东营有能做网站优化,Wordpress西联,广州越秀区发布紧急通告,做网站的组要具备哪些素质三 矩阵乘法和逆矩阵 1. 矩阵乘法1.1 常规方法1.2 列向量组合1.3 行向量组合1.4 单行和单列的乘积和1.5 块乘法 2. 逆矩阵2.1 逆矩阵的定义2.2 奇异矩阵2.3 Gauss-Jordan 求逆矩阵2.3.1 求逆矩阵 ⟺ \Longleftrightarrow ⟺解方程组2.3.2 Gauss-Jordan求逆矩阵 1. 矩阵乘法
1.… 三 矩阵乘法和逆矩阵 1. 矩阵乘法1.1 常规方法1.2 列向量组合1.3 行向量组合1.4 单行和单列的乘积和1.5 块乘法 2. 逆矩阵2.1 逆矩阵的定义2.2 奇异矩阵2.3 Gauss-Jordan 求逆矩阵2.3.1 求逆矩阵 ⟺ \Longleftrightarrow ⟺解方程组2.3.2 Gauss-Jordan求逆矩阵 1. 矩阵乘法
1.1 常规方法 [ . . . . . . . . . . . . a 31 a 32 a 33 a 34 . . . . . . . . . . . . ] ⏟ A m ∗ n [ . . . . . . . . . b 14 . . . . . . . . . b 24 . . . . . . . . . b 34 . . . . . . . . . b 44 ] ⏟ B n ∗ p [ . . . . . . . . . . . . . . . . . . . . . C 34 . . . . . . . . . . . . ] ⏟ C m ∗ p \underbrace{\begin{bmatrix} ............\\ a_{31}a_{32}a_{33}a_{34}\\ ............\\ \end{bmatrix}}_{A_{m*n}} \underbrace{\begin{bmatrix} .........b_{14}\\ .........b_{24}\\ .........b_{34}\\ .........b_{44} \end{bmatrix}}_{B_{n*p}} \underbrace{\begin{bmatrix} ............\\ .........C_{34}\\ ............ \end{bmatrix}}_{C_{m*p}} Am∗n ...a31......a32......a33......a34... Bn∗p ....................................b14b24b34b44 Cm∗p ..............................C34... C 34 A r o w 3 ∗ B c o l 4 ∑ i 1 n a 3 i ∗ b i 4 C_{34} A_{row_3}*B_{col_4} \sum\limits_{i1}^{n}a_{3i}*b_{i4} C34Arow3∗Bcol4i1∑na3i∗bi4
1.2 列向量组合
已知 [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 11 B 21 B 31 ] B 11 ∗ A c o l 1 B 21 ∗ A c o l 2 B 31 ∗ A c o l 3 [ B 11 ∗ A 11 B 21 ∗ A 12 B 31 ∗ A 13 B 11 ∗ A 21 B 21 ∗ A 22 B 31 ∗ A 23 B 11 ∗ A 31 B 21 ∗ A 32 B 31 ∗ A 33 ] \begin{aligned} \begin{bmatrix} A_{11}A_{12}A_{13}\\ A_{21}A_{22}A_{23}\\ A_{31}A_{32}A_{33} \end{bmatrix} \begin{bmatrix} B_{11}\\ B_{21}\\ B_{31} \end{bmatrix} B_{11}*A_{col1}B_{21}*A_{col2}B_{31}*A_{col3} \newline \begin{bmatrix} B_{11}*A_{11}B_{21}*A_{12}B_{31}*A_{13}\\ B_{11}*A_{21}B_{21}*A_{22}B_{31}*A_{23}\\ B_{11}*A_{31}B_{21}*A_{32}B_{31}*A_{33} \end{bmatrix}\end{aligned} A11A21A31A12A22A32A13A23A33 B11B21B31 B11∗Acol1B21∗Acol2B31∗Acol3 B11∗A11B21∗A12B31∗A13B11∗A21B21∗A22B31∗A23B11∗A31B21∗A32B31∗A33 那么 [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] ⏟ A [ B 