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基础部分
方程组{ a x + y + z = 1 x + a y + z = 1 x + y + a z = − 2 \begin{cases}ax+y+z=1\\x+ay+z=1\\x+y+az=-2\end{cases} ⎩ ⎨ ⎧ a x + y + z = 1 x + a y + z = 1 x + y + a z = − 2 a = a= a =
设3维列向量组α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α 1 , α 2 , α 3 k , l k,l k , l β 1 = k α 1 + l α 2 \beta_1=k\alpha_1+l\alpha_2 β 1 = k α 1 + l α 2 β 2 = k α 2 + l α 3 \beta_2=k\alpha_2+l\alpha_3 β 2 = k α 2 + l α 3 β 3 = k α 3 + l α 1 \beta_3=k\alpha_3+l\alpha_1 β 3 = k α 3 + l α 1 B = [ β 1 , β 2 , β 3 ] B=[\beta_1,\beta_2,\beta_3] B = [ β 1 , β 2 , β 3 ] B x = 0 Bx=0 B x = 0 k − l = 0 k-l=0 k − l = 0 k + l = 0 k+l=0 k + l = 0 k − l ≠ 0 k-l\neq0 k − l = 0 k + l ≠ 0 k+l\neq0 k + l = 0
设A A A m × n m\times n m × n B B B n × m n\times m n × m m > n m>n m > n ∣ A B ∣ = 0 |AB|=0 ∣ A B ∣ = 0 m > n m>n m > n A B AB A B n > m n>m n > m A B x = 0 ABx=0 A B x = 0 n > m n>m n > m r ( A B ) < m r(AB)<m r ( A B ) < m
设A A A n ( n > 2 ) n(n>2) n ( n > 2 ) r ( A ∗ ) = 1 r(A^*)=1 r ( A ∗ ) = 1 α 1 , α 2 \alpha_1,\alpha_2 α 1 , α 2 A x = b Ax=b A x = b k k k A x = b Ax=b A x = b ( k − 1 ) α 1 + k α 2 (k-1)\alpha_1+k\alpha_2 ( k − 1 ) α 1 + k α 2 ( k − 1 ) α 1 − k α 2 (k-1)\alpha_1-k\alpha_2 ( k − 1 ) α 1 − k α 2 ( k + 1 ) α 1 + k α 2 (k+1)\alpha_1+k\alpha_2 ( k + 1 ) α 1 + k α 2 ( k + 1 ) α 1 − k α 2 (k+1)\alpha_1-k\alpha_2 ( k + 1 ) α 1 − k α 2
设α 1 = [ 4 1 2 ] \alpha_1=\begin{bmatrix}4\\1\\2\end{bmatrix} α 1 = 4 1 2 α 2 = [ 1 1 − 1 ] \alpha_2=\begin{bmatrix}1\\1\\-1\end{bmatrix} α 2 = 1 1 − 1 α 3 = [ − 2 0 4 ] \alpha_3=\begin{bmatrix}-2\\0\\4\end{bmatrix} α 3 = − 2 0 4 α 4 = [ 7 2 − 3 ] \alpha_4=\begin{bmatrix}7\\2\\-3\end{bmatrix} α 4 = 7 2 − 3 A = [ α 1 , α 2 , α 3 , α 4 ] A=[\alpha_1,\alpha_2,\alpha_3,\alpha_4] A = [ α 1 , α 2 , α 3 , α 4 ] A x = α 2 + 3 α 4 Ax=\alpha_2+3\alpha_4 A x = α 2 + 3 α 4
设A A A n ( n ≥ 2 ) n(n\geq2) n ( n ≥ 2 ) A ∗ A^* A ∗ A A A n n n α \alpha α A ∗ α = 0 A^*\alpha=0 A ∗ α = 0 A x = 0 Ax=0 A x = 0 k k k
