Find a basis for the set of all solutions to the difference equation.
-Let $B = \left\{ \mathbf { b } _ { 1 } , \mathbf { b } _ { 2 } \right\}$ and $C = \left\{ \mathbf { c } _ { 1 } , \mathbf { c } _ { 2 } \right\}$ be bases for $\mathcal { R } ^ { 2 }$, where
$$\mathbf { b } _ { 1 } = \left[ \begin{array} { l }
2 \
4
\end{array} \right] , \mathbf { b } _ { 2 } = \left[ \begin{array} { l }
1 \
3
\end{array} \right] , \mathbf { c } _ { 1 } = \left[ \begin{array} { l }
1 \
3
\end{array} \right] , \mathbf { c } _ { 2 } = \left[ \begin{array} { r }
- 4 \
- 10
\end{array} \right]$$
Find the change-of-coordinates matrix from $B$ to $C$.
A)
$\left[ \begin{array} { r r } 2 & 1 \ - 1 & 0 \end{array} \right]$

B)
$\left[ \begin{array} { l l } 0 & - 1 \ 1 & - 2 \end{array} \right]$

C)
$\left[ \begin{array} { r r } 1 & - 4 \ 3 & - 10 \end{array} \right]$

D)

$\left[ \begin{array} { l l } - 2 & 1 \ - 1 & 0 \end{array} \right]$

Question

Find a basis for the set of all solutions to the difference equation.
-Let $B = \left\{ \mathbf { b } _ { 1 } , \mathbf { b } _ { 2 } \right\}$ and $C = \left\{ \mathbf { c } _ { 1 } , \mathbf { c } _ { 2 } \right\}$ be bases for $\mathcal { R } ^ { 2 }$, where
$$\mathbf { b } _ { 1 } = \left[ \begin{array} { l } 
2 \
4
\end{array} \right] , \mathbf { b } _ { 2 } = \left[ \begin{array} { l } 
1 \
3
\end{array} \right] , \mathbf { c } _ { 1 } = \left[ \begin{array} { l } 
1 \
3
\end{array} \right] , \mathbf { c } _ { 2 } = \left[ \begin{array} { r } 
- 4 \
- 10
\end{array} \right]$$
Find the change-of-coordinates matrix from $B$ to $C$.
A)
$\left[ \begin{array} { r r } 2 & 1 \ - 1 & 0 \end{array} \right]$

B)
$\left[ \begin{array} { l l } 0 & - 1 \ 1 & - 2 \end{array} \right]$

C)
$\left[ \begin{array} { r r } 1 & - 4 \ 3 & - 10 \end{array} \right]$

D)

$\left[ \begin{array} { l l } - 2 & 1 \ - 1 & 0 \end{array} \right]$

Quizplus · Accepted Answer

The change-of-coordinates matrix from $B$ to $C$ is found by expressing each vector of $C$ as a linear combination of vectors in $B$. For $\mathbf{c}_1$, we can see that it is the same as $\mathbf{b}_2$, so it can be represented as $0\mathbf{b}_1 + 1\mathbf{b}_2$. For $\mathbf{c}_2$, we need to solve the equation $a\mathbf{b}_1 + b\mathbf{b}_2 = \mathbf{c}_2$, which leads to $a[2, 4]^T + b[1, 3]^T = [-4, -10]^T$. Solving this system of equations gives $a = -1$ and $b = -2$. Thus, the change-of-coordinates matrix is $\left[ \begin{array} { l l } 0 & - 1 \ 1 & - 2 \end{array} \right]$.

Find a Basis for the Set of All Solutions to the Difference