查看全集:🏸线性代数/Linear Algebra
计算下列矩阵乘积 ABABAB 和 BABABA:
A=(1234),B=(5678)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix},\quad B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}A=(1324),B=(5768)
求矩阵 C=(2111)C = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}C=(2111) 的逆矩阵
证明:若 AAA 和 BBB 是对称矩阵,则 A−BA - BA−B 也是对称矩阵
设 N=(0100)N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}N=(0010),验证 N2=0N^2 = 0N2=0,并求 (I−N)−1(I - N)^{-1}(I−N)−1
构造一个实数矩阵 AAA 满足 A2=−IA^2 = -IA2=−I,并验证其性质
解矩阵方程 AX=BAX = BAX=B,其中:
AB=(1×5+2×71×6+2×83×5+4×73×6+4×8)=(19224350)AB = \begin{pmatrix} 1×5+2×7 & 1×6+2×8 \\ 3×5+4×7 & 3×6+4×8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}AB=(1×5+2×73×5+4×71×6+2×83×6+4×8)=(19432250)
BA=(5×1+6×35×2+6×47×1+8×37×2+8×4)=(23343146)BA = \begin{pmatrix} 5×1+6×3 & 5×2+6×4 \\ 7×1+8×3 & 7×2+8×4 \end{pmatrix} = \begin{pmatrix} 23 & 34 \\ 31 & 46 \end{pmatrix}BA=(5×1+6×37×1+8×35×2+6×47×2+8×4)=(23313446)
通过行变换 [C∣I]→[I∣C−1][C|I] \rightarrow [I|C^{-1}][C∣I]→[I∣C−1]:
[21101101]→[101−101−12]\left[\begin{array}{cc|cc} 2 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right] \rightarrow \left[\begin{array}{cc|cc} 1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 2 \end{array}\right][21111001]→[10011−1−12]
C−1=(1−1−12) C^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} C−1=(1−1−12)
(A−B)T=AT−BT=A−B(因A,B对称)(A - B)^T = A^T - B^T = A - B \quad \text{(因$A,B$对称)}(A−B)T=AT−BT=A−B(因A,B对称)
计算 N2=(0000)N^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}N2=(0000)
逆矩阵计算:
(I−N)−1=I+N=(1101)(I - N)^{-1} = I + N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} (I−N)−1=I+N=(1011)
例子:
A=(01−10)A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}A=(0−110)
验证:
A2=(−100−1)=−IA^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I A2=(−100−1)=−I
先求 A−1=1−2(4−2−31)A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}A−1=−21(4−3−21)
解为:
X=A−1B=(−4−54.56)X = A^{-1}B = \begin{pmatrix} -4 & -5 \\ 4.5 & 6 \end{pmatrix}X=A−1B=(−44.5−56)