Problem # A.1:
(a) Check that the vectors \(\mathbf{v}=(1,2,1)\)
and \(\mathbf{w}=(8,4,-1)\) are solutions to the linear equation
\[
2x_1 - 3x_2 + 4x_3 = 0. \qquad\text{(1)}
\]
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(b) The vector \(\mathbf{u}=(-37,-14,8)\) is also a solution to equation (1).
Find real numbers \(a,b\in\mathbf{R}\) so that
\[
\mathbf{u} = a\mathbf{v}+b\mathbf{w}.
\]
(c) Prove the following general result.
If the vector \(\mathbf{z}=(z_1,z_2,z_3)\in\mathbf{R}^3\) is
a solution to equation (1), then there are scalars
\(a,b\in\mathbf{R}\) so that
\[
\mathbf{z} = a\mathbf{v}+b\mathbf{w}. \qquad\text{(2)}
\]
(d) In (c), prove that for a given vector \(\mathbf{z}\),
there is only one choice for \(a\) and \(b\) that makes equation (2) true.
Solution:
(b)
We need to find \(a\) and \(b\) so that
\[\begin{aligned}
\mathbf{u} &= a \mathbf{v} + b \mathbf{w} \\
(-37,-14,8) &= a(1,2,1) + b(8,4,-1) \\
(-37,-14,8) &= (a+8b,2a+4b,a-b). \\
\end{aligned}
\]
So we need to solve the simultaneous equations
\[
-37=a+8b,\qquad -14=2a+4b,\qquad 8=a-b.
\]
There are lots of ways to do this.
For example, subtracting the third equation from the first one,
we get
\[
\begin{aligned}
-37-8 &= (a+8b)-(a-b) \\
-45 &= 9b \\
-5 &= b. \\
\end{aligned}
\]
Substituting \(b=-5\) into the last equation gives \(a=3\). Then
one should check that \((a,b)=(3,-5)\) is a solution to the three
simultaneous equations. It is, so the solution to (b) is
\[
\mathbf{u} = 3 \mathbf{v} - 5 \mathbf{w}.
\]
(c)
We are given that \(\mathbf{z}=(z_1,z_2,z_3)\) satisfies equation (1),
so we know that
\[
2z_1 - 3z_2 + 4z_3 = 0.
\]
We want to find \(a\) and \(b\) so that
\(\mathbf{z} = a\mathbf{v}+b\mathbf{w}\). This is more-or-less the same
problem as in (b), we need to solve
\[\begin{aligned}
\mathbf{z} &= a \mathbf{v} + b \mathbf{w} \\
(z_1,z_2,z_3) &= a(1,2,1) + b(8,4,-1) \\
(z_1,z_2,z_3) &= (a+8b,2a+4b,a-b). \\
\end{aligned}
\]
So we need to solve
\[\begin{aligned}
a+8b &= z_1 \\
2a+4b &= z_2 \\
a-b &= z_3.\\
\end{aligned}
\]
Proceeding as in (b), we subtract the third equation from the first to
get
\[
9b = z_1-z_3,
\quad\text{so}\quad
b = \frac{z_1-z_3}{9}.
\]
Then substituting in to the first equation gives
\[
a = z_1 - 8b = z_1 - \frac{8z_1-8z_3}{9}
= \frac{z_1+8z_3}{9}.
\]
We now need to check that these values of \(a\) and \(b\) actually work.
(Note that so far, we haven't used the fact that \(\mathbf{z}\) satisfies
equation (1).) So we compute
\[\begin{align*}
a \mathbf{v} + b \mathbf{w}
&= \frac{z_1+8z_3}{9}(1,2,1) + \frac{z_1-z_3}{9}(8,4,-1) \\
&= \left( z_1, \frac{2z_1+4z_3}{3}, z_3 \right) \\
&= \left( z_1, \frac{3z_2}{3}, z_3 \right)
\quad\text{because \(\mathbf{z}\) satisfies equation (1).} \\
&= (z_1,z_2,z_3) \; \checkmark
\end{align*}
\]
(d)
The computation that we did in (c) essentially shows that the only
possible choices for \(a\) and \(b\) are \(a = \frac{z_1+8z_3}{9}\) and \(b =
\frac{z_1-z_3}{9}\).
Alternatively, here is a direct proof, if we wanted to prove (d) without
having first done (c). We assume that
\[
\mathbf{z} = a\mathbf{v}+b\mathbf{w}
= a'\mathbf{v}+b'\mathbf{w},
\]
and we need to show that \(a=a'\) and \(b=b'\). Subtracting, we have
\[
(a-a')\mathbf{v}+(b-b')\mathbf{w} = \mathbf{z}-\mathbf{z}= \mathbf{0},
\]
so what we really need to do is show that \(\mathbf{v}\)
and \(\mathbf{w}\) are linearly independent. So we suppose that
\[
c_1\mathbf{v}+c_1\mathbf{w} = \mathbf{0},
\]
and we will prove that \(c_1=c_2=0\). Thus
\[\begin{aligned}
c_1\mathbf{v}+c_1\mathbf{w} &= \mathbf{0} \quad\text{by assumption,} \\
c_1(1,2,1)+c_1(8,4,-1) &= (0,0,0) \quad\text{these are the values
of \(\mathbf{v}\) and \(\mathbf{w}\),} \\
(c_1+8c_2,2c_1+4c_2,c_1-c_2) &= (0,0,0).\\
\end{aligned}
\]
So
\[
c_1+8c_2=0,\qquad 2c_1+4c_2=0,\qquad c_1-c_2=0.
\]
Subtracting the third equation from the first gives \(9c_2=0\), so
\(c_2=0\), and then substituting \(c_2=0\) into the first equation gives
\(c_1=0\). This completes the proof that \(\mathbf{v}\) and \(\mathbf{w}\)
are linearly independent, so there is at most one way to write any
given vector \(\mathbf{z}\) as a linear combination of \(\mathbf{v}\) and
\(\mathbf{w}\).