matrix theory_basic results and techniques_exercise_1.2.2,1.2.3
2012-12-10 20:36
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Evaluate the determinant
$$
\begin{vmatrix}
1+x&1&1\\
1&1+y&1\\
1&1&1+z\\
\end{vmatrix}
$$
Answer:将矩阵的第一行乘以-1加到第二行,得到
$$
\begin{vmatrix}
1+x&1&1\\
-x&y&0\\
1&1&1+z\\
\end{vmatrix}
$$
将矩阵的第一行乘以-1加到第三行,得到
$$
\begin{vmatrix}
1+x&1&1\\
-x&y&0\\
-x&0&z\\
\end{vmatrix}=xyz+xy+yz+zx
$$
Show the $3\times 3$ Vandermonde determinant identity
$$
\begin{vmatrix}
1&1&1\\
a_1&a_2&a_3\\
a_1^2&a_2^2&a_3^2\\
\end{vmatrix}=(a_1-a_2)(a_2-a_3)(a_3-a_1)
$$
我们先看该矩阵的转置:
$$
\begin{vmatrix}
1&a_1&a_1^2\\
1&a_2&a_2^2\\
1&a_3&a_3^2\\
\end{vmatrix}
$$
然后第一行乘以-1加到第二行上,将第一行乘以-1加到第三行上:
$$
\begin{vmatrix}
1&a_1&a_1^2\\
0&a_2-a_1&a_2^2-a_1^2\\
0&a_3-a_1&a_3^2-a_1^2\\
\end{vmatrix}=(a_2-a_1)(a_3^2-a_1^2)-(a_3-a_1)(a_2^2-a_1^2)=(a_1-a_2)(a_3-a_1)(a_2-a_3)
$$
And evaluate the determinant
$$
\begin{vmatrix}
1&a&a^2-bc\\
1&b&b^2-ca\\
1&c&c^2-ab\\
\end{vmatrix}
$$
Answer:将矩阵的第一行乘以-1加到第二行上,将矩阵的第二行乘以-1加到第三行上,将矩阵的第三行乘以-1加到第一行上:
$$
\begin{vmatrix}
0&a-c&(a-c)(a+b+c)\\
0&b-a&(b-a)(a+b+c)\\
0&c-b&(c-b)(a+b+c)\\
\end{vmatrix}
$$
可见,最后的结果是0.
$$
\begin{vmatrix}
1+x&1&1\\
1&1+y&1\\
1&1&1+z\\
\end{vmatrix}
$$
Answer:将矩阵的第一行乘以-1加到第二行,得到
$$
\begin{vmatrix}
1+x&1&1\\
-x&y&0\\
1&1&1+z\\
\end{vmatrix}
$$
将矩阵的第一行乘以-1加到第三行,得到
$$
\begin{vmatrix}
1+x&1&1\\
-x&y&0\\
-x&0&z\\
\end{vmatrix}=xyz+xy+yz+zx
$$
Show the $3\times 3$ Vandermonde determinant identity
$$
\begin{vmatrix}
1&1&1\\
a_1&a_2&a_3\\
a_1^2&a_2^2&a_3^2\\
\end{vmatrix}=(a_1-a_2)(a_2-a_3)(a_3-a_1)
$$
我们先看该矩阵的转置:
$$
\begin{vmatrix}
1&a_1&a_1^2\\
1&a_2&a_2^2\\
1&a_3&a_3^2\\
\end{vmatrix}
$$
然后第一行乘以-1加到第二行上,将第一行乘以-1加到第三行上:
$$
\begin{vmatrix}
1&a_1&a_1^2\\
0&a_2-a_1&a_2^2-a_1^2\\
0&a_3-a_1&a_3^2-a_1^2\\
\end{vmatrix}=(a_2-a_1)(a_3^2-a_1^2)-(a_3-a_1)(a_2^2-a_1^2)=(a_1-a_2)(a_3-a_1)(a_2-a_3)
$$
And evaluate the determinant
$$
\begin{vmatrix}
1&a&a^2-bc\\
1&b&b^2-ca\\
1&c&c^2-ab\\
\end{vmatrix}
$$
Answer:将矩阵的第一行乘以-1加到第二行上,将矩阵的第二行乘以-1加到第三行上,将矩阵的第三行乘以-1加到第一行上:
$$
\begin{vmatrix}
0&a-c&(a-c)(a+b+c)\\
0&b-a&(b-a)(a+b+c)\\
0&c-b&(c-b)(a+b+c)\\
\end{vmatrix}
$$
可见,最后的结果是0.
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