Find the dimension and a basis for the general solution \(W\) of the following
homogeneous system using matrix notation:
$$
\begin{array}{r}
x_{1}+2 x_{2}+3 x_{3}-2 x_{4}+4 x_{5}=0 \\
2 x_{1}+4 x_{2}+8 x_{3}+x_{4}+9 x_{5}=0 \\
3 x_{1}+6 x_{2}+13 x_{3}+4 x_{4}+14 x_{5}=0
\end{array}
$$
Show how the basis gives the parametric form of the general solution of the
system.
When a system is homogeneous, we represent the system by its coefficient
matrix \(A\) rather than by its augmented matrix \(M\), because the last column of
the augmented matrix \(M\) is a zero column, and it will remain a zero column
during any row-reduction process.
Reduce the coefficient matrix \(A\) to echelon form, obtaining
$$
A=\left[\begin{array}{rrrrr}
1 & 2 & 3 & -2 & 4 \\
2 & 4 & 8 & 1 & 9 \\
3 & 6 & 13 & 4 & 14
\end{array}\right] \sim\left[\begin{array}{rrrrr}
1 & 2 & 3 & -2 & 4 \\
0 & 0 & 2 & 5 & 1 \\
0 & 0 & 4 & 10 & 2
\end{array}\right] \sim\left[\begin{array}{rrrrr}
1 & 2 & 3 & -2 & 4 \\
0 & 0 & 2 & 5 & 1
\end{array}\right]
$$
(The third row of the second matrix is deleted, because it is a multiple of
the second row and will result in a zero row.) We can now proceed in one of
two ways.
(a) Write down the corresponding homogeneous system in echelon form:
$$
\begin{array}{r}
x_{1}+2 x_{2}+3 x_{3}-2 x_{4}+4 x_{5}=0 \\
2 x_{3}+5 x_{4}+x_{5}=0
\end{array}
$$
The system in echelon form has three free variables, \(x_{2}, x_{4}, x_{5}\), so
\(\operatorname{dim} W=3\). A basis \(\left[u_{1}, u_{2}, u_{3}\right]\) for \(W\)
may be obtained as follows:
(1) Set \(x_{2}=1, x_{4}=0, x_{5}=0\). Back-substitution yields \(x_{3}=0\), and
then \(x_{1}=-2\). Thus, \(u_{1}=(-2,1,0,0,0)\)
(2) Set \(x_{2}=0, x_{4}=1, x_{5}=0\). Back-substitution yields
\(x_{3}=-\frac{5}{2}\), and then \(x_{1}=\frac{19}{2}\). Thus,
\(u_{2}=\left(\frac{19}{2}, 0,-\frac{5}{2}, 1,0\right)\)
(3) Set \(x_{2}=0, x_{4}=0, x_{5}=1\). Back-substitution yields
\(x_{3}=-\frac{1}{2}\), and then \(x_{1}=-\frac{5}{2}\). Thus,
\(u_{3}=\left(-\frac{5}{2}, 0,-\frac{1}{2}, 0,1\right)\)
[One could avoid fractions in the basis by choosing \(x_{4}=2\) in (2) and
\(x_{5}=2\) in (3), which yields multiples of \(u_{2}\) and \(u_{3}\).] The
parametric form of the general solution is obtained from the following linear
combination of the basis vectors using parameters \(a, b, c\) :
$$
a u_{1}+b u_{2}+c u_{3}=\left(-2 a+\frac{19}{2} b-\frac{5}{2} c,
a,-\frac{5}{2} b-\frac{1}{2} c, b, c\right)
$$
(b) Reduce the echelon form of \(A\) to row canonical form:
$$
A \sim\left[\begin{array}{ccccc}
1 & 2 & 3 & -2 & 4 \\
0 & 0 & 1 & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \sim\left[\begin{array}{rrrrr}
1 & 2 & 3 & -\frac{19}{2} & \frac{5}{2} \\
0 & 0 & 1 & \frac{5}{2} & \frac{1}{2}
\end{array}\right]
$$
Write down the corresponding free-variable solution:
$$
\begin{aligned}
&x_{1}=-2 x_{2}+\frac{19}{2} x_{4}-\frac{5}{2} x_{5} \\
&x_{3}=-\frac{5}{2} x_{4}-\frac{1}{2} x_{5}
\end{aligned}
$$
Using these equations for the pivot variables \(x_{1}\) and \(x_{3}\), repeat the
above process to obtain a basis \(\left[u_{1}, u_{2}, u_{3}\right]\) for \(W\).
That is, set \(x_{2}=1, x_{4}=0, x_{5}=0\) to get \(u_{1} ;\) set \(x_{2}=0,
x_{4}=1, x_{5}=0\) to get \(u_{2}\); and set \(x_{2}=0\), \(x_{4}=0, x_{5}=1\) to get
\(u_{3}\).
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