Week 2 Discussion

Exercise S.C15

Working within the vector space C³, determine if

\[b = \left[\begin{array}{r} 4\\ 3\\ 1\\ \end{array} \right] \]

is in the subspace W,

\[W = <\left[\begin{array}{r} 3\\ 2\\ 3\\ \end{array} \right], \left[\begin{array}{r} 1\\ 0\\ 3\\ \end{array} \right], \left[\begin{array}{r} 1\\ 1\\ 0\\ \end{array} \right], \left[\begin{array}{r} 2\\ 1\\ 3\\ \end{array} \right]>\]

This span and constraint vector b can be combined to form an augmented matrix.

\[ \left[\begin{array}{rrrr|r} 3 & 1 & 1 & 2 & 4 \\ 2 & 0 & 1 & 1 & 3 \\ 3 & 3 & 0 & 3 & 1 \\ \end{array} \right] \]

Let’s create the reduced echelon form matrix by hand.

1. -R1 + R3

\[ \left[\begin{array}{rrrr|r} 3 & 1 & 1 & 2 & 4 \\ 2 & 0 & 1 & 1 & 3 \\ 0 & 2 & -1 & 1 & -3 \\ \end{array} \right] \]

2. -R1 + R2

\[ \left[\begin{array}{rrrr|r} 3 & 1 & 1 & 2 & 4 \\ -1 & -1 & 0 & -1 & -1 \\ 0 & 2 & -1 & 1 & -3 \\ \end{array} \right] \]

3. 3 * R2 + R1

\[ \left[\begin{array}{rrrr|r} 0 & -2 & 1 & -1 & 1 \\ -1 & -1 & 0 & -1 & -1 \\ 0 & 2 & -1 & 1 & -3 \\ \end{array} \right] \]

4. -1 * R2, then switch R1 & R2

\[ \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & 1 \\ 0 & -2 & 1 & -1 & 1 \\ 0 & 2 & -1 & 1 & -3 \\ \end{array} \right] \]

5. R2 + R3

\[ \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & 1 \\ 0 & -2 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 & -2 \\ \end{array} \right] \]

6. (-1/2) * R3

\[ \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & 1 \\ 0 & -2 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \]

7. (-1/2) * R2

\[ \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & -1/2 & 1/2 & -1/2 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \]

8. -R2 + R1

\[ \left[\begin{array}{rrrr|r} 1 & 0 & 1/2 & 1/2 & 3/2 \\ 0 & 1 & -1/2 & 1/2 & -1/2 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \]

9. (1/2) * R3 + R2

\[ \left[\begin{array}{rrrr|r} 1 & 0 & 1/2 & 1/2 & 3/2 \\ 0 & 1 & -1/2 & 1/2 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \]

10. (-3/2) * R3 + R1

\[ \left[\begin{array}{rrrr|r} 1 & 0 & 1/2 & 1/2 & 0 \\ 0 & 1 & -1/2 & 1/2 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] \]

As you can see, their is a nonzero column in this augmented matrix, thus making the last row a pivot column. Given the theories that were posted in the “First Course in Linear Algebra”, if the last column is a pivot column in an augmented matrix, this system is inconsistent and does not have a solution. Therefore b is not an element in the subspace.