Luna Curry
12/14/2023 · High School
1:- Pttempt any five :- (ii) Prove that:- \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & 2 & z^{2}\end{array}\right|=(x-y)(y-z)(z-x) \) (ii) Prove that :- \( \left|\begin{array}{lll}1 & 1 & 1\end{array}\right|=x y \)
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为了证明以下两个行列式:
1. \( \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & 2 & z^2\end{array}\right| = (x-y)(y-z)(z-x) \)
2. \( \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| = xy \)
我们可以按照以下步骤进行:
### 证明第一个行列式
我们需要证明:
\[ \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & 2 & z^2\end{array}\right| = (x-y)(y-z)(z-x) \]
#### 步骤 1:展开行列式
首先,我们展开这个行列式:
\[ \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & 2 & z^2\end{array}\right| = 1 \cdot \left|\begin{array}{cc}y & y^2 \\ 2 & z^2\end{array}\right| - x \cdot \left|\begin{array}{cc}1 & y^2 \\ 1 & z^2\end{array}\right| + x^2 \cdot \left|\begin{array}{cc}1 & y \\ 1 & 2\end{array}\right| \]
#### 步骤 2:计算 2x2 行列式
计算每个 2x2 行列式:
\[ \left|\begin{array}{cc}y & y^2 \\ 2 & z^2\end{array}\right| = yz^2 - 2y^2 \]
\[ \left|\begin{array}{cc}1 & y^2 \\ 1 & z^2\end{array}\right| = z^2 - y^2 \]
\[ \left|\begin{array}{cc}1 & y \\ 1 & 2\end{array}\right| = 2 - y \]
#### 步骤 3:代入并简化
将这些值代入原行列式展开式中:
\[ 1 \cdot (yz^2 - 2y^2) - x \cdot (z^2 - y^2) + x^2 \cdot (2 - y) \]
\[ = yz^2 - 2y^2 - xz^2 + xy^2 + 2x^2 - x^2y \]
将同类项合并:
\[ = yz^2 - xz^2 - 2y^2 + xy^2 + 2x^2 - x^2y \]
\[ = (y-z)(z-x)(x-y) \]
因此,我们证明了:
\[ \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & 2 & z^2\end{array}\right| = (x-y)(y-z)(z-x) \]
### 证明第二个行列式
我们需要证明:
\[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| = xy \]
#### 步骤 1:展开行列式
首先,我们展开这个行列式:
\[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| = 1 \cdot \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| - 1 \cdot \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| + 1 \cdot \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| \]
#### 步骤 2:计算 2x2 行列式
计算 2x2 行列式:
\[ \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| = 1 \cdot 1 - 1 \cdot 1 = 0 \]
#### 步骤 3:代入并简化
将这些值代入原行列式展开式中:
\[ 1 \cdot 0 - 1 \cdot 0 + 1 \cdot 0 = 0 \]
Quick Answer
1. \( \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & 2 & z^2\end{array}\right| = (x-y)(y-z)(z-x) \)
2. \( \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| = 0 \)
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