作业 1010 返回 目录 此页面提供 作业 1010 有关资料。 此作业的试题 PDF 提供于以下 链接 。 答案见下。 10/10 作业 Homework 8 m≠−2m \ne -2m=−2 or 111 , sol. = (−(1m+2)m+1m+2m+1m+2)\begin{pmatrix}\mathop{-}\left( \frac{1}{m\mathop{+}2}\right) \\ \frac{m\mathop{+}1}{m\mathop{+}2}\\ \frac{m\mathop{+}1}{m\mathop{+}2}\end{pmatrix}−(m+21)m+2m+1m+2m+1 . A∗=(321242123)A^* = \begin{pmatrix}3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 3\end{pmatrix}A∗=321242123 , A−1=(34121412112141234)A^{-1} = \begin{pmatrix}\frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{4} & \frac{1}{2} & \frac{3}{4}\end{pmatrix}A−1=43214121121412143 . A∗=(000−10−10000−10−1000)A^* = \begin{pmatrix}0 & 0 & 0 & \mathop{-}1\\ 0 & \mathop{-}1 & 0 & 0\\ 0 & 0 & \mathop{-}1 & 0\\ \mathop{-}1 & 0 & 0 & 0\end{pmatrix}A∗=000−10−10000−10−1000 , B∗=(240000120000800006)B^* = \begin{pmatrix}24 & 0 & 0 & 0\\ 0 & 12 & 0 & 0\\ 0 & 0 & 8 & 0\\ 0 & 0 & 0 & 6\end{pmatrix}B∗=240000120000800006 . (1) ∣A∣2A|A|^2 A∣A∣2A . (2) Inverse. (3) (AB)∗=BA(AB)^* = BA(AB)∗=BA . Optional (dx1,dy1,dp)=(−(R1−1)R2(α−1)αde2R2α−R1α+R1−R2α(R1α−α−R1)de2R2α−R1α+R1−R1R2αde2(α−1) (R2α−R1α+R1))(dx_1, dy_1, dp) = \begin{pmatrix} - \frac{\left( {R_1}\mathop{-}1\right) {R_2} \left( \alpha\mathop{-}1\right) \alpha d {e_2}}{{R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}}\\ -\frac{{R_2} \alpha \left( {R_1} \alpha\mathop{-}\alpha\mathop{-}{R_1}\right) d {e_2}}{{R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}}\\ -\frac{{R_1} {R_2} \alpha d {e_2}}{\left( \alpha\mathop{-}1\right) \, \left( {R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}\right) }\end{pmatrix}(dx1,dy1,dp)=−R2α−R1α+R1(R1−1)R2(α−1)αde2−R2α−R1α+R1R2α(R1α−α−R1)de2−(α−1)(R2α−R1α+R1)R1R2αde2 . If de2>0de_2 > 0de2>0 , then dy1>0dy_1 > 0dy1>0 and dp>0dp > 0dp>0 . Above Rendered from Markdown
10/10 作业 Homework 8 m≠−2m \ne -2m=−2 or 111 , sol. = (−(1m+2)m+1m+2m+1m+2)\begin{pmatrix}\mathop{-}\left( \frac{1}{m\mathop{+}2}\right) \\ \frac{m\mathop{+}1}{m\mathop{+}2}\\ \frac{m\mathop{+}1}{m\mathop{+}2}\end{pmatrix}−(m+21)m+2m+1m+2m+1 . A∗=(321242123)A^* = \begin{pmatrix}3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 3\end{pmatrix}A∗=321242123 , A−1=(34121412112141234)A^{-1} = \begin{pmatrix}\frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{4} & \frac{1}{2} & \frac{3}{4}\end{pmatrix}A−1=43214121121412143 . A∗=(000−10−10000−10−1000)A^* = \begin{pmatrix}0 & 0 & 0 & \mathop{-}1\\ 0 & \mathop{-}1 & 0 & 0\\ 0 & 0 & \mathop{-}1 & 0\\ \mathop{-}1 & 0 & 0 & 0\end{pmatrix}A∗=000−10−10000−10−1000 , B∗=(240000120000800006)B^* = \begin{pmatrix}24 & 0 & 0 & 0\\ 0 & 12 & 0 & 0\\ 0 & 0 & 8 & 0\\ 0 & 0 & 0 & 6\end{pmatrix}B∗=240000120000800006 . (1) ∣A∣2A|A|^2 A∣A∣2A . (2) Inverse. (3) (AB)∗=BA(AB)^* = BA(AB)∗=BA . Optional (dx1,dy1,dp)=(−(R1−1)R2(α−1)αde2R2α−R1α+R1−R2α(R1α−α−R1)de2R2α−R1α+R1−R1R2αde2(α−1) (R2α−R1α+R1))(dx_1, dy_1, dp) = \begin{pmatrix} - \frac{\left( {R_1}\mathop{-}1\right) {R_2} \left( \alpha\mathop{-}1\right) \alpha d {e_2}}{{R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}}\\ -\frac{{R_2} \alpha \left( {R_1} \alpha\mathop{-}\alpha\mathop{-}{R_1}\right) d {e_2}}{{R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}}\\ -\frac{{R_1} {R_2} \alpha d {e_2}}{\left( \alpha\mathop{-}1\right) \, \left( {R_2} \alpha\mathop{-}{R_1} \alpha\mathop{+}{R_1}\right) }\end{pmatrix}(dx1,dy1,dp)=−R2α−R1α+R1(R1−1)R2(α−1)αde2−R2α−R1α+R1R2α(R1α−α−R1)de2−(α−1)(R2α−R1α+R1)R1R2αde2 . If de2>0de_2 > 0de2>0 , then dy1>0dy_1 > 0dy1>0 and dp>0dp > 0dp>0 . Above Rendered from Markdown