Linear Combination: Can You Express Vectors?

Hey guys! Today, we're diving deep into the fascinating world of linear algebra with a cool exercise that'll really get your brains buzzing. We're talking about linear combinations, a fundamental concept that pops up everywhere from computer graphics to quantum mechanics. Think of it like this: imagine you have a set of building blocks (our vectors a₁, a₂, and a₃), and you want to see if you can build a specific structure (our vector y) using only those blocks. The trick is, you can use each block as many times as you want, but you can also use them in reverse (negative multiples!). Our challenge is to figure out if we can perfectly construct vector y by adding up some multiples of a₁, a₂, and a₃. If we can, great! We express y as a linear combination. If not, well, we gotta show why it's impossible. Let's get started with our specific vectors!

Our Building Blocks and Target Structures

So, we've got our main building blocks, which are vectors in four-dimensional space (yep, four dimensions!):

• a₁ = [-1, 3, 0, 1]
• a₂ = [3, 1, 2, 0]
• a₃ = [1, 1, 1, 1]

And our target structures, the vectors we want to build, are:

• (a) y = [1, 2, 4, 0]
• (b) y = [-1, 9, 2, 6]

Our mission, should we choose to accept it (and we totally should!), is to determine for each 'y' whether it can be written as c₁a₁ + c₂a₂ + c₃a₃ for some scalar numbers c₁, c₂, and c₃. If it can, we need to find those magical numbers. If not, we need to prove it's a no-go.

This is super important because understanding linear combinations helps us grasp concepts like vector spaces, spanning sets, and independence. It's the backbone of so much cool math!

Let's tackle part (a) first. We want to see if [1, 2, 4, 0] can be expressed as c₁[-1, 3, 0, 1] + c₂[3, 1, 2, 0] + c₃[1, 1, 1, 1].

This translates into a system of linear equations. For each component of the vectors, we get an equation:

• 1st component: -1c₁ + 3c₂ + 1c₃ = 1
• 2nd component: 3c₁ + 1c₂ + 1c₃ = 2
• 3rd component: 0c₁ + 2c₂ + 1c₃ = 4
• 4th component: 1c₁ + 0c₂ + 1c₃ = 0

So we have this system:

1. -c₁ + 3c₂ + c₃ = 1
2. 3c₁ + c₂ + c₃ = 2
3. 2c₂ + c₃ = 4
4. c₁ + c₃ = 0

This is where the real work begins, guys. We need to solve this system. The best way to do this is often by using matrices and Gaussian elimination, or just good old-fashioned substitution and elimination. Let's try substitution first, using equation (4) which looks pretty simple.

From equation (4), c₁ + c₃ = 0, we can easily get c₁ = -c₃. This is a huge help!

Now, let's plug this c₁ = -c₃ into equation (3): 2c₂ + c₃ = 4. This equation doesn't involve c₁ directly, so it remains as it is.

Let's substitute c₁ = -c₃ into equation (1) and (2):

• Equation (1) becomes: -(-c₃) + 3c₂ + c₃ = 1 which simplifies to c₃ + 3c₂ + c₃ = 1, leading to 3c₂ + 2c₃ = 1.
• Equation (2) becomes: 3(-c₃) + c₂ + c₃ = 2 which simplifies to -3c₃ + c₂ + c₃ = 2, leading to c₂ - 2c₃ = 2.

Now, we have a smaller, simpler system with just c₂ and c₃:

• 3c₂ + 2c₃ = 1
• c₂ - 2c₃ = 2

This is much easier to solve! Let's add these two equations together. The 2c₃ and -2c₃ terms will cancel out:

(3c₂ + 2c₃) + (c₂ - 2c₃) = 1 + 2 4c₂ = 3 c₂ = 3/4

Awesome! We found c₂. Now we can substitute this value back into one of the c₂ and c₃ equations to find c₃. Let's use c₂ - 2c₃ = 2:

(3/4) - 2c₃ = 2 -2c₃ = 2 - 3/4 -2c₃ = 8/4 - 3/4 -2c₃ = 5/4 c₃ = (5/4) / (-2) c₃ = -5/8

We've got c₂ and c₃! Finally, we can find c₁ using c₁ = -c₃:

c₁ = -(-5/8) c₁ = 5/8

So, for part (a), we found the coefficients: c₁ = 5/8, c₂ = 3/4, and c₃ = -5/8. This means that yes, y = [1, 2, 4, 0] is a linear combination of a₁, a₂, and a₃, and we can express it as:

[1, 2, 4, 0] = (5/8)a₁ + (3/4)a₂ + (-5/8)a₃

Let's quickly check this to be absolutely sure:

(5/8)[-1, 3, 0, 1] + (3/4)[3, 1, 2, 0] + (-5/8)[1, 1, 1, 1] = [-5/8, 15/8, 0, 5/8] + [9/4, 3/4, 6/4, 0] + [-5/8, -5/8, -5/8, -5/8] = [-5/8, 15/8, 0, 5/8] + [18/8, 6/8, 12/8, 0] + [-5/8, -5/8, -5/8, -5/8]

Now, add the components:

• 1st: -5/8 + 18/8 - 5/8 = (18 - 10)/8 = 8/8 = 1 (Matches!)
• 2nd: 15/8 + 6/8 - 5/8 = (15 + 6 - 5)/8 = 16/8 = 2 (Matches!)
• 3rd: 0 + 12/8 - 5/8 = 7/8 ... Wait a minute! Something is wrong here. Let's recheck the math. It seems I made a mistake in the calculation or in the system setup. Let's re-evaluate the system of equations and solve it more rigorously, perhaps using matrices.

Okay, deep breaths! Let's revisit the system of equations for part (a):

1. -c₁ + 3c₂ + c₃ = 1
2. 3c₁ + c₂ + c₃ = 2
3. 2c₂ + c₃ = 4
4. c₁ + c₃ = 0

From (4), c₁ = -c₃. From (3), c₃ = 4 - 2c₂. Substitute c₃ into (4): c₁ + (4 - 2c₂) = 0 => c₁ = 2c₂ - 4.

Now we have c₁ and c₃ in terms of c₂. Let's substitute these into equation (1):

-(2c₂ - 4) + 3c₂ + (4 - 2c₂) = 1 -2c₂ + 4 + 3c₂ + 4 - 2c₂ = 1 (-2 + 3 - 2)c₂ + (4 + 4) = 1 -c₂ + 8 = 1 -c₂ = 1 - 8 -c₂ = -7 c₂ = 7

Now that we have c₂ = 7, we can find c₃ and c₁:

c₃ = 4 - 2c₂ = 4 - 2(7) = 4 - 14 = -10 c₁ = 2c₂ - 4 = 2(7) - 4 = 14 - 4 = 10

So the proposed coefficients are c₁ = 10, c₂ = 7, and c₃ = -10.

Let's check these in equation (2) to see if they hold true: 3c₁ + c₂ + c₃ = 3(10) + 7 + (-10) = 30 + 7 - 10 = 27. But equation (2) requires this to be equal to 2. Since 27 ≠ 2, the system of equations is inconsistent. This means there are no values of c₁, c₂, and c₃ that satisfy all four equations simultaneously.

Therefore, for part (a), y = [1, 2, 4, 0] is not a linear combination of a₁, a₂, and a₃.

Why did this happen? It means that the vector y lies outside the