已知非齐次线性方程组{ x 1 + x 2 + x 3 + x 4 = − 1 4 x 1 + 3 x 2 + 5 x 3 − x 4 = − 1 a x 1 + x 2 + 3 x 3 + b x 4 = 1 \begin{cases}x_1+x_2+x_3+x_4=-1\\4x_1+3x_2+5x_3-x_4=-1\\ax_1+x_2+3x_3+bx_4=1\end{cases} ⎩ ⎨ ⎧ x 1 + x 2 + x 3 + x 4 = − 1 4 x 1 + 3 x 2 + 5 x 3 − x 4 = − 1 a x 1 + x 2 + 3 x 3 + b x 4 = 1 A A A a , b a,b a , b A T A x = 0 A^TAx=0 A T A x = 0
设A = [ α 1 , α 2 , ⋯ , α n ] A=[\alpha_1,\alpha_2,\cdots,\alpha_n] A = [ α 1 , α 2 , ⋯ , α n ] B = [ β 1 , β 2 , ⋯ , β n ] B=[\beta_1,\beta_2,\cdots,\beta_n] B = [ β 1 , β 2 , ⋯ , β n ] A A A B B B k k k A x = b Ax=b A x = b B x = b Bx=b B x = b
设A A A A 2 = O A^2=O A 2 = O A x = b Ax=b A x = b
已知线性方程组A x = k β 1 + β 2 Ax=k\beta_1+\beta_2 A x = k β 1 + β 2 A = [ 1 1 − 1 1 − 2 1 1 − 1 − 1 ] A=\begin{bmatrix}1&1&-1\\1&-2&1\\1&-1&-1\end{bmatrix} A = 1 1 1 1 − 2 − 1 − 1 1 − 1 β 1 = [ 2 1 3 ] \beta_1=\begin{bmatrix}2\\1\\3\end{bmatrix} β 1 = 2 1 3 β 2 = [ 1 3 − 1 ] \beta_2=\begin{bmatrix}1\\3\\-1\end{bmatrix} β 2 = 1 3 − 1 k k k
已知A A A A A A A x = 0 Ax=0 A x = 0 k 1 [ 1 , 2 , − 2 ] T + k 2 [ 2 , 1 , 2 ] T k_1[1,2,-2]^T+k_2[2,1,2]^T k 1 [ 1 , 2 , − 2 ] T + k 2 [ 2 , 1 , 2 ] T α = [ 1 , 1 , 1 ] T \alpha=[1,1,1]^T α = [ 1 , 1 , 1 ] T k 1 , k 2 k_1,k_2 k 1 , k 2 β \beta β A β A\beta A β α \alpha α β = [ 3 , 6 , − 3 ] T \beta=[3,6,-3]^T β = [ 3 , 6 , − 3 ] T A β A\beta A β
设3维列向量组α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α 1 , α 2 , α 3 β 1 , β 2 , β 3 \beta_1,\beta_2,\beta_3 β 1 , β 2 , β 3 A = [ α 1 , α 2 , α 3 ] A=[\alpha_1,\alpha_2,\alpha_3] A = [ α 1 , α 2 , α 3 ] B = [ β 1 , β 2 , β 3 ] B=[\beta_1,\beta_2,\beta_3] B = [ β 1 , β 2 , β 3 ] A x = 0 Ax=0 A x = 0 B x = 0 Bx=0 B x = 0 A T x = 0 A^Tx=0 A T x = 0 B T x = 0 B^Tx=0 B T x = 0 [ A B ] x = 0 \begin{bmatrix}A\\B\end{bmatrix}x=0 [ A B ] x = 0 A x = 0 Ax=0 A x = 0 [ A T B T ] x = 0 \begin{bmatrix}A^T\\B^T\end{bmatrix}x=0 [ A T B T ] x = 0 A T x = 0 A^Tx=0 A T x = 0