11 B 12 B 21 B 22 B 31 B 32 ] ⏟ B [ B 11 ∗ A c o l 1 B 21 ∗ A c o l 2 B 31 ∗ A c o l 3 B 12 ∗ A c o l 1 B 22 ∗ A c o l 2 B 32 ∗ A c o l 3 ] ⏟ C [ B 11 ∗ A 11 B 21 ∗ A 12 B 31 ∗ A 13 B 12 ∗ A 11 B 22 ∗ A 12 B 32 ∗ A 13 B 11 ∗ A 21 B 21 ∗ A 22 B 31 ∗ A 23 B 12 ∗ A 21 B 22 ∗ A 22 B 32 ∗ A 23 B 11 ∗ A 31 B 21 ∗ A 32 B 31 ∗ A 33 B 12 ∗ A 31 B 22 ∗ A 32 B 32 ∗ A 33 ] \begin{aligned} \underbrace{\begin{bmatrix} A_{11}A_{12}A_{13}\\ A_{21}A_{22}A_{23}\\ A_{31}A_{32}A_{33} \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} B_{11}B_{12}\\ B_{21}B_{22}\\ B_{31}B_{32} \end{bmatrix}}_{B} \underbrace{\begin{bmatrix}B_{11}*A_{col1}B_{21}*A_{col2}B_{31}*A_{col3} B_{12}*A_{col1}B_{22}*A_{col2}B_{32}*A_{col3}\end{bmatrix}}_{C} \newline \begin{bmatrix} B_{11}*A_{11}B_{21}*A_{12}B_{31}*A_{13} B_{12}*A_{11}B_{22}*A_{12}B_{32}*A_{13}\\ B_{11}*A_{21}B_{21}*A_{22}B_{31}*A_{23} B_{12}*A_{21}B_{22}*A_{22}B_{32}*A_{23}\\ B_{11}*A_{31}B_{21}*A_{32}B_{31}*A_{33} B_{12}*A_{31}B_{22}*A_{32}B_{32}*A_{33} \end{bmatrix}\end{aligned} A A11A21A31A12A22A32A13A23A33 B B11B21B31B12B22B32 C [B11∗Acol1B21∗Acol2B31∗Acol3B12∗Acol1B22∗Acol2B32∗Acol3] B11∗A11B21∗A12B31∗A13B11∗A21B21∗A22B31∗A23B11∗A31B21∗A32B31∗A33B12∗A11B22∗A12B32∗A13B12∗A21B22∗A22B32∗A23B12∗A31B22∗A32B32∗A33 C矩阵是A矩阵的列向量组合
1.3 行向量组合
已知 [ A 11 A 12 A 13 ] [ B 11 B 12 B 21 B 22 B 31 B 32 ] A 11 ∗ B r o w 1 A 12 ∗ B r o w 2 A 13 ∗ B r o w 3 [ A 11 ∗ B 11 A 11 ∗ B 12 A 12 ∗ B 21 A 12 ∗ B 22 A 13 ∗ B 31 A 13 ∗ B 32 ] \begin{aligned} \begin{bmatrix} A_{11}A_{12}A_{13} \end{bmatrix} \begin{bmatrix} B_{11}B_{12}\\ B_{21}B_{22}\\ B_{31}B_{32} \end{bmatrix} A_{11}*B_{row1}A_{12}*B_{row2}A_{13}*B_{row3} \newline \begin{bmatrix} A_{11}*B_{11}A_{11}*B_{12}\\ \\ A_{12}*B_{21}A_{12}*B_{22}\\ \\ A_{13}*B_{31}A_{13}*B_{32} \end{bmatrix}\end{aligned} [A11A12A13] B11B21B31B12B22B32 A11∗Brow1A12∗Brow2A13∗Brow3 A11∗B11A12∗B21A13∗B31A11∗B12A12∗B22A13∗B32 那么 [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] ⏟ A [ B 11 B 12 B 21 B 22 B 31 B 32 ] ⏟ B [ A 11 ∗ B r o w 1 A 12 ∗ B r o w 2 A 13 ∗ B r o w 3 A 21 ∗ B r o w 1 A 22 ∗ B r o w 2 A 23 ∗ B r o w 3 A 31 ∗ B r o w 1 A 32 ∗ B r