解答题 （共 11 题 ），解答过程应写出必要的文字说明、证明过程或演算步骤
计算$\left|\begin{array}{llll}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0
\end{array}\right|$
计算 $\left|\begin{array}{ccccc}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 2 & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 0 & \cdots & n-1 \\
n & 0 & 0 & \cdots & 0
\end{array}\right|$
计算 $\left|\begin{array}{llll}
1 & 1 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0
\end{array}\right|$
计算 $\left|\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\
a_{31} & a_{32} & 0 & 0 & 0 \\
a_{41} & a_{42} & 0 & 0 & 0 \\
a_{51} & a_{52} & 0 & 0 & 0
\end{array}\right|$
计算 $\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|$
用化成三角形行列式的方法, 计算三阶行列式 $\left|\begin{array}{ccc}1+x & 2 & 3 \\ 1 & 2+y & 3 \\ 1 & 2 & 3+z\end{array}\right|$, 其中 $x y z \neq 0$.
计算$\left|\begin{array}{cccccc}
1 & 2 & 3 & \cdots & n-1 & n \\
-1 & 0 & 3 & \cdots & n-1 & n \\
-1 & -2 & 0 & \cdots & n-1 & n \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
-1 & -2 & -3 & \cdots & 0 & n \\
-1 & -2 & -3 & \cdots & -(n-1) & 0
\end{array}\right|$
计算$n$阶行列式 $\left|\begin{array}{ccccc}
0 & x & x & \cdots & x \\
x & 0 & x & \cdots & x \\
x & x & 0 & \cdots & x \\
\vdots & \vdots & \vdots & & \vdots \\
x & x & x & \cdots & 0
\end{array}\right|$
计算$\left|\begin{array}{cccccc}
x & a_1 & a_2 & \cdots & a_{n-1} & 1 \\
a_1 & x & a_2 & \cdots & a_{n-1} & 1 \\
a_1 & a_2 & x & \cdots & a_{n-1} & 1 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
a_1 & a_2 & a_3 & \cdots & x & 1 \\
a_1 & a_2 & a_3 & \cdots & a_n & 1
\end{array}\right|$
计算$\left|\begin{array}{cccccc}
a_0 & 1 & 1 & \cdots & 1 & 1 \\
1 & a_1 & 0 & \cdots & 0 & 0 \\
1 & 0 & a_2 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
1 & 0 & 0 & \cdots & a_{n-1} & 0 \\
1 & 0 & 0 & \cdots & 0 & a_n
\end{array}\right|, a_i \neq 0 \quad(i=1,2, \cdots, n)$
计算$\left|\begin{array}{rrrlrr}
-a_1 & a_1 & 0 & \cdots & 0 & 0 \\
0 & -a_2 & a_2 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & -a_n & a_n \\
1 & 1 & 1 & \cdots & 1 & 1
\end{array}\right|$
证明题 （共 3 题 ），解答过程应写出必要的文字说明、证明过程或演算步骤
用行列式的性质证明: $\left|\begin{array}{lll}a_1+k b_1 & b_1+c_1 & c_1 \\ a_2+k b_2 & b_2+c_2 & c_2 \\ a_3+k b_3 & b_3+c_3 & c_3\end{array}\right|=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$
用行列式的性质证明: $\left|\begin{array}{lll}
b_1+c_1 & c_1+a_1 & a_1+b_1 \\
b_2+c_2 & c_2+a_2 & a_2+b_2 \\
b_3+c_3 & c_3+a_3 & a_3+b_3
\end{array}\right|=2\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|$
证明：$\left|\begin{array}{cccc}
a_1-b_1 & a_1-b_2 & \cdots & a_1-b_n \\
a_2-b_1 & a_2-b_2 & \cdots & a_2-b_n \\
\vdots & \vdots & & \vdots \\
a_n-b_1 & a_n-b_2 & \cdots & a_n-b_n
\end{array}\right|=0 \quad(n>2)$