设A A A m × n m\times n m × n e = [ 1 , 1 , ⋯ , 1 ] T e=[1,1,\cdots,1]^T e = [ 1 , 1 , ⋯ , 1 ] T A y = e Ay=e A y = e ( I ) A T x = 0 (I)A^Tx=0 ( I ) A T x = 0 ( I I ) { A T x = 0 e T x = 0 (II)\begin{cases}A^Tx=0\\e^Tx=0\end{cases} ( I I ) { A T x = 0 e T x = 0
设齐次线性方程组( I ) { x 1 + 3 x 3 + 5 x 4 = 0 x 1 − x 2 − 2 x 3 + 2 x 4 = 0 (I)\begin{cases}x_1+3x_3+5x_4=0\\x_1-x_2-2x_3+2x_4=0\end{cases} ( I ) { x 1 + 3 x 3 + 5 x 4 = 0 x 1 − x 2 − 2 x 3 + 2 x 4 = 0 a x 1 + b x 2 + c x 3 + d x 4 = 0 ax_1+bx_2+cx_3+dx_4=0 a x 1 + b x 2 + c x 3 + d x 4 = 0 ( I I ) { x 1 + 3 x 3 + 5 x 4 = 0 x 1 − x 2 − 2 x 3 + 2 x 4 = 0 a x 1 + b x 2 + c x 3 + d x 4 = 0 (II)\begin{cases}x_1+3x_3+5x_4=0\\x_1-x_2-2x_3+2x_4=0\\ax_1+bx_2+cx_3+dx_4=0\end{cases} ( I I ) ⎩ ⎨ ⎧ x 1 + 3 x 3 + 5 x 4 = 0 x 1 − x 2 − 2 x 3 + 2 x 4 = 0 a x 1 + b x 2 + c x 3 + d x 4 = 0 a , b , c , d a,b,c,d a , b , c , d
设平面π 1 : x + a y = a \pi_1:x+ay=a π 1 : x + a y = a π 2 : a x + z = 1 \pi_2:ax+z=1 π 2 : a x + z = 1 π 3 : a y + z = 1 \pi_3:ay+z=1 π 3 : a y + z = 1 a = a= a =
在空间直角坐标系O − x y z O-xyz O − x y z π 1 : a x + y − z = 1 \pi_1:ax+y-z=1 π 1 : a x + y − z = 1 π 2 : x + y + b z = a \pi_2:x+y+bz=a π 2 : x + y + b z = a π 3 : x + a y − z = 1 \pi_3:x+ay-z=1 π 3 : x + a y − z = 1 a = − 2 , b = 2 a=-2,b=2 a = − 2 , b = 2 a ≠ − 2 , b = 2 a\neq-2,b=2 a = − 2 , b = 2 a = 1 , b = − 1 a=1,b=-1 a = 1 , b = − 1 a = 1 , b ≠ − 1 a=1,b\neq-1 a = 1 , b = − 1
设B B B B x = 0 Bx=0 B x = 0 2 2 2 A = [ 1 2 − 2 4 a 3 3 − 1 1 ] A=\begin{bmatrix}1&2&-2\\4&a&3\\3&-1&1\end{bmatrix} A = 1 4 3 2 a − 1 − 2 3 1 A B = O AB=O A B = O A x = 0 Ax=0 A x = 0
强化部分
设4阶矩阵A = [ a i j ] A=[a_{ij}] A = [ a ij ] a 12 a_{12} a 12 A 12 ≠ 0 A_{12}\neq0 A 12 = 0 A A A α 1 , α 2 , α 3 , α 4 \alpha_1,\alpha_2,\alpha_3,\alpha_4 α 1 , α 2 , α 3 , α 4 k 1 , k 2 , k 3 k_1,k_2,k_3 k 1 , k 2 , k 3 A ∗ x = 0 A^*x=0 A ∗ x = 0 k 1 α 1 + k 2 α 2 + k 3 α 3 k_1\alpha_1+k_2\alpha_2+k_3\alpha_3 k 1 α 1 + k 2 α 2 + k 3 α 3 k 1 α 1 + k 2 α 2 + k 3 α 4 k_1\alpha_1+k_2\alpha_2+k_3\alpha_4 k 1 α 1 + k 2 α 2 + k 3 α 4 k 1 α 1 + k 2 α 3 + k 3 α 4 k_1\alpha_1+k_2\alpha_3+k_3\alpha_4 k 1 α 1 + k 2 α 3 + k 3 α 4 k 1 α 2 + k 2 α 3 + k 3 α 4 k_1\alpha_2+k_2\alpha_3+k_3\alpha_4 k 1 α 2 + k 2 α 3 + k 3 α 4
