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# Tutor profile: Luke B.

Luke B.
PhD student, former Teaching Assistant

## Questions

### Subject:Linear Algebra

TutorMe
Question:

Use the Cayley-Hamilton Theorem to calculate the matrix $$A^{16}$$ where $$A$$ is the 2x2 matrix $$A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}.$$

Luke B.

First calculate the characteristic polynomial of $$A$$: $$\begin{vmatrix} x-2 & -1 \\ 0 & x-3 \end{vmatrix} = \left(x-2\right) \, \left(x-3\right) .$$ By the Cayley-Hamilton Theorem $$\left(A-2I\right) \, \left(A-3I\right)=O$$. This means that $$A^{16}$$ can be calculated by finding an affine function $$bx + c$$ such that $$x^{16} - bx - c$$ has $$(x-2)(x-3)$$ as a factor. In that case, $$A^{16} = bA + cI$$. Consider $$x^{16}-bx-c$$ at $$x=2$$ and at $$x=3$$, where any product of $$(x-2)(x-3)$$ vanishes, so that it suffices to consider $$x^{16}=bx + c$$ at those points. This gives the pair of linear equations $$2^{16} =2b+c$$ and $$3^{16}=3b+c=0$$. Subtracting the first equation from the second gives $$b=3^{16}-2^{16}$$ and solving for $$c$$ gives $$c=3\cdot 2^{16} - 2\cdot 3^{16}$$. Therefore, $$A^{16} = \left(3^{16}-2^{16}\right) \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} - \left(2 \cdot 3^{16} - 3\cdot 2^{16}\right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2^{16} & 3^{16}-2^{16} \\ 0 & 3^{16} \end{bmatrix}.$$

### Subject:Calculus

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Question:

Evaluate the definite integral $$\int_{x=1}^3 \frac{\left(x+1\right)\,dx}{x^2 + 2x+2}.$$

Luke B.

First use the substitution $$t=x+1$$, and apply this to the limits (so $x=1$ becomes $t=2$, for instance) and rewrite the integrand: $$\int_{t=2}^4 \frac{t \, dt}{t^2 + 1}.$$ Next use the substitution $$u=t^2 +1$$ (so that $$du=2t\,dt$$): $$\frac{1}{2} \int_{u=5}^{17} \frac{du}{u}.$$ Now use the fact that $$\ln u$$ is an antiderivative of $$du/u$$ and evaluate the definite integral: $$\frac{1}{2} \int_{u=5}^{17} \frac{du}{u} = \left. \ln u \right|^{17}_5 = \ln {17} - \ln {5} .$$

### Subject:Differential Equations

TutorMe
Question:

Find the general solution $$y = f(x)$$ to the differential equation $$dy + y^3 \, dx = 0$$.

Luke B.

It looks like a good candidate for separation of variables, so first divide it by $$y^3$$ to obtain $$y^{-3} \, dy + dx = 0.$$ Next integrate the equation (remembering the integration constant $$C$$) to obtain $$\frac{y^{-2}}{-2} + x = C.$$ Finally, carry out some algebraic manipulations to put it into the desired form. $$y^{-2} -2x = -2C$$ $$y^{-2} = 2 \left(x - C\right)$$ $$y^2 = \frac{1}{2 \left( x - C\right)}$$ Square roots generally need to be treated carefully. In this case, either the square root or its negative is a solution. (This can be seen by considering $$\left(-y\right)^{-3} \, d(-y) = y^{-3} \, dy$$.) So the general solution to the differential equation can be written as $$y(x) = \frac{\pm 1}{2 \left(x - C\right)},$$ where a specific sign would be chosen later when a specific initial value problem needed to be solved.

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