A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{ \begin{array}{cccc} a_{11}& a_{12} &\cdots & a_{1n} \\ a_{21}& a_{22} &\cdots & a_{2n} \\ \cdots& \cdots &\cdots & \cdots \\ a_{m1}& a_{m2} &\cdots & a_{mn} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
B = [ 0 0 ⋯ 0 0 0 ⋯ 0 ⋯ ⋯ ⋯ ⋯ 0 0 ⋯ 0 ] B = \left [ \begin{array}{cccc} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \cdots & \cdots &\cdots & \cdots \\ 0 & 0 & \cdots & 0 \\ \end{array} \right ] B=⎣⎢⎢⎡00⋯000⋯0⋯⋯⋯⋯00⋯0⎦⎥⎥⎤
A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{ \begin{array}{cccc} a_{11} & a_{12} &\cdots & a_{1n} \\ a_{21} & a_{22} &\cdots & a_{2n} \\ \cdots & \cdots &\cdots & \cdots \\ a_{m1} & a_{m2} &\cdots & a_{mn} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
B = { b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋯ ⋯ ⋯ ⋯ b m 1 b m 2 ⋯ b m n } B = \left \{ \begin{array}{cccc} b_{11} & b_{12} &\cdots & b_{1n} \\ b_{21} & b_{22} &\cdots & b_{2n} \\ \cdots & \cdots &\cdots & \cdots \\ b_{m1} & b_{m2} &\cdots & b_{mn} \\ \end{array} \right \} B=⎩⎪⎪⎨⎪⎪⎧b11b21⋯bm1b12b22⋯bm2⋯⋯⋯⋯b1nb2n⋯bmn⎭⎪⎪⎬⎪⎪⎫
C = A ± B = { a 11 ± b 11 a 12 ± b 12 ⋯ a 1 n ± b 1 n a 21 ± b 21 a 22 ± b 22 ⋯ a 2 n ± b 2 n ⋯ ⋯ ⋯ ⋯ a m 1 ± b m 1 a m 2 ± b m 2 ⋯ a m n ± b m n } C = A \pm B = \left \{ \begin{array}{cccc} a_{11} \pm b_{11} & a_{12} \pm b_{12} & \cdots & a_{1n} \pm b_{1n} \\ a_{21} \pm b_{21} & a_{22} \pm b_{22} &\cdots & a_{2n} \pm b_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{m1} \pm b_{m1} & a_{m2} \pm b_{m2} &\cdots & a_{mn} \pm b_{mn} \\ \end{array} \right \} C=A±B=⎩⎪⎪⎨⎪⎪⎧a11±b11a21±b21⋯am1±bm1a12±b12a22±b22⋯am2±bm2⋯⋯⋯⋯a1n±b1na2n±b2n⋯amn±bmn⎭⎪⎪⎬⎪⎪⎫
[ 1 2 5 6 ] + [ 3 4 7 8 ] = [ 4 6 12 14 ] \left [ \begin{array}{cccc} 1 & 2 \\ 5 & 6 \end{array} \right ] + \left [ \begin{array}{cccc} 3 & 4 \\ 7 & 8 \end{array} \right ] = \left [ \begin{array}{cccc} 4 & 6 \\ 12 & 14 \end{array} \right ] [1526]+[3748]=[412614]
[ 6 8 11 14 ] − [ 3 4 7 8 ] = [ 3 4 4 6 ] \left [ \begin{array}{cccc} 6 & 8 \\ 11 & 14 \end{array} \right ] - \left [ \begin{array}{cccc} 3 & 4 \\ 7 & 8 \end{array} \right ] = \left [ \begin{array}{cccc} 3 & 4 \\ 4 & 6 \end{array} \right ] [611814]−[3748]=[3446]
A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
λ A = { λ a 11 λ a 12 ⋯ λ a 1 n λ a 21 λ a 22 ⋯ λ a 2 n ⋯ ⋯ ⋯ ⋯ λ a m 1 λ a m 2 ⋯ λ a m n } \lambda A = \left \{ \begin{array}{cccc} \lambda a_{11} & \lambda a_{12} &\cdots & \lambda a_{1n} \\ \lambda a_{21} & \lambda a_{22} &\cdots & \lambda a_{2n} \\ \cdots & \cdots &\cdots & \cdots \\ \lambda a_{m1} & \lambda a_{m2} &\cdots & \lambda a_{mn} \\ \end{array} \right \} λA=⎩⎪⎪⎨⎪⎪⎧λa11λa21⋯λam1λa12λa22⋯λam2⋯⋯⋯⋯λa1nλa2n⋯λamn⎭⎪⎪⎬⎪⎪⎫
3 × [ 1 2 5 6 ] = [ 3 × 1 3 × 2 3 × 5 3 × 6 ] = [ 3 6 15 18 ] 3 × \left [ \begin{array}{cccc} 1 & 2 \\ 5 & 6 \end{array} \right ] = \left [ \begin{array}{cccc} 3 × 1 & 3 × 2 \\ 3 × 5 & 3 × 6 \end{array} \right ] = \left [ \begin{array}{cccc} 3 & 6 \\ 15 & 18 \end{array} \right ] 3×[1526]=[3×13×53×23×6]=[315618]
数乘运算律
例子: A = [ 1 0 2 2 − 1 3 ] A = \left [\begin{array}{cccc}1 & 0 & 2 \\ 2 & -1 & 3 \end{array} \right] A=[120−123], B = [ 1 3 4 1 0 5 ] B = \left [\begin{array}{cccc}1 & 3 & 4 \\ 1 & 0 & 5 \end{array} \right] B=[113045], 求:-3A+2B