A A A A A A A ∗ = [ 1 0 1 0 0 2 0 2 0 0 1 1 0 0 0 4 ] A^*=\begin{bmatrix}1&0&1&0\\0&2&0&2\\0&0&1&1\\0&0&0&4\end{bmatrix} A ∗ = 1 0 0 0 0 2 0 0 1 0 1 0 0 2 1 4 b = [ 1 , 1 , 1 , 1 ] T b=[1,1,1,1]^T b = [ 1 , 1 , 1 , 1 ] T A x = b Ax=b A x = b 设方程组{ x 1 + a x 2 − 2 x 3 = 4 x 1 + 2 x 2 + x 3 = 1 2 x 1 + 3 x 2 + ( a + 2 ) x 3 = 3 \begin{cases}x_1+ax_2-2x_3=4\\x_1+2x_2+x_3=1\\2x_1+3x_2+(a+2)x_3=3\end{cases} ⎩ ⎨ ⎧ x 1 + a x 2 − 2 x 3 = 4 x 1 + 2 x 2 + x 3 = 1 2 x 1 + 3 x 2 + ( a + 2 ) x 3 = 3 A A A b b b A x = b Ax=b A x = b A T A x = A T b A^TAx=A^Tb A T A x = A T b a = a= a =
设A A A α , β \alpha,\beta α , β B = [ A β α T 1 ] B=\begin{bmatrix}A&\beta\\\alpha^T&1\end{bmatrix} B = [ A α T β 1 ] r ( B ) = 2 r(B)=2 r ( B ) = 2 { A x = β α T x = 1 \begin{cases}Ax=\beta\\\alpha^Tx=1\end{cases} { A x = β α T x = 1
设四元齐次线性方程组( I ) { 2 x 1 + 3 x 2 − x 3 = 0 x 1 + 2 x 2 + x 3 − x 4 = 0 (I)\begin{cases}2x_1+3x_2-x_3=0\\x_1+2x_2+x_3-x_4=0\end{cases} ( I ) { 2 x 1 + 3 x 2 − x 3 = 0 x 1 + 2 x 2 + x 3 − x 4 = 0 ( I I ) (II) ( I I ) ξ 1 = [ 2 , − 1 , k + 2 , 1 ] T \xi_1=[2,-1,k+2,1]^T ξ 1 = [ 2 , − 1 , k + 2 , 1 ] T ξ 2 = [ − 1 , 2 , 4 , k + 8 ] T \xi_2=[-1,2,4,k+8]^T ξ 2 = [ − 1 , 2 , 4 , k + 8 ] T ( I ) (I) ( I ) ( I I ) (II) ( I I ) k k k
已知A , B A,B A , B 2 × 4 2\times4 2 × 4 A x = 0 Ax=0 A x = 0 α 1 = [ 1 , 1 , 2 , 1 ] T \alpha_1=[1,1,2,1]^T α 1 = [ 1 , 1 , 2 , 1 ] T α 2 = [ 0 , − 3 , 1 , 0 ] T \alpha_2=[0,-3,1,0]^T α 2 = [ 0 , − 3 , 1 , 0 ] T B x = 0 Bx=0 B x = 0 β 1 = [ 1 , 3 , 0 , 2 ] T \beta_1=[1,3,0,2]^T β 1 = [ 1 , 3 , 0 , 2 ] T β 2 = [ 1 , 2 , − 1 , a ] T \beta_2=[1,2,-1,a]^T β 2 = [ 1 , 2 , − 1 , a ] T A A A A x = 0 Ax=0 A x = 0 B x = 0 Bx=0 B x = 0 a a a
设3阶矩阵A , B A,B A , B r ( B A ) < r ( A B ) r(BA)<r(AB) r ( B A ) < r ( A B ) A B x = 0 ABx=0 A B x = 0 B A x = 0 BAx=0 B A x = 0 A B A x = 0 ABAx=0 A B A x = 0 B A B x = 0 BABx=0 B A B x = 0
