%P.M. Wong home number 272 7109

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\newcommand{\docid}{Quiz 1}
\renewcommand{\today}{September 1, 2000}

\input localmacs

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\doctitle

\noindent Consider the following system of linear equations:
\begin{center}
  $\begin{matrix}
    2x_1 & + & 4x_2 & + & 6x_3 & = & 6 \\
    x_1 & + & 3x_2 & + & 4x_3 & = & 4 \\
    &     & x_2 & + & ax_3 & = & 1 \\
\end{matrix}$
\end{center}

 \begin{problem}(2 points)
   Find $A$ and $\vec{b}$ such that the above system in the form $A
   \vec{x} = \vec{b}$ (i.e.\ an matrix equation).
 \end{problem}

$A = \begin{bmatrix}
  2 &  4 &  6  \\
  1 &  3 &  4  \\
  0 &  1 &  a  \\
\end{bmatrix}$ and 
$\vec{b} = \begin{bmatrix}
  6 \\
  4 \\
  1 \\
\end{bmatrix}$.

\begin{problem}(5 points)
  Explicitly showing \emph{all} the steps row reduce the
  \emph{augmented matrix} for the above system into echelon form.
\end{problem}

The augmented matrix is 
$M = \begin{bmatrix}
  2 &  4 &  6 &  6 \\
  1 &  3 &  4 & 4 \\
  0 &  1 &  a & 1 \\
\end{bmatrix}$.  First lets divide the top row by $2$ to get: 
$\begin{bmatrix}
  1 &  2 &  3 &  3 \\
  1 &  3 &  4 & 4 \\
  0 &  1 &  a & 1 \\
\end{bmatrix}$.  
Now multiple the top row by $-1$ and add to the second row to get
$\begin{bmatrix}
  1 &  2 &  3 &  3 \\
  0 &  1 &  1 & 1 \\
  0 &  1 &  a & 1 \\
\end{bmatrix}$.  Subtract the second row from the third row to get 
$N = \begin{bmatrix}
  1 &  2 &  3 &  3 \\
  0 &  1 &  1 & 1 \\
  0 &  0 &  a-1  & 0 \\
\end{bmatrix}$.   This is now in echelon form.

\begin{problem}(2 points)
  For which $a$ does the augmented matrix have 3 pivots?
\end{problem}

When $a \neq 1$.

\begin{problem}(4 points)
  Explicitly showing \emph{all} the steps reduce the augmented matrix
  into reduce echelon form.  (Hint: there are 2 cases.)
\end{problem}
First lets assume $a \neq 1$.  Then we can divide the last row by $a-1
\neq 0$ to get:
$\begin{bmatrix}
  1 &  2 &  3 &  3 \\
  0 &  1 &  1 & 1 \\
  0 &  0 &  1  & 0 \\
\end{bmatrix}$.  
Then subtract 3 times the last row from the first row and the last row
from the second row to get:  
$\begin{bmatrix}
  1 &  2 &  0 &  3 \\
  0 &  1 &  0 & 1 \\
  0 &  0 &  1  & 0 \\
\end{bmatrix}$.  
Now subtract 2 times the second from the first row to get:
$\begin{bmatrix}
  1 &  0 &  0 &  1\\
  0 &  1 &  0 & 1 \\
  0 &  0 &  1  & 0 \\
\end{bmatrix}$ 
which is in reduce echelon form.

Now assume $a = 1$. Hence 
$N = \begin{bmatrix}
  1 &  2 &  3 &  3 \\
  0 &  1 &  1 & 1 \\
  0 &  0 &  0  & 0 \\
\end{bmatrix}$.  
Subtract two times the second from the first row to get
$\begin{bmatrix}
  1 &  0 &  1 &  1 \\
  0 &  1 &  1 & 1 \\
  0 &  0 &  0  & 0 \\
\end{bmatrix}$ 
which is in reduce echelon form.
\begin{problem}(2 points)
  Is there an $a$ such that the system does not have a solution?  Why
  or why not?
\end{problem}         

No!  It is never the case that the last column contains a pivot.
\newpage

\begin{problem} (5 points)
  Find the general solution to the above system (Hint: there are two
  cases).
\end{problem}


There are two cases: either the reduce echelon from is
$\begin{bmatrix}
  1 &  0 &  0 &  1\\
  0 &  1 &  0 & 1 \\
  0 &  0 &  1  & 0 \\
\end{bmatrix}$ 
or
$\begin{bmatrix}
  1 &  0 &  1 &  1 \\
  0 &  1 &  1 & 1 \\
  0 &  0 &  0  & 0 \\
\end{bmatrix}$.  
In the first case, the general solution is $x_1 = 0, x_2 = 1$ and $x_2
= 1$ or $\vec{x} =\begin{bmatrix}
  1 \\
  1 \\
  0 \\
\end{bmatrix}$.  In the second case, the general solution is $x_1 = 1
- x_3, x_2 = 1 - x_3$ and $x_3$ is free.  Another way to write this is:
$\vec{x} =\begin{bmatrix}
  1 - x_3 \\
  1 - x_3\\
  x_3 \\
\end{bmatrix} = \begin{bmatrix}
  1  \\
  1 \\
  0  \\
\end{bmatrix} + \begin{bmatrix}
  - x_3 \\
  - x_3\\
  x_3 \\
\end{bmatrix} = \begin{bmatrix}
  1  \\
  1 \\
  0  \\
\end{bmatrix} + x_3\begin{bmatrix}
  - 1 \\
  - 1\\
  1 \\
\end{bmatrix}$.  

 


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