A homogeneous linear system of equations has a coefficient matrix A which
is row equivalent to the following matrix R in reduced echelon form:
\begin{bmatrix}
1 & 2 & 0 & 3 & 0 &5\\
0 & 0 & 1 &4&0&2 \\
0 & 0 & 0&0&1 &-3
\end{bmatrix}
(a) Describe the solution set in parametric vector form.
(b) Suppose the first column of A is:
\begin{bmatrix}
2\\
3 \\
5
\end{bmatrix}
What is the second column of A?
(c) If, in addition to the above, the third column of A is:
\begin{bmatrix}
3\\
-1 \\
4
\end{bmatrix}
what is the fourth column of A? Hint: Use special solutions of Ax = 0
obtained by setting one free variable to 1, and the others to 0. Each
solution of Ax = 0 gives a linear combination of the columns of A that
is 0. The special solutions make it easy to solve for the free columns
in terms of the pivot columns.
I already did part (a) but I am unsure on how to get the columns of A. Could I just obtain them through performing row operations on R, since they are row equivalent? And how can I use special solutions to solve for the fourth column?
Best Answer
In part (a), we find that every solution to $A\vec{x}=\vec{O}$ is of the form \begin{align*} \vec{x} &= \left[\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \end{array}\right] = \left[\begin{array}{r} -2 \, x_{2} - 3 \, x_{4} - 5 \, x_{6} \\ x_{2} \\ -4 \, x_{4} - 2 \, x_{6} \\ x_{4} \\ 3 \, x_{6} \\ x_{6} \end{array}\right] \\ &= x_2\left[\begin{array}{r} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]+x_4\left[\begin{array}{r} -3 \\ 0 \\ -4 \\ 1 \\ 0 \\ 0 \end{array}\right]+x_6\left[\begin{array}{r} -5 \\ 0 \\ -2 \\ 0 \\ 3 \\ 1 \end{array}\right] \end{align*} This means that the three vectors \begin{align*} \vec{v}_1 &= \left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle & \vec{v}_2 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle & \vec{v}_3 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle \end{align*} form a basis of $\operatorname{Null}(A)$.
In particular, the vector $\vec{v}_1=\left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle$ satisfies $A\vec{v}_1=\vec{O}$. This means that $$ (-2)\cdot\vec{a}_1+1\cdot\vec{a}_2+0\cdot\vec{a}_3+0\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O} $$ where $\{\vec{a}_1, \vec{a}_2, \vec{a}_3, \vec{a}_4, \vec{a}_5, \vec{a}_6 \}$ are the columns of $A$. This gives $\vec{a}_2=2\cdot\vec{a}_1$.
So, if the first column of $A$ is $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$, then the second column of $A$ is $\vec{a}_2=2\cdot\vec{a}_1=\left\langle4,\,6,\,10\right\rangle$.
Also note that the vector $\vec{v}_2=\left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle$ satisfies $A\vec{v}_2=\vec{O}$. This means that $$ 3\cdot\vec{a}_1+0\cdot\vec{a}_2+(-4)\cdot\vec{a}_3+1\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O} $$ Solving for $\vec{a}_4$ gives $$ \vec{a}_4 = -3\cdot\vec{a}_1+4\cdot\vec{a}_3 $$ So, if $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$ and $\vec{a}_3=\left\langle3,\,-1,\,4\right\rangle$, then $$ \vec{a}_4 = -3\cdot\vec{a}_1+4\cdot\vec{a}_3 = -3\cdot\left\langle2,\,3,\,5\right\rangle+4\cdot\left\langle3,\,-1,\,4\right\rangle = \left\langle6,\,-13,\,1\right\rangle $$