We read your photo as:

Let G be the set of all 3 x 3 matrices in the form [[1,0,0],[a,1,0],[b,c,1]] where a, b, c are in Z+. Prove that G forms a group under matrix multiplication.

Mathematics

ScanSolve AI Answer

Answer

3θ=2(π−sin⁡θ)3\theta = 2(\pi - \sin\theta)

Step-by-step solution

5 steps

  1. Define the areas involved.

    Let rr be the radius of the semicircle.
    The angle ∠ROQ=θ\angle ROQ = \theta radians.
    Since PQ is the diameter, ∠POQ=π\angle POQ = \pi radians.
    Therefore, ∠POR=π−θ\angle POR = \pi - \theta radians.

  2. Calculate the area of sector ROQ.

    The formula for the area of a sector is 12r2α\frac{1}{2} r^2 \alpha, where α\alpha is the angle in radians.

    Area(sector ROQ)=12r2θ\text{Area(sector ROQ)} = \frac{1}{2} r^2 \theta

  3. Calculate the area of the unshaded segment PR.

    The area of segment PR is the area of sector POR minus the area of triangle POR.

    Area(sector POR)=12r2(π−θ)\text{Area(sector POR)} = \frac{1}{2} r^2 (\pi - \theta)

    The area of triangle POR is 12r2sin⁡(∠POR)\frac{1}{2} r^2 \sin(\angle POR).
    Since ∠POR=π−θ\angle POR = \pi - \theta, and sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta:

    Area(triangle POR)=12r2sin⁡(π−θ)=12r2sin⁡θ\text{Area(triangle POR)} = \frac{1}{2} r^2 \sin(\pi - \theta) = \frac{1}{2} r^2 \sin\theta

    So, the area of segment PR is:

    Area(segment PR)=12r2(π−θ)−12r2sin⁡θ=12r2(π−θ−sin⁡θ)\text{Area(segment PR)} = \frac{1}{2} r^2 (\pi - \theta) - \frac{1}{2} r^2 \sin\theta = \frac{1}{2} r^2 (\pi - \theta - \sin\theta)

  4. Use the given condition to form an equation.

    The problem states that the area of sector ROQ is equal to twice the area of the unshaded segment PR.

    Area(sector ROQ)=2×Area(segment PR)\text{Area(sector ROQ)} = 2 \times \text{Area(segment PR)}

    Substitute the expressions from Step 2 and Step 3:

    12r2θ=2[12r2(π−θ−sin⁡θ)]\frac{1}{2} r^2 \theta = 2 \left[ \frac{1}{2} r^2 (\pi - \theta - \sin\theta) \right]

  5. Simplify the equation.

    12r2θ=r2(π−θ−sin⁡θ)\frac{1}{2} r^2 \theta = r^2 (\pi - \theta - \sin\theta)

    Divide both sides by r2r^2 (assuming r≠0r \neq 0):

    12θ=π−θ−sin⁡θ\frac{1}{2} \theta = \pi - \theta - \sin\theta

    Multiply the entire equation by 2:

    θ=2(π−θ−sin⁡θ)\theta = 2(\pi - \theta - \sin\theta)

    Distribute the 2 on the right side:

    θ=2π−2θ−2sin⁡θ\theta = 2\pi - 2\theta - 2\sin\theta

    Add 2θ2\theta to both sides:

    θ+2θ=2π−2sin⁡θ\theta + 2\theta = 2\pi - 2\sin\theta 3θ=2π−2sin⁡θ3\theta = 2\pi - 2\sin\theta

    Factor out 2 from the right side:

    3θ=2(π−sin⁡θ)3\theta = 2(\pi - \sin\theta)
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QGiven matrices G = [[6, 0], [−4, 2]] and H = [[9, 0], [−6, 3]], prove or disprove that 5G − 2H = 0.QProve that the determinant of the matrix [[b+c, a, a], [b, c+a, b], [c, c, a+b]] is 2abc.QIf multiplication of complex numbers is defined as binary operation in set G = z ∈ C : |z|= 1 then…QLet z be the centre of a group G, then the set of all inner automorphisms on the group G to the…QIf x = b c, y = b a and z = _b a. Prove that xyz = xy + yz + zx
Browse by topic:Linear Algebra
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.