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Tutor profile: Carlos V.

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Carlos V.
Tutor for more than 10 years.
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Questions

Subject: Linear Algebra

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Question:

Suppose that $$\{ \boldsymbol{v_1},\boldsymbol{v_2} \}$$ is a basis for $$\mathbb{R}^2$$ and $$A$$ is a $$2\times 2$$ matrix such that $$A\boldsymbol{v_1}=2\boldsymbol{v_1}-\boldsymbol{v_2}\\ A\boldsymbol{v_2}=-\boldsymbol{v_1}+2\boldsymbol{v_2}$$ Find all the eigenvalues of $$A$$.

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Carlos V.
Answer:

Let's remind that $$\lambda$$ is an eigenvalue of $$A$$ if and only if there exists a nonzero vector $$\boldsymbol{v}$$ such that $$A\boldsymbol{v}=\lambda\boldsymbol{v}$$ Now back to our problem, let's notice the following $$A(\boldsymbol{v_1}+\boldsymbol{v_2})=A\boldsymbol{v_1}+A\boldsymbol{v_2}$$ $$=2\boldsymbol{v_1}-\boldsymbol{v_2}-\boldsymbol{v_1}+2\boldsymbol{v_2}$$ $$=\boldsymbol{v_1}+\boldsymbol{v_2}$$ $$=1(\boