1

I am trying to do put three vectors in a set for my math homework, but when i run it it shows me some errors and i have no idea how to work with it. Please help! Here is the code and its error

\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}

and the errors are shown

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
7
  • Don't activate math mode $...$ inside an equation. It's already in math mode.
    – Werner
    May 11 '16 at 1:03
  • You can't have $ ... $ within equation. equation is a math mode environment, you're already in math mode. The first $ of $ ... $ will take you out of math mode, just when you need it, and the second $ will put you back in math mode. But more generally, you should be in math mode for the entirety of your equations, not just for the matrices. If you're writing a single mathematical object/expression/equation, don't hop in and out of math mode, you need to stay in it
    – Au101
    May 11 '16 at 1:04
  • so i should take off all the $ in equation?
    – Peter Zeng
    May 11 '16 at 1:27
  • Certainly you should remove the $s you have within \begin{equation} ... \end{equation}
    – Au101
    May 11 '16 at 1:32
  • what about the missing \right?
    – Peter Zeng
    May 11 '16 at 1:48
1

You are already in mathmode inside the matrixes, so you don't need to start it again with $. That's where Display math should end with $$. comes from.

The error Missing \right. inserted. is because you seem to be leaving mathmode with the misplaced $ in the matrix but you have started braces with \left and they are not closed with \right before the $.

This solved the issues:

\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}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.