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2025-2026学年上学期期中试卷

一、

  1. 计算四阶行列式 D=1232120510124312D = \begin{vmatrix} 1 & 2 & 3 & 2 \\ 1 & 2 & 0 & -5 \\ 1 & 0 & 1 & 2 \\ 4 & 3 & 1 & 2 \end{vmatrix}
  2. 证明abcdaa+ba+b+ca+b+c+da2a+b3a+2b+c4a+3b+2c+da3a+b6a+3b+c10a+6b+3c+d=a4\begin{vmatrix} a & b & c & d \\ a & a + b & a + b + c & a + b + c + d \\ a & 2a + b & 3a + 2b + c & 4a + 3b + 2c + d \\ a & 3a + b & 6a + 3b + c & 10a + 6b + 3c + d \\ \end{vmatrix} = a^4

  1. 计算四阶行列式 D4=a13a23a33a43a12b1a22b2a32b3a42b4a1b12a2b22a3b32a4b42b13b23b33b43,ai0,(i=1,2,3,4)D_4 = \begin{vmatrix} a_1^3 & a_2^3 & a_3^3 & a_4^3 \\ a_1^2b_1 & a_2^2b_2 & a_3^2b_3 & a_4^2b_4 \\ a_1b_1^2 & a_2b_2^2 & a_3b_3^2 & a_4b_4^2 \\ b_1^3 & b_2^3 & b_3^3 & b_4^3 \end{vmatrix}, a_i \not= 0,(i = 1,2,3,4)
  2. 计算 f(x+1)f(x)f(x+1) - f(x), 其中
f(x)=10000x12000x213300x31nCn2Cn3Cnn1xn1n+1Cn+12Cn+13Cn+1n1nn+1f(x) = \begin{vmatrix} 1 & 0 & 0 & 0 & \cdots & 0 & x \\ 1 & 2 & 0 & 0 & \cdots & 0 & x^2 \\ 1 & 3 & 3 & 0 & \cdots & 0 & x^3 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & n & C_n^2 & C_n^3 & \cdots & C_n^{n-1} & x^n \\ 1 & n + 1 & C_{n+1}^2 & C_{n+1}^3 & \cdots & C_{n+1}^{n-1} & n^{n+1} \end{vmatrix}
  1. 计算以下 n+1n+1 阶行列式:
Dn+1=aaxax2axn1axn1aaxaxn2axn101aaxn1axn2000aax0001aD_{n+1} = \begin{vmatrix} a & ax & ax^2 & \cdots & ax^{n-1} & ax^n \\ -1 & a & ax & \cdots & ax^{n-2} & ax^{n-1} \\ 0 & -1 & a & \cdots & ax^{n-1} & ax^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a & ax \\ 0 & 0 & 0 & \cdots & -1 & a \end{vmatrix}

  1. 设行列式
D=3040222207005322D = \begin{vmatrix} 3 & 0 & 4 & 0 \\ 2 & 2 & 2 & 2 \\ 0 & -7 & 0 & 0 \\ 5 & 3 & -2 & 2 \end{vmatrix}

DD 的第 44 行元素的余子式之和 M41+M42+M43+M44的值。M_{41}+M_{42}+M_{43}+M_{44}的值。

2.设a1a2an10,n>1a_1a_2\cdots a_{n-1} \not= 0, n > 1。试求行列式

D=xb1b2bn11a10010a20100an1D = \begin{vmatrix} x & b_1 & b_2 & \cdots & b_{n-1} \\ 1 & a_1 & 0 & \cdots & 0 \\ 1 & 0 & a_2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & a_{n-1} \end{vmatrix}

的第一行中诸元素的代数余子式之和 A11+A12++A1nA_{11} + A_{12} + \cdots +A_{1n}

  1. 已知齐次线性方程组
{(3λ)x1+x2+x3=0,(2λ)x2x3=0,4x12x2+(1λ)x3=0\left \{ \begin{aligned} (3-\lambda)x_1 + x_2 + x_3 &= 0, \\ (2-\lambda)x_2 - x_3 &= 0, \\ 4x_1 - 2x_2 + (1-\lambda)x_3 &= 0 \end{aligned} \right .

有非零解,求λ\lambda的值。

  1. 已知 a2b2a^2 \not = b^2,试证方程组
{ax1+bx2n=1,ax2+bx2n1=1,axn+bxn+1=1,bxn+axn+1=1,bxn1+axn+2=1,bx1+ax2n=1\left \{ \begin{aligned} ax_1 + bx_{2n} &= 1, \\ ax_2 + bx_{2n-1} &= 1, \\ \vdots \\ ax_n + bx_{n+1} &= 1, \\ bx_n+ax_{n+1} &= 1, \\ bx_{n-1} + ax_{n+2} &= 1, \\ \vdots \\ bx_1+ax_{2n} &= 1 \\ \end{aligned} \right .

有唯一解,并求解。

A=[1111111111111111]A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}

  1. A2A^2
  2. AA 可逆,且求 A1A^{-1}
  3. (A)1(A^*)^{-1}

  1. ABA=CABA=C, 其中 A=[100113011],C=[101010001]A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 3 \\ 0 & 1 & -1 \end{bmatrix},C = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, 求 BB 的伴随矩阵 BB^*

  2. 已知矩阵 A=(101014205),B=(2112),C=(103102)A = \begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 4 \\ 2 & 0 & 5\end{pmatrix}, B = \begin{pmatrix}2 & 1 \\1 & 2\end{pmatrix}, C = \begin{pmatrix}1 & 0 \\ 3 & 1 \\ 0 & 2\end{pmatrix}, 且 AXB=CAXB = C, 求矩阵 XX

已知 AA 为三阶可逆矩阵,BB 为三阶矩阵,且满足 2A1B=B4E2A^{-1}B=B-4E

  1. 证明:(A2E)(A-2E) 为可逆矩阵,且写出 (A2E)1(A-2E)^{-1}
  2. B=(120120002)B = \begin{pmatrix}1 & -2 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}, 求矩阵 AA

{(2λ)x1+2x22x3=12x1+(5λ)x24x3=22x14x2+(5λ)x3=λ1\left \{ \begin{aligned} (2-\lambda)x_1+2x_2-2x_3 &= 1 \\ 2x_1+(5-\lambda)x_2-4x_3 &= 2 \\ -2x_1-4x_2+(5-\lambda)x_3 &= -\lambda - 1 \end{aligned} \right .λ\lambda 为何值时,此方程组有唯一解,无解或有无穷多解?并在有无穷多解时求其解。

已知矩阵 A=(1111111111111111)A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix}

  1. An(n=2,3,)A^n(n=2,3,\cdots)
  2. 若方阵BB满足A3+A2+AB3A2E=0A^3+A^2+AB-3A-2E=0,求BB