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\( \begin{array}{l}\text { If } A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] \\ A^{5} B^{7}+A^{5} B^{8}=\end{array} \quad B=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] \) then |
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Answer» A = \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) A2 = \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\)\(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) = \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) = A B = \(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) Then B2 = \(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\)\(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) = \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 &0 &1 \end{bmatrix}\) = I Now, A5B7 + A5B8 = (A2)2 A(B2)3B + (A2)2 A(B2)4 = A2 A I3B + A2 A(I)4 (∵ A2 = A and B2 = I)a = A2 B + A2 I (∵ I3 = I4 = I, A2 = A) = AB + A (∵ A2 I = A2 = A) Now, AB = \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\)\(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) = \(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 0 &0 &1 \end{bmatrix}\) And AB + A = \(\begin{bmatrix} 0 &0 & 1 \\ 0 & 1 & 0 \\ 0 &0 &1 \end{bmatrix}\) + \(\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 1 &0 &0 \end{bmatrix}\) = \(\begin{bmatrix} 1 &0 &1 \\ 0 & 1 & 0 \\ 1 &0 & 1 \end{bmatrix}\) Therefore, A5B7 + A5B8 = AB + A = \(\begin{bmatrix} 1 &0 &1 \\ 0 & 1 & 0 \\ 1 &0 & 1 \end{bmatrix}\) |
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