Consider the following bases of \(\mathbf{R}^{2}\) :
$$
S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,-2),(3,-4)\\} \quad \text { and } \quad
S^{\prime}=\left\\{v_{1}, v_{2}\right\\}=\\{(1,3),(3,8)\\}
$$
(a) Find the coordinates of \(v=(a, b)\) relative to the basis \(S\).
(b) Find the change-of-basis matrix \(P\) from \(S\) to \(S^{\prime}\).
(c) Find the coordinates of \(v=(a, b)\) relative to the basis \(S^{\prime}\).
(d) Find the change-of-basis matrix \(Q\) from \(S^{\prime}\) back to \(S\).
(e) Verify \(Q=P^{-1}\).
(f) Show that, for any vector \(v=(a, b)\) in \(\mathbf{R}^{2},
P^{-1}[v]_{S}=[v]_{S^{\prime}}\). (See Theorem 6.6.)
(a) Let \(v=x u_{1}+y u_{2}\) for unknowns \(x\) and \(y\); that is,
$$
\left[\begin{array}{l}
a \\
b
\end{array}\right]=x\left[\begin{array}{r}
1 \\
-2
\end{array}\right]+y\left[\begin{array}{r}
3 \\
-4
\end{array}\right] \quad \text { or } \begin{aligned}
x+3 y=a \\
-2 x-4 y=b
\end{aligned} \quad \text { or } \quad \begin{aligned}
x+3 y &=a \\
2 y &=2 a+b
\end{aligned}
$$
Solve for \(x\) and \(y\) in terms of \(a\) and \(b\) to get \(x=-2 a-\frac{3}{2} b\)
and \(y=a+\frac{1}{2} b\). Thus,
$$
(a, b)=\left(-2 a-\frac{3}{2}\right) u_{1}+\left(a+\frac{1}{2} b\right) u_{2}
\quad \text { or } \quad[(a, b)]_{S}=\left[-2 a-\frac{3}{2} b, a+\frac{1}{2}
b\right]^{T}
$$
(b) Use part (a) to write each of the basis vectors \(v_{1}\) and \(v_{2}\) of
\(S^{\prime}\) as a linear combination of the basis vectors \(u_{1}\) and \(u_{2}\)
of \(S\); that is,
$$
\begin{aligned}
&v_{1}=(1,3)=\left(-2-\frac{9}{2}\right) u_{1}+\left(1+\frac{3}{2}\right)
u_{2}=-\frac{13}{2} u_{1}+\frac{5}{2} u_{2} \\
&v_{2}=(3,8)=(-6-12) u_{1}+(3+4) u_{2}=-18 u_{1}+7 u_{2}
\end{aligned}
$$
Then \(P\) is the matrix whose columns are the coordinates of \(v_{1}\) and
\(v_{2}\) relative to the basis \(S ;\) that is,
$$
P=\left[\begin{array}{rr}
-\frac{13}{2} & -18 \\
\frac{5}{2} & 7
\end{array}\right]
$$
(c) Let \(v=x v_{1}+y v_{2}\) for unknown scalars \(x\) and \(y\) :
$$
\left[\begin{array}{l}
a \\
b
\end{array}\right]=x\left[\begin{array}{l}
1 \\
3
\end{array}\right]+y\left[\begin{array}{l}
3 \\
8
\end{array}\right] \quad \text { or } \quad \begin{array}{r}
x+3 y=a \\
3 x+8 y=b
\end{array} \quad \text { or } \quad \begin{array}{r}
x+3 y=a \\
-y=b-3 a
\end{array}
$$
Solve for \(x\) and \(y\) to get \(x=-8 a+3 b\) and \(y=3 a-b\). Thus,
$$
(a, b)=(-8 a+3 b) v_{1}+(3 a-b) v_{2} \quad \text { or } \quad[(a,
b)]_{S^{\prime}}=[-8 a+3 b, \quad 3 a-b]^{T}
$$
(d) Use part ( \(c)\) to express each of the basis vectors \(u_{1}\) and \(u_{2}\)
of \(S\) as a linear combination of the basis vectors \(v_{1}\) and \(v_{2}\) of
\(S^{\prime}\)
$$
\begin{aligned}
&u_{1}=(1,-2)=(-8-6) v_{1}+(3+2) v_{2}=-14 v_{1}+5 v_{2} \\
&u_{2}=(3,-4)=(-24-12) v_{1}+(9+4) v_{2}=-36 v_{1}+13 v_{2}
\end{aligned}
$$
Write the coordinates of \(u_{1}\) and \(u_{2}\) relative to \(S^{\prime}\) as
columns to obtain \(Q=\left[\begin{array}{rr}-14 & -36 \\ 5 &
13\end{array}\right]\).
(e) \(Q P=\left[\begin{array}{rr}-14 & -36 \\ 5 &
13\end{array}\right]\left[\begin{array}{rr}-\frac{13}{2} & -18 \\ \frac{5}{2}
& 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 &
1\end{array}\right]=I\)
(f) Use parts \((\mathrm{a}),(\mathrm{c})\), and (d) to obtain
$$
P^{-1}[v]_{S}=Q[v]_{S}=\left[\begin{array}{rr}
-14 & -36 \\
5 & 13
\end{array}\right]\left[\begin{array}{r}
-2 a-\frac{3}{2} b \\
a+\frac{1}{2} b
\end{array}\right]=\left[\begin{array}{c}
-8 a+3 b \\
3 a-b
\end{array}\right]=[v]_{S}
$$
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