− 3 A + 2 B = [ − 3 0 − 6 − 6 3 − 9 ] + [ 2 6 8 2 0 10 ] = [ − 1 6 2 − 4 3 1 ] -3A + 2B = \left [ \begin{array}{cccc} -3 & 0 & -6 \\ -6 & 3 & -9 \end{array} \right ] + \left [ \begin{array}{cccc} 2 & 6 & 8 \\ 2 & 0 & 10 \end{array} \right ] = \left [ \begin{array}{cccc} -1 & 6 & 2 \\ -4 & 3 & 1 \end{array} \right ] −3A+2B=[−3−603−6−9]+[2260810]=[−1−46321]
A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{ \begin{array}{cccc} a_{11} & a_{12} &\cdots & a_{1n} \\ a_{21} & a_{22} &\cdots & a_{2n} \\ \cdots & \cdots &\cdots & \cdots \\ a_{m1} & a_{m2} &\cdots & a_{mn} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
x ⃗ = ( x 1 x 2 ⋯ x n ) \vec{x} = \left ( \begin{array}{cccc} x_1 \\ x_2 \\ \cdots \\ x_n \\ \end{array} \right ) x=⎝⎜⎜⎛x1x2⋯xn⎠⎟⎟⎞
y ⃗ = A x ⃗ = ( y 1 y 2 ⋯ y m ) \vec{y} = A \vec{x} = \left ( \begin{array}{cccc} y_1 \\ y_2 \\ \cdots \\ y_m \\ \end{array} \right ) y=Ax=⎝⎜⎜⎛y1y2⋯ym⎠⎟⎟⎞
y i = ∑ j = 1 n a i j x j y_i = \sum_{j=1}^n a_{ij} x_j yi=j=1∑naijxj
A = { a 11 a 12 ⋯ a 1 s a 21 a 22 ⋯ a 2 s ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m s } A = \left \{ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1s} \\ a_{21} & a_{22} & \cdots & a_{2s} \\ \cdots & \cdots & \cdots & \cdots \\ a_{m1} & a_{m2} & \cdots & a_{ms} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1sa2s⋯ams⎭⎪⎪⎬⎪⎪⎫
B = { b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋯ ⋯ ⋯ ⋯ b s 1 b s 2 ⋯ b s n } B = \left \{ \begin{array}{cccc} b_{11} & b_{12} & \cdots & b_{1n} \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ b_{s1} & b_{s2} & \cdots & b_{sn} \\ \end{array} \right \} B=⎩⎪⎪⎨⎪⎪⎧b11b21⋯bs1b12b22⋯bs2⋯⋯⋯⋯b1nb2n⋯bsn⎭⎪⎪⎬⎪⎪⎫
C = A ∗ B = { c 11 c 12 ⋯ c 1 n c 21 c 22 ⋯ c 2 n ⋯ ⋯ ⋯ ⋯ c m 1 c m 2 ⋯ c m n } C=A*B= \left \{ \begin{array}{cccc} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ c_{m1} & c_{m2} & \cdots & c_{mn} \\ \end{array} \right \} C=A∗B=⎩⎪⎪⎨⎪⎪⎧c11c21⋯cm1c12c22⋯cm2⋯⋯⋯⋯c1nc2n⋯cmn⎭⎪⎪⎬⎪⎪⎫
( A B ) C = A ( B C ) (AB)C = A(BC) (AB)C=A(BC)
( A + B ) C = A C + B C (A+B)C = AC+BC (A+B)C=AC+BC
C ( A + B ) = C A + C B C(A+B) = CA + CB C(A+B)=CA+CB
A B ≠ B A AB \neq BA AB=BA 注意这里不满足交换律,重点
例: A = [ 1 0 − 1 2 − 1 1 3 0 0 5 − 1 4 ] A = \left [\begin{array}{cccc}1 & 0 & -1 & 2 \\-1 & 1 & 3 & 0 \\0 & 5 & -1 & 4\end{array} \right ] A=⎣⎡1−10015−13−1204⎦⎤, B = [ 0 3 4 1 2 1 3 1 − 1 − 1 5 1 ] B = \left [\begin{array}{cccc}0 & 3 & 4 \\1 & 2 & 1 \\3 & 1 & -1 \\-1 & 5 & 1 \\\end{array} \right ] B=⎣⎢⎢⎡013−1321541−11⎦⎥⎥⎤,求C=AB
例: A = [ 1 1 − 1 − 1 ] A = \left [\begin{array}{cccc}1 & 1 \\-1 & -1\end{array} \right ] A=[1−11−1], B = [ 1 1 1 1 ] B = \left [\begin{array}{cccc}1 & 1 \\1 & 1\end{array} \right ] B=[1111], C = [ − 1 2 3 0 ] C = \left [\begin{array}{cccc}-1 & 2 \\3 & 0\end{array} \right ] C=[−1320]
注意:
A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
A T = { a 11 a 21 ⋯ a m 1 a 12 a 22 ⋯ a m 2 ⋯ ⋯ ⋯ ⋯ a 1 n a 2 n ⋯ a m n } A^T = \left \{ \begin{array}{cccc} a_{11} & a_{21} & \cdots & a_{m1} \\ a_{12} & a_{22} & \cdots & a_{m2} \\ \cdots & \cdots & \cdots & \cdots \\ a_{1n} & a_{2n} & \cdots & a_{mn} \\ \end{array} \right \} AT=⎩⎪⎪⎨⎪⎪⎧a11a12⋯a1na21a22⋯a2n⋯⋯⋯⋯am1am2⋯amn⎭⎪⎪⎬⎪⎪⎫