o w 2 A 33 ∗ B r o w 3 ] ⏟ C \begin{aligned} \underbrace{\begin{bmatrix} A_{11}A_{12}A_{13}\\ A_{21}A_{22}A_{23}\\ A_{31}A_{32}A_{33} \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} B_{11}B_{12}\\ B_{21}B_{22}\\ B_{31}B_{32} \end{bmatrix}}_{B} \underbrace{\begin{bmatrix} A_{11}*B_{row1}A_{12}*B_{row2}A_{13}*B_{row3}\\ A_{21}*B_{row1}A_{22}*B_{row2}A_{23}*B_{row3}\\ A_{31}*B_{row1}A_{32}*B_{row2}A_{33}*B_{row3} \end{bmatrix}}_{C} \newline \end{aligned} A A11A21A31A12A22A32A13A23A33 B B11B21B31B12B22B32 C A11∗Brow1A12∗Brow2A13∗Brow3A21∗Brow1A22∗Brow2A23∗Brow3A31∗Brow1A32∗Brow2A33∗Brow3 C矩阵是B矩阵的行向量组合
1.4 单行和单列的乘积和 [ 2 7 3 8 4 9 ] [ 1 6 1 1 ] [ 2 3 4 ] [ 1 6 ] [ 7 8 9 ] [ 1 1 ] [ 9 19 11 26 13 33 ] \begin{aligned} \begin{bmatrix} 27\\ 38\\ 49 \end{bmatrix} \begin{bmatrix} 16\\ 11\\ \end{bmatrix} \begin{bmatrix} 2\\ 3\\ 4 \end{bmatrix} \begin{bmatrix} 16\\ \end{bmatrix} \begin{bmatrix} 7\\ 8\\ 9 \end{bmatrix} \begin{bmatrix} 11\\ \end{bmatrix} \newline \begin{bmatrix} 919\\ 1126\\ 1333 \end{bmatrix} \end{aligned} 234789 [1161] 234 [16] 789 [11] 91113192633
1.5 块乘法 [ A 1 ∣ A 2 —— —— —— A 3 ∣ A 4 ] [ B 1 ∣ B 2 —— —— —— B 3 ∣ B 4 ] [ A 1 ∗ B 1 A 2 ∗ B 3 ∣ A 1 ∗ B 2 A 2 ∗ B 4 ———————— —— ———————— A 3 ∗ B 1 A 4 ∗ B 3 ∣ A 3 ∗ B 2 A 4 ∗ B 4 ] \begin{bmatrix} A_{1}|A_{2}\\ ——————\\ A_{3}|A_{4} \end{bmatrix} \begin{bmatrix} B_{1}|B_{2}\\ ——————\\ B_{3}|B_{4} \end{bmatrix} \begin{bmatrix} A_{1}*B_{1}A_2*B_{3}|A_{1}*B_{2}A_2*B_{4}\\ ——————————————————\\ A_{3}*B_{1}A_4*B_{3}|A_{3}*B_{2}A_4*B_{4} \end{bmatrix} A1——A3∣——∣A2——A4 B1——B3∣——∣B2——B4 A1∗B1A2∗B3————————A3∗B1A4∗B3∣——∣A1∗B2A2∗B4————————A3∗B2A4∗B4
2. 逆矩阵
2.1 逆矩阵的定义
存在 A − 1 A I A^{-1}A I A−1AI 那么称 A − 1 A^{-1} A−1为A的逆矩阵A是可逆的记为非奇异矩阵
当A为方阵行数列数时左逆矩阵右逆矩阵 A − 1 A I A A − 1 A^{-1}A IAA^{-1} A−1AIAA−1
2.2 奇异矩阵
存在 A x 0 ( x 非零向量 ) ⇒ A 不可逆 Ax0(x非零向量)\Rightarrow A不可逆 Ax0(x非零向量)⇒A不可逆 证明如下 A x 0 ⇒ A − 1 A I A − 1 A x 0 ⇒ x 0 与 x 为非零向量冲突 \begin{aligned} Ax 0 \newline\xRightarrow{A^{-1}AI} A^{-1}Ax0\newline \xRightarrow{} x0 与x为非零向量冲突 \end{aligned} Ax0A−1AI A−1Ax0 x0与x为非零向量冲突
