Question

1. write (mathbf{w} = \begin{bmatrix} 8 \\ 21 \\ 15 end{bmatrix}) as a linear combination of (mathbf{u} = \begin{bmatrix} 2 \\ 3 \\ 5 end{bmatrix}) and (mathbf{v} = \begin{bmatrix} 4 \\ -3 \\ 5 end{bmatrix}), or show that it is not possible to do so.

Explanation:

Step1: Define Linear Combination

We want to find scalars \( x \) and \( y \) such that \( w = x u + y v \). So,

$$
LATEXBLOCK0
= x
LATEXBLOCK1
+ y
LATEXBLOCK2
$$

This gives the system of equations:

1. \( 2x + 4y = 8 \)
2. \( 3x - 3y = 21 \)
3. \( 5x + 5y = 15 \)

Step2: Simplify Equations

• From equation 1: Divide by 2, \( x + 2y = 4 \) (Equation 1a)
• From equation 2: Divide by 3, \( x - y = 7 \) (Equation 2a)
• From equation 3: Divide by 5, \( x + y = 3 \) (Equation 3a)

Step3: Solve Equations 1a and 2a

Subtract Equation 2a from Equation 1a:
\( (x + 2y) - (x - y) = 4 - 7 \)
\( 3y = -3 \)
\( y = -1 \)

Substitute \( y = -1 \) into Equation 2a:
\( x - (-1) = 7 \)
\( x + 1 = 7 \)
\( x = 6 \)

Step4: Check with Equation 3a

Substitute \( x = 6 \) and \( y = -1 \) into Equation 3a:
\( 6 + (-1) = 5
eq 3 \)

Answer:

Since substituting \( x = 6 \) and \( y = -1 \) into the third simplified equation does not satisfy it, \( w \) cannot be written as a linear combination of \( u \) and \( v \).