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\begin{center}
\textbf{Exercises on the geometry of linear equations}
\end{center}

\begin{problem}
(1.3 \#4.  {\em Introduction to Linear Algebra:} Strang) Find a combination $x_1 \vb w_1 + x_2 \vb w_2 + x_3 \vb w_3$ that gives the zero vector:
$$\vb w_1 = \threebyone {1} {2} {3}
\vb w_2 = \threebyone {4} {5} {6}
\vb w_3 = \threebyone {7} {8} {9}.$$
Those vectors are (independent)(dependent).

The three vectors lie in a  \rule{20mm}{0.4pt}. The matrix $W$ with those columns is \textit{not invertible}.
\end{problem}

\solution{
We might observe that $\vb w_1 + \vb w_3 - 2\vb w_2 = 0,$ or we might simultaneously solve the system of equations:
\begin{eqnarray*}
1x_1 + 4x_2 + 7x_3 &=& 0\\
2x_1 + 5x_2 + 8x_3 &=& 0\\
3x_1 + 6x_2 + 9x_3 &=& 0
\end{eqnarray*}

Subtracting twice equation 1 from equation 2 gives us $-3x_2 -6x_3 = 0$.  Subtracting thrice equation 1 from equation 3 gives us $-6x_2-12x_3 = 0$, which is equivalent to the previous equation and so leads us to suspect that the vectors are dependent.  At this point we might guess $x_2 = -2$ and $x_3 = 1$ which would lead us to the answer we observed above:
$$x_1 = 1,\; x_2 = -2, \; x_3 = 1 \; \text{and}\; \vb w_1 - 2\vb w_2+\vb w_3 = 0.$$
Those vectors are \textbf{dependent} because there is a combination of the vectors that gives the zero vector. 

\vspace{10 pt}

The three vectors lie in a \textbf{plane.}
}

\begin{problem}Multiply:
$\threebythree
1  2  0 
2  0  3 
4  1  1 
\threebyone
3  {-2}  1 $. 
\end{problem}

\solution{ $\threebyonec 
{1\cdot 3 + 2\cdot{(-2)} + 0 \cdot 1}
{6+0+3}
{12-2+1}
=
\threebyone
{-1}  9  {11}$.}

\vspace{10pt}

\problem{ True or false: A 3 by 2 matrix $\mb A$ times a 2 by 3 matrix $\mb B$ equals a 3 by 3 matrix $\mb A \mb B$. If this is false, write a similar sentence which is correct.}

\solution{ The statement is true.  In order to multiply two matrices,
  the number of columns of $\mb A$ must equal the number of rows of
  $\mb B$.  The product $\mb A \mb B$ will have the same number of
  rows as the first matrix and the same number of columns as the
  second:
$$\mb A (m \text{ by }n) \text{ times } \mb B (n \text{ by } p) \text{ equals } \mb A \mb B (m \text{ by } p).$$
}

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