延伸学习了后面的列向量等 A x Ax Ax是A的列向量的线性组合 A x 0 有解 Ax0有解 Ax0有解说明存在A的列向量的组合为0A不是满秩矩阵。那么奇异矩阵不是满秩矩阵 那能不能说明由此推导出满秩矩阵可逆 好像不是很充分除非能推导出 A x 0 ( x 非零向量 ) 无解 ⇒ A 可逆 Ax0(x非零向量)无解\Rightarrow A可逆 Ax0(x非零向量)无解⇒A可逆
2.3 Gauss-Jordan 求逆矩阵
2.3.1 求逆矩阵 ⟺ \Longleftrightarrow ⟺解方程组 [ 1 3 2 7 ] ⏟ A [ a c b d ] ⏟ A − 1 [ 1 0 0 1 ] ⏟ I ⟺ { a 3 b 1 2 c 7 d 1 \underbrace{\begin{bmatrix} 13\\ 27 \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} ac\\ bd \end{bmatrix}}_{A^{-1}} \underbrace{\begin{bmatrix} 10\\ 01 \end{bmatrix}}_{I} \Longleftrightarrow \begin{cases} a3b1 \\ 2c7d1\\ \end{cases} A [1237]A−1 [abcd]I [1001]⟺{a3b12c7d1
2.3.2 Gauss-Jordan求逆矩阵 A A − 1 I AA^{-1}I AA−1I 可写为 { [ 1 3 2 7 ] [ a b ] [ 1 0 ] [ 1 3 2 7 ] [ c d ] [ 0 1 ] \begin{cases} \begin{bmatrix} 13\\ 27 \end{bmatrix} \begin{bmatrix} a\\b \end{bmatrix} \begin{bmatrix} 1\\0 \end{bmatrix} \\\\ \begin{bmatrix} 13\\ 27 \end{bmatrix} \begin{bmatrix} c\\d \end{bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix} \end{cases} ⎩ ⎨ ⎧[1237][ab][10][1237][cd][01] [ 1 3 1 0 2 7 0 1 ] ⏟ 增广矩阵[A|I] ⇒ r o w 2 − 2 r o w 1 [ 1 3 1 0 0 1 − 2 1 ] ⇒ r o w 1 − 3 r o w 2 [ 1 0 7 − 3 0 1 − 2 1 ] ⏟ [ I ∣ E ] \begin{aligned} \underbrace{\begin{bmatrix} 1310\\ 2701 \end{bmatrix}}_{\text{增广矩阵[A|I]}} \xRightarrow{row_{2}-2row_{1}} \begin{bmatrix} 1310\\ 01-21 \end{bmatrix} \newline\xRightarrow{row_{1}-3row_{2}} \underbrace{\begin{bmatrix} 107-3\\ 01-21 \end{bmatrix}}_{[I|E]} \end{aligned} 增广矩阵[A|I] [12371001]row2−2row1 [10311−201]row1−3row2 [I∣E] [10017−2−31] 第一种老师上课讲的公式推导 E [ A I ] [ I E ] ⇒ E A I ⇒ E A − 1 E\begin{bmatrix} AI \end{bmatrix} \begin{bmatrix} IE \end{bmatrix} \Rightarrow EAI \Rightarrow E A^{-1} E[AI][IE]⇒EAI⇒EA−1 ps:
从矩阵A经过消元变成了单位矩阵, 那么A满秩不然变不成单位矩阵。所以说如果A可逆那么A一定是满秩矩阵。如果A满秩那么A一定可逆。
第二种回代到方程组中也能求出解 { [ 1 0 0 1 ] [ a b ] [ 7 − 2 ] [ 1 0 0 1 ] [ c d ] [ − 3 1 ] ⇒ { a 7 b − 2 c − 3 d 1 \begin{cases} \begin{bmatrix} 10\\ 01 \end{bmatrix} \begin{bmatrix} a\\b \end{bmatrix} \begin{bmatrix} 7\\-2 \end{bmatrix} \\\\ \begin{bmatrix} 10\\ 01 \end{bmatrix} \begin{bmatrix} c\\d \end{bmatrix} \begin{bmatrix} -3\\1 \end{bmatrix} \end{cases} \Rightarrow \begin{cases} a 7\\ b-2\\ c-3\\ d1 \end{cases} ⎩ ⎨ ⎧[1001][ab][7−2][1001][cd][−31]⇒⎩ ⎨ ⎧a7b−2c−3d1
