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Next: Problems on Row Reduction Up: Solutions to Supplementary Homework Previous: Solutions to Supplementary Homework

Problems on Bases

1.
The coefficient matrix is

\begin{displaymath}\left[\begin{array}{rr}3&-2\\ 6&-4\end{array}\right]
\end{displaymath}

which row reduces to

\begin{displaymath}\left[\begin{array}{rr}1&-\frac23\\ 0&0\end{array}\right].
\end{displaymath}

This gives

\begin{eqnarray*}x_1 - {\textstyle\frac23} x_2 &=& 0\\ 0&=&0.
\end{eqnarray*}


Solving for x1 gives

\begin{displaymath}x_1={\textstyle\frac23} x_2.
\end{displaymath}

Assigning the value t to x2 gives $x_1=\frac23t$, so the general solution is

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\end{array}\right] =
\left[\b...
...eft[\begin{array}{c}{\textstyle\frac23}\\ 1\end{array}\right].
\end{displaymath}

Thus the single vector

\begin{displaymath}\left[\begin{array}{c}{\textstyle\frac23}\\ 1\end{array}\right]
\end{displaymath}

forms a basis for the solution space. Any multiple of this vector will also do.

2.
As above, row reduce the coefficient matrix:

\begin{displaymath}\left[\begin{array}{rrr}1&1&1\\ 2&2&2\end{array}\right]
\rig...
...row
\left[\begin{array}{rrr}1&1&1\\ 0&0&0\end{array}\right].
\end{displaymath}

This gives the system

\begin{eqnarray*}x_1+x_2+x_3&=&0\\ 0&=&0
\end{eqnarray*}


Solving for x1 gives

x1=-x2-x3

assigning the values s and t to x2 and x3 respectively gives x1=-s-t, so the general solution has the form

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\\ x_3\end{array}\right]
= \l...
...ight]
+ t \left[\begin{array}{r}-1\\ 0\\ 1\end{array}\right].
\end{displaymath}

The set

\begin{displaymath}\left\{\left[\begin{array}{r}-1\\ 1\\ 0\end{array}\right],
\left[\begin{array}{r}-1\\ 0\\ 1\end{array}\right]\right\}
\end{displaymath}

forms a basis for the solution space.

3.
The coefficient matrix is invertible, so the system has only the trivial solution x1=x2=0. The empty set forms a basis for the solution space.


next up previous
Next: Problems on Row Reduction Up: Solutions to Supplementary Homework Previous: Solutions to Supplementary Homework
Frank Beatrous
1998-09-21