Question

question 7
let ( b = left{ \begin{bmatrix} 1 \\ 1 end{bmatrix}, \begin{bmatrix} -1 \\ 1 end{bmatrix}
ight} ) be an ordered basis for ( mathbb{r}^2 ). find the coordinate vector ( c_b(vec{v}) ) of ( vec{v} = \begin{bmatrix} 2 \\ 4 end{bmatrix} ).
(a) ( \begin{bmatrix} 1 \\ 3 end{bmatrix} )
(b) ( \begin{bmatrix} 2 \\ 4 end{bmatrix} )
(c) ( \begin{bmatrix} 3 \\ 1 end{bmatrix} )
(d) ( \begin{bmatrix} 3 \\ -1 end{bmatrix} )
(e) none of the above.

Explanation:

Step1: Define the equation

Let \( \vec{v} = a

$$\begin{bmatrix}1\\1\end{bmatrix}$$

+ b

$$\begin{bmatrix}-1\\1\end{bmatrix}$$

\), where \( a \) and \( b \) are the scalars we need to find (the coordinates in basis \( B \)). So we have the system of equations from the components:
For the first component: \( a - b = 2 \)
For the second component: \( a + b = 4 \)

Step2: Solve the system

Add the two equations: \( (a - b) + (a + b) = 2 + 4 \)
Simplify: \( 2a = 6 \), so \( a = 3 \)
Substitute \( a = 3 \) into \( a - b = 2 \): \( 3 - b = 2 \), so \( b = 1 \)

Step3: Form the coordinate vector

The coordinate vector \( C_B(\vec{v}) \) is \(

$$\begin{bmatrix}a\\b\end{bmatrix}$$

=

$$\begin{bmatrix}3\\1\end{bmatrix}$$

\)

Answer:

C. \(

$$\begin{bmatrix}3\\1\end{bmatrix}$$

\)