zlotus/notes-linear-algebra

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gaufung opened this issue · 3 comments

  • chapter 08
    The prof. takes $b_1=1,b_2=5,b_3=6 $ for an example, so
    $
    \begin{eqnarray*}
    x_1 & + & 2x_3 & = & 1 \
    & & 2x_3 & = & 3 \
    \end{eqnarray*}
    $
    $
    \begin{eqnarray*}
    x_1 & = & -2 \
    x_3 & = & \frac{3}{2} \
    \end{eqnarray*}
    $

  • chapter 16

$Pb$将会把向量投影在$A$的行空间中

Should be
$Pb$将会把向量投影在$A$的列空间中

Thanks, I've the error in chapter 16.
But I don't get the equations mentioned in chapter 08, could you make a more specific description?

Thank you~

解法:另所有自由变量取0,则有
X1+2X3 = 0
2X3 = 0
解得X1 = -2 X3=3/2

The Prof. took b1=1,b2=5,b3=6 for example which subjects to b3-b2-b1 =0 condition. So if we set free varibales =0 , we should get X1+2X3 = 1;2X3=3,and solve this equations.

Sorry, my mistake.
Thanks a lot~