一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
已知 $\boldsymbol{A} ^ { – 1 } = \left[ \begin{array} { c c c } 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right]$, 则:
$$
\begin{aligned}
\left( 3 \boldsymbol{A} ^ {*} \right) ^ { – 1 } & = ? \\
\left( 2 \boldsymbol {A} \right) ^ {*} & = ?
\end{aligned}
$$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
考研题目有时候不光考察考生的知识储备,还会利用心理的弱点,设置一些“坑”——
看完这道题目,大家首先想一个问题:题目所给的具体矩阵是 $\boldsymbol{A}$, 还是 $\boldsymbol{A} ^{-1}$ ?
如果回答是 “$\boldsymbol{A}$”, 那么,这道题目大概率就会产生低级错误。而且,在下面的计算中,我们也很容易一不小心就陷入这个低级错误之中。
首先,根据 $\boldsymbol{A} ^{*}$ $=$ $\begin{vmatrix} \boldsymbol{A} \end{vmatrix} \boldsymbol{A} ^{-1}$ 可得:
$$
\begin{aligned}
\left( 3 \boldsymbol{A} ^ {*} \right) ^ { – 1 } \\ \\
& = \left( 3 \begin{vmatrix} \boldsymbol{A} \end{vmatrix} \cdot \boldsymbol{A} ^{-1} \right) ^{-1} \\ \\
& = \frac{1}{3 \begin{vmatrix} \boldsymbol{\textcolor{orangered}{A} } \end{vmatrix}} \cdot \boldsymbol{ \textcolor{orangered}{A} }
\end{aligned}
$$
由于计算到上面这一步出现了两个矩阵 “$\boldsymbol{A}$”, 我们就会很容易觉得题目中所给的已知条件就是矩阵 “$\boldsymbol{A}$”, 但实际上,题目所给的已知条件是矩阵 $\boldsymbol{A}$ 的逆矩阵 “$\boldsymbol{A} ^{-1}$”.
因此,我们首先需要根据求逆矩阵的方法求解出来矩阵 $\textcolor{orangered}{\boldsymbol{A} ^{-1}}$ 的逆矩阵,也就是矩阵 $\textcolor{yellow}{\boldsymbol{A}}$:
$$
\begin{aligned}
\begin{bmatrix}
\textcolor{orangered}{1} & \textcolor{orangered}{0} & \textcolor{orangered}{0} & \vdots & \textcolor{pink}{1} & \textcolor{pink}{0} & \textcolor{pink}{0} \\
\textcolor{orangered}{2} & \textcolor{orangered}{1} & \textcolor{orangered}{0} & \vdots & \textcolor{pink}{0} & \textcolor{pink}{1} & \textcolor{pink}{0} \\
\textcolor{orangered}{3} & \textcolor{orangered}{2} & \textcolor{orangered}{1} & \vdots & \textcolor{pink}{0} & \textcolor{pink}{0} & \textcolor{pink}{1}
\end{bmatrix} & \Rightarrow \\ \\
\begin{bmatrix}
1 & 0 & 0 & \vdots & 1 & 0 & 0 \\
0 & 1 & 0 & \vdots & -2 & 1 & 0 \\
3 & 2 & 1 & \vdots & 0 & 0 & 1
\end{bmatrix} & \Rightarrow \\ \\
\begin{bmatrix}
\textcolor{pink}{1} & \textcolor{pink}{0} & \textcolor{pink}{0} & \vdots & \textcolor{yellow}{1} & \textcolor{yellow}{0} & \textcolor{yellow}{0} \\
\textcolor{pink}{0} & \textcolor{pink}{1} & \textcolor{pink}{0} & \vdots & \textcolor{yellow}{-2} & \textcolor{yellow}{1} & \textcolor{yellow}{0} \\
\textcolor{pink}{0} & \textcolor{pink}{0} & \textcolor{pink}{1} & \vdots & \textcolor{yellow}{1} & \textcolor{yellow}{-2} & \textcolor{yellow}{1}
\end{bmatrix} & \Rightarrow
\end{aligned}
$$
于是可知:
$$
\textcolor{yellow}{
\boldsymbol{A} = \begin{bmatrix}
1 & 0 & 0 \\
-2 & 1 & 0 \\
1 & -2 & 1
\end{bmatrix}
}
$$
因此:
$$
\textcolor{springgreen}{
\left( 3 \boldsymbol{A} ^ {*} \right) ^ { – 1 } = \frac { 1 } { 3 } \begin{bmatrix}
1 & 0 & 0 & \\
– 2 & 1 & 0 \\
1 & – 2 & 1
\end{bmatrix}
}
$$
类似的,由于:
$$
\begin{aligned}
\left( 2 \boldsymbol {A} \right) ^ {*} \\ \\
& = \begin{vmatrix} 2 \boldsymbol{A} \end{vmatrix} \cdot \left( 2 \boldsymbol{A} \right) ^{-1} \\ \\
& = 2 ^{\boldsymbol{\textcolor{orangered}{3}}} \cdot \frac{1}{2} \boldsymbol{A} ^{-1} \\ \\
& = 4 \boldsymbol{A} ^{-1}
\end{aligned}
$$
Tip
需要注意的是,由于 $\boldsymbol{A}$ 是三阶矩阵,所以 $\begin{vmatrix} 2 \boldsymbol{A} \end{vmatrix}$ $=$ $2 ^{\boldsymbol{\textcolor{orangered}{3}}} \cdot \begin{vmatrix} \boldsymbol{A} \end{vmatrix}$
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于是:
$$
\textcolor{springgreen}{
\left( 2 \boldsymbol {A} \right) ^ {*} = 4 \begin{bmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{bmatrix}
}
$$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。