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\newcommand{\docid}{Quiz 3} \renewcommand{\today}{October 2, 2000}

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\begin{document}
\doctitle
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\noindent

\begin{problem}
  Is $A = 
  \begin{bmatrix}
    1 & 2 & 3\\
    1 & 3 & 4\\
    0 & {-2} & -2
  \end{bmatrix}
$ invertible?  If so find an inverse. Find all the solutions to
  $A\vec{x} = \vec{e}_2$.
\end{problem}

The matrix $\begin{bmatrix}
  1 & 2 & 3 & 1 & 0 & 0\\
  1 & 3 & 4 & 0 & 1 & 0\\
  0 & {-2} & -2 & 0 & 0 & 1
  \end{bmatrix}$ row reduces to 
  $\begin{bmatrix}
    1& 0& 1& 0& 1& \frac{3}{2}\\
    0& 1& 1& 0& 0& \frac12 \\
    0& 0& 0& 1& -1 & -\frac12 \end{bmatrix}$ so $A$ is not invertible.
  From this we can also see that $\begin{bmatrix}
    1 & 2 & 3 & 0 \\
    1 & 3 & 4 & 1\\
    0 & {-2} & -2 & 0
  \end{bmatrix}$ row reduces to $\begin{bmatrix}
    1& 0& 1&  1& \\
    0& 1& 1&  0&  \\
    0& 0& 0& -1 & \end{bmatrix}$ which has a pivot in the last column
  and therefore $A\vec{x} = \vec{e}_2$ is not consistent (has no
  solutions).


\begin{problem}
  $B =
  \begin{bmatrix}
    1 & 2 & 0 \\
    0 & \frac{1}{2} & 0 \\
    1 & 0 & 3
  \end{bmatrix}$ is invertible. Write $B$ and $B^{-1}$ as the product of
  elementary matrices.
\end{problem}

Since we know $B$ is invertible we will reduce $B$ to $I$ keeping
track of the elementary row operations.  We will write this out as
multiplication by elementary matrices.

$$  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    -1 & 0 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 2 & 0 \\
    0 & \frac{1}{2} & 0 \\
    1 & 0 & 3
  \end{bmatrix} =  \begin{bmatrix}
    1 & 2 & 0 \\
    0 & \frac{1}{2} & 0 \\
    0 & -2 & 3
  \end{bmatrix}$$


$$ 
 \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 1
  \end{bmatrix}
\begin{bmatrix}
    1 & 2 & 0 \\
    0 & \frac{1}{2} & 0 \\
    0 & -2 & 3
  \end{bmatrix} = \begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & -2 & 3
  \end{bmatrix}$$
  
  $$
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 2 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & -2 & 3
  \end{bmatrix} = \begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 3
  \end{bmatrix}$$
  
  $$\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & \frac13
  \end{bmatrix}\begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 3
  \end{bmatrix} =\begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{bmatrix}$$
  
  $$\begin{bmatrix}
  1 & 2 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{bmatrix}
= \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{bmatrix}$$
  
  Hence 
  
  $$
  B^{-1} = \begin{bmatrix}
    1 & 2 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & \frac13
  \end{bmatrix}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 2 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 1
  \end{bmatrix} \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    -1 & 0 & 1
  \end{bmatrix}$$
  
  (You might have a slightly different order with slightly different
  matrices.)  $(B^{-1})^{-1}= B$, the inverse of an elementary matrix
  is also an elementary matrix and the inverse of $AC$ is
  $C^{-1}A^{-1}$ so
  $$B = \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    -1 & 0 & 1
  \end{bmatrix}^{-1} \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 1
  \end{bmatrix}^{-1}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 2 & 1
  \end{bmatrix}^{-1}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & \frac13
  \end{bmatrix}^{-1} \begin{bmatrix}
  1 & 2 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
  \end{bmatrix}^{-1}$$
\newpage
$$B = \begin{bmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  1 & 0 & 1
  \end{bmatrix} \begin{bmatrix}
    1 & 0 & 0 \\
    0 & \frac12 & 0 \\
    0 & 0 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & -2 & 1
  \end{bmatrix}\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 3
  \end{bmatrix} \begin{bmatrix}
  1 & -2 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
  \end{bmatrix}$$

\begin{problem}
  Assume that $AB$ exists.  Show that if the columns of $B$ are
  linearly dependent then the columns of $AB$ are also.
\end{problem}