设A A A n n n [ A O E A T A ] x = 0 \begin{bmatrix}A&O\\E&A^TA\end{bmatrix}x=0 [ A E O A T A ] x = 0 [ O A A T A A A T A ] x = 0 \begin{bmatrix}O&A\\A^TA&AA^TA\end{bmatrix}x=0 [ O A T A A A A T A ] x = 0 [ A A T A O A T A ] x = 0 \begin{bmatrix}A&A^TA\\O&A^TA\end{bmatrix}x=0 [ A O A T A A T A ] x = 0 [ A T A A O A ] x = 0 \begin{bmatrix}A^TA&A\\O&A\end{bmatrix}x=0 [ A T A O A A ] x = 0 [ A A T A A T A O A ] x = 0 \begin{bmatrix}AA^TA&A^TA\\O&A\end{bmatrix}x=0 [ A A T A O A T A A ] x = 0 [ A T A 2 A O A T A ] x = 0 \begin{bmatrix}A^TA^2&A\\O&A^TA\end{bmatrix}x=0 [ A T A 2 O A A T A ] x = 0
设A , B A,B A , B n n n A A A A 2 − A = 3 E A^2-A=3E A 2 − A = 3 E [ A B ] x = 0 \begin{bmatrix}A\\B\end{bmatrix}x=0 [ A B ] x = 0 [ A − B A + A B ] x = 0 \begin{bmatrix}A-B\\A+AB\end{bmatrix}x=0 [ A − B A + A B ] x = 0 [ A + B A + A B − B ] x = 0 \begin{bmatrix}A+B\\A+AB-B\end{bmatrix}x=0 [ A + B A + A B − B ] x = 0 [ A − B 2 A + B ] x = 0 \begin{bmatrix}A-B\\2A+B\end{bmatrix}x=0 [ A − B 2 A + B ] x = 0 [ A + B B A + B 2 ] x = 0 \begin{bmatrix}A+B\\BA+B^2\end{bmatrix}x=0 [ A + B B A + B 2 ] x = 0
设α 1 = [ 1 , − 2 , 1 , 0 , 0 ] T \alpha_1=[1,-2,1,0,0]^T α 1 = [ 1 , − 2 , 1 , 0 , 0 ] T α 2 = [ 1 , − 2 , 0 , 1 , 0 ] T \alpha_2=[1,-2,0,1,0]^T α 2 = [ 1 , − 2 , 0 , 1 , 0 ] T α 3 = [ 0 , 0 , 1 , − 1 , 0 ] T \alpha_3=[0,0,1,-1,0]^T α 3 = [ 0 , 0 , 1 , − 1 , 0 ] T α 4 = [ 1 , − 2 , 3 , − 2 , 0 ] T \alpha_4=[1,-2,3,-2,0]^T α 4 = [ 1 , − 2 , 3 , − 2 , 0 ] T { x 1 + x 2 + x 3 + x 4 + x 5 = 0 3 x 1 + 2 x 2 + x 3 + x 4 − 3 x 5 = 0 x 2 + 2 x 3 + 2 x 4 + 6 x 5 = 0 5 x 1 + 4 x 2 + 3 x 3 + 3 x 4 − x 5 = 0 \begin{cases}x_1+x_2+x_3+x_4+x_5=0\\3x_1+2x_2+x_3+x_4-3x_5=0\\x_2+2x_3+2x_4+6x_5=0\\5x_1+4x_2+3x_3+3x_4-x_5=0\end{cases} ⎩ ⎨ ⎧ x 1 + x 2 + x 3 + x 4 + x 5 = 0 3 x 1 + 2 x 2 + x 3 + x 4 − 3 x 5 = 0 x 2 + 2 x 3 + 2 x 4 + 6 x 5 = 0 5 x 1 + 4 x 2 + 3 x 3 + 3 x 4 − x 5 = 0 α 1 , α 2 , α 3 , α 4 \alpha_1,\alpha_2,\alpha_3,\alpha_4 α 1 , α 2 , α 3 , α 4
