+C)=AB+ACA(B+C)=AB+ACA(B+C)=AB+ACSolutionTheoryQuestion 4If AAA and BBB are square matrices of the same order, then explain why in general:(i)(A+B)2≠A2+2AB+B2(A+B)^2\ne A^2+2AB+B^2(A+B)2=A2+2AB+B2(ii)(A−B)2≠A2−2AB+B2(A-B)^2\ne A^2-2AB+B^2(A−B)2=A2−2AB+B2(iii)(A+B)(A−B)≠A2−B2(A+B)(A-B)\ne A^2-B^2(A+B)(A−B)=A2−B2SolutionTheoryQuestion 5If A=[−123102−353]A=\begin{bmatrix}-1&2&3\\1&0&2\\-3&5&3\end{bmatrix}A=−11−3205323, then find A+ATA+A^TA+AT, A−ATA-A^TA−AT, AATAA^TAAT, ATAA^TAATA and (AT)T(A^T)^T(AT)T.SolutionTheoryQuestion 6Solve the matrix equation A2−5A+4I−X=0A^2-5A+4I-X=0A2−5A+4I−X=0 if A=[2012131−10]A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}A=22101−1130.SolutionTheoryQuestion 7If AAA and BBB are two matrices such that AB=BAB=BAB=B and BA=ABA=ABA=A, show that A2+B2=A+BA^2+B^2=A+BA2+B2=A+B.SolutionTheory