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段落第 14-38 行缺少 \right 和 Underfull \hbox（badness 10000）

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段落第 14-38 行缺少 \right 和 Underfull \hbox（badness 10000）

我正在尝试将三个向量放在一组中以完成我的数学作业，但当我运行它时，它显示一些错误，我不知道如何处理它。请帮忙！这是代码及其错误

\begin{document}
1.  To find basis for the four fundemental subspaces, we need to get the
RREF of the matrix
A = $\begin{bmatrix}
               -3 & 0 & 0 & -2 & 1 \\
               0 & -1 & 1 & 2 & -1 \\
               1 & 2 & -3 & -4 & 2 \\
               -2 & 0 & -1 & -2 & 1 \\
               0 & 2 & 1 & -2 & 1 \\
\end{bmatrix}$ = 
$\begin {bmatrix} 
             1 & 2 & -3 & -4 & 2 \\
             0 & 1 & -1 & -2 & 1 \\
             0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} \\
             0 & 0 & 0 & 0 & 0 \\
             0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$  = $\begin {bmatrix}
             1 & 0 & 0 & \frac{2}{3} & \frac{1}{3} \\
             0 & 1 & 0 & -\frac{4}{3} & \frac{4}{3} \\
             0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} \\
             0 & 0 & 0 & 0 & 0 \\
             0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$\\
\\
now, we need to find the basis fo Col(A) and Row(A)\\
recall that a basis for Col(A) is the columns in the original matrix
with all the columns in RREF that has a leading one\\
thus, a basis for Col(A) is \\
\begin{equation}
Col(A) = \left \{$\begin{bmatrix} -3 \\ 0 \\ 1 \\ -2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -1 \\ 2 \\ 0 \\ 2 \end{bmatrix}$ , $\begin{bmatrix} 0 \\ 1 \\ -3 \\ -1 \\ 1 \end{bmatrix}$ \right\}
\end{equation}
\end{document}

并显示错误

39 Missing \right. inserted.
<inserted text> 
            \right .
l.39 Col(A) = \left \{$
                   \begin{bmatrix} -3 \\ 0 \\ 1 \\ -2 \\ 0 \end{bmatrix}...

39 Display math should end with $$.
<to be read again> 
\def 
l.39 Col(A) = \left \{$\begin{bmatrix}


14
Underfull \hbox (badness 10000) in paragraph at lines 14--38

答案1

您已经在矩阵中处于数学模式，因此无需使用 重新启动它$。这就是 的Display math should end with $$.来源。

出现此错误的Missing \right. inserted.原因在于，您似乎在离开数学模式时将 放错了矩阵$，但您已使用 开始使用大括号，并且它们在 之前\left没有用 结束。\right$

这解决了以下问题：

\begin{document}
        1.  To find basis for the four fundemental subspaces, we need to get the
        RREF of the matrix\\
        $A = \begin{bmatrix}
        -3 & 0 & 0 & -2 & 1 \\
        0 & -1 & 1 & 2 & -1 \\
        1 & 2 & -3 & -4 & 2 \\
        -2 & 0 & -1 & -2 & 1 \\
        0 & 2 & 1 & -2 & 1 \\
        \end{bmatrix} = 
        \begin {bmatrix} 
        1 & 2 & -3 & -4 & 2 \\
        0 & 1 & -1 & -2 & 1 \\
        0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} \\
        0 & 0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 & 0 \\
        \end{bmatrix}  = \begin {bmatrix}
        1 & 0 & 0 & \frac{2}{3} & \frac{1}{3} \\
        0 & 1 & 0 & -\frac{4}{3} & \frac{4}{3} \\
        0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} \\
        0 & 0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 & 0 \\
        \end {bmatrix}$\\
        \\
        now, we need to find the basis for $Col(A)$ and $Row(A)$\\
        recall that a basis for $Col(A)$ is the columns in the original matrix
        with all the columns in RREF that has a leading one\\
        thus, a basis for $Col(A)$ is \\
        \begin{equation}
            Col(A) = \left \{\begin{bmatrix} -3 \\ 0 \\ 1 \\ -2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 2 \\ 0 \\ 2 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ -3 \\ -1 \\ 1 \end{bmatrix} \right\}
        \end{equation}
\end{document}

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