已知a a a A = [ 1 2 a 1 3 0 2 7 − a ] A=\begin{bmatrix}1&2&a\\1&3&0\\2&7&-a\end{bmatrix} A = 1 1 2 2 3 7 a 0 − a B = [ 1 a 2 0 1 1 − 1 1 1 ] B=\begin{bmatrix}1&a&2\\0&1&1\\-1&1&1\end{bmatrix} B = 1 0 − 1 a 1 1 2 1 1 a a a A P = B AP=B A P = B P P P
设A = [ 1 − 2 3 0 1 − 1 1 2 0 ] A=\begin{bmatrix}1&-2&3\\0&1&-1\\1&2&0\end{bmatrix} A = 1 0 1 − 2 1 2 3 − 1 0 β = [ − 4 1 − 3 ] \beta=\begin{bmatrix}-4\\1\\-3\end{bmatrix} β = − 4 1 − 3 B = [ − 2 1 3 1 0 − 1 2 1 0 ] B=\begin{bmatrix}-2&1&3\\1&0&-1\\2&1&0\end{bmatrix} B = − 2 1 2 1 0 1 3 − 1 0 [ A , β ] = B C [A,\beta]=BC [ A , β ] = B C A x = β Ax=\beta A x = β C C C [ A , β ] Y = E [A,\beta]Y=E [ A , β ] Y = E Y Y Y
设A 3 × 3 = [ α 1 , α 2 , α 3 ] A_{3\times3}=[\alpha_1,\alpha_2,\alpha_3] A 3 × 3 = [ α 1 , α 2 , α 3 ] A x = β Ax=\beta A x = β k ξ + η = k [ 1 , 2 , − 3 ] T + [ 2 , − 1 , 1 ] T k\xi+\eta=k[1,2,-3]^T+[2,-1,1]^T k ξ + η = k [ 1 , 2 , − 3 ] T + [ 2 , − 1 , 1 ] T k k k B = [ α 1 + α 2 + α 3 + β , α 1 , α 2 , α 3 ] B=[\alpha_1+\alpha_2+\alpha_3+\beta,\alpha_1,\alpha_2,\alpha_3] B = [ α 1 + α 2 + α 3 + β , α 1 , α 2 , α 3 ] B y = β By=\beta B y = β
如图所示有三张平面，其中有两张平面平行，第三张平面与它们相交，其方程a i 1 x + a i 2 y + a i 3 z = d i ( i = 1 , 2 , 3 ) a_{i1}x+a_{i2}y+a_{i3}z=d_i(i=1,2,3) a i 1 x + a i 2 y + a i 3 z = d i ( i = 1 , 2 , 3 ) A A A A ‾ \overline{A} A r ( A ) = 2 , r ( A ‾ ) = 3 r(A)=2,r(\overline{A})=3 r ( A ) = 2 , r ( A ) = 3 r ( A ) = 2 , r ( A ‾ ) = 2 r(A)=2,r(\overline{A})=2 r ( A ) = 2 , r ( A ) = 2 r ( A ) = 1 , r ( A ‾ ) = 2 r(A)=1,r(\overline{A})=2 r ( A ) = 1 , r ( A ) = 2 r ( A ) = 1 , r ( A ‾ ) = 1 r(A)=1,r(\overline{A})=1 r ( A ) = 1 , r ( A ) = 1
设α i = [ a i , b i , c i ] T ( i = 1 , 2 , 3 ) \alpha_i=[a_i,b_i,c_i]^T(i=1,2,3) α i = [ a i , b i , c i ] T ( i = 1 , 2 , 3 ) x − a 1 a 2 = y − b 1 b 2 = z − c 1 c 2 \frac{x-a_1}{a_2}=\frac{y-b_1}{b_2}=\frac{z-c_1}{c_2} a 2 x − a 1 = b 2 y − b 1 = c 2 z − c 1 ( a 3 , b 3 , c 3 ) (a_3,b_3,c_3) ( a 3 , b 3 , c 3 ) π i : α i T [ x y z ] = 1 ( i = 1 , 2 , 3 ) \pi_i:\alpha_i^T\begin{bmatrix}x\\y\\z\end{bmatrix}=1(i=1,2,3) π i : α i T x y z = 1 ( i = 1 , 2 , 3 )