Merge remote-tracking branch 'origin/develop' into develop

pull/5/head
刘柯妤 3 weeks ago
commit 031bb9c3ae

@ -54,14 +54,17 @@ tags:
10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$
其中常数 $a_1, a_2, a_3, a_4$ 互不相等,则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$
11. 矩阵$$A=\begin{bmatrix}
0 & 0 & \cdots & 1 & 1 & \cdots & 1 & 1 \\
0 & 0 & \cdots & 0 & 1 & \cdots & 1 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 0 & \cdots & 1 & 1 \\
0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0
0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\
0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0
\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____.
12. $\underline{\qquad\qquad\qquad\qquad}$

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