Example Question - linear combination
Here are examples of questions we've helped users solve.
Calculation of a Linear Combination
<p>The given expression is a linear combination of numbers:</p> <p>To solve the expression, perform the operations as indicated:</p> <p>\( 3(0.6) + 3(7) - 6 \left( \frac{2}{3} \right) \)</p> <p>First, multiply each term inside the parentheses:</p> <p>\( 3 \times 0.6 = 1.8 \)</p> <p>\( 3 \times 7 = 21 \)</p> <p>\( 6 \times \frac{2}{3} = 4 \)</p> <p>Combine the results of the multiplications:</p> <p>\( 1.8 + 21 - 4 \)</p> <p>Now, add and subtract the numbers:</p> <p>\( 1.8 + 21 = 22.8 \)</p> <p>\( 22.8 - 4 = 18.8 \)</p> <p>So the solution to the expression is:</p> <p>\( 18.8 \)</p>
Generating the yz-Plane with Vectors
The question is asking to show that the yz-plane, given by the set W = {(0, b, c) | b, c ∈ ℝ}, can be generated by the indicated sets of vectors. The yz-plane in ℝ³ is the set of all points where the x-coordinate is 0. The general point in this plane can be written as (0, y, z), where y and z can take any real values. (i) To show that W is generated by the vectors (0, 1, 1) and (0, 2, -1), we need to express any vector (0, b, c) in W as a linear combination of (0, 1, 1) and (0, 2, -1). Let's try to express any point (0, b, c) as a linear combination of (0, 1, 1) and (0, 2, -1): (0, b, c) = α(0, 1, 1) + β(0, 2, -1) Expanding this, we have: (0, b, c) = (0, α + 2β, α - β) We want to solve for α and β such that the second and third components of the vectors match. This gives us two equations: α + 2β = b α - β = c We can solve these equations simultaneously to find α and β in terms of b and c. Adding the two equations, we get: 2α + β = b + c Subtracting the second equation from the first one, we get: 3β = b - c Thus, β = (b - c) / 3. Substitute β back into one of the original equations to get α. For example, using the first equation: α = b - 2β α = b - 2(b - c) / 3 α = (3b - 2b + 2c) / 3 α = (b + 2c) / 3 So any vector (0, b, c) can be represented as a linear combination of (0, 1, 1) and (0, 2, -1), with the coefficients α = (b + 2c) / 3 and β = (b - c) / 3. (ii) Similarly, to show that W is generated by the vectors (0, 1, 2), (0, 2, 3) and (0, 3, 1), we need to express any vector (0, b, c) in W as a linear combination of these vectors. (0, b, c) = α(0, 1, 2) + β(0, 2, 3) + γ(0, 3, 1) Expanding this, we have: (0, b, c) = (0, α + 2β + 3γ, 2α + 3β + γ) Again, set up a system of equations to solve for α, β, and γ: α + 2β + 3γ = b 2α + 3β + γ = c This system has two equations with three unknowns, making it underdetermined. However, since the third component isn't given (as there is no x-component in the vectors generating W), we can freely choose γ (for example, γ = 0), and the other two can be determined from the equations above. This way, every vector in the yz-plane can be expressed as a combination of these three vectors. For any given b and c, we can always find α and β (and choose γ) such that (0, b, c) is a linear combination of (0, 1, 2), (0, 2, 3), and (0, 3, 1), confirming that these vectors generate the yz-plane.
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