$AB = A
\begin{bmatrix}
  \vec{b}_1 & \vec{b}_2 & \ldots & \vec{b}_n
\end{bmatrix} =
\begin{bmatrix}
  A\vec{b}_1 & A\vec{b}_2 & \ldots & A\vec{b}_n
\end{bmatrix}$.
The columns of $B$ are linearly dependent iff there are $c_i$'s not
all zero such that $c_1\vec{b}_1 + c_1\vec{b}_2 \ldots + c_n\vec{b}_n
= 0$.  Then $A(c_1\vec{b}_1 + c_1\vec{b}_2 \ldots + c_n\vec{b}_n) = 0$
and hence $Ac_1\vec{b}_1 + Ac_1\vec{b}_2 \ldots + Ac_n\vec{b}_n =
c_1A\vec{b}_1 + c_1A\vec{b}_2 \ldots + c_nA\vec{b}_n = 0$. Therefore
the columns of $AB$ are linearly dependent.



\begin{problem}
  Assume $AB$ exists.  If $A$ is invertible and $AB = 0$ what is $B$?
  Give an example where $A \neq 0$ and $B \neq 0$ but $AB = 0$.
\end{problem}

If $A$ is invertible then we can multiple on the left by $A^{-1}$ to
get $$A^{-1}AB = A 0 $$
$$IB = 0 $$
$$B = 0. $$  In our example neither $A$ nor $B$ can be invertible.  Let
$ A = B =
\begin{bmatrix}
  0 & 0 \\
  1 & 0 
\end{bmatrix}$.  $AB =\begin{bmatrix}
  0 & 0 \\
  1 & 0 
\end{bmatrix} \begin{bmatrix}
  0 & 0 \\
  1 & 0 
\end{bmatrix}= 0$.


\begin{problem}  Find the determinant of
  $ A = \begin{bmatrix}
    2 & 1 & 0 & -1 & 0 \\
    2 & -1& 1 & -3 & 0 \\
    1 & 2 & 0 & -2 & 3 \\
    0 & 2 & 0 & 0  & 0 \\
    1 & 2 & 2 & 3 & 0
     \end{bmatrix}$.
\end{problem}


Start by taking the cofactor expansion across the fourth row.  Only
$a_{4,2}$ is nonzero, so $\det(A) = (2)(-1)^{4+2}\det(A_{4,2}) =
(2)\det(A_{4,2})$. Let $B = A_{4,2} =\begin{bmatrix}
  2 & 0 & -1 & 0 \\
  2 & 1 & -3 & 0 \\
  1 & 0 & -2 & 3 \\
  1 & 2 & 3 & 0 \end{bmatrix}$. To take the determinant of $B$ lets
use the cofactor expansion down the last column.  Only $b_{3,4}$ is
nonzero, so $\det(B) = (3) (-1)^{3+4} \det(B_{3,4}) = -3
\det(B_{3,4})$.  $B_{3,4} = \begin{bmatrix}
  2 & 0 & -1  \\
  2 & 1 & -3 \\
  1 & 2 & 3 \end{bmatrix}$.  $\det(B_{3,4}) = 2\det( \begin{bmatrix}
  1 & -3 \\
  2 & 3 \end{bmatrix}) + (-1) \det (\begin{bmatrix}
  2 & 1  \\
  1 & 2 \end{bmatrix}) = (2) (9) + (-1)(3) = 15$. So $\det(A) =
(2)(-3)(15) = -90$.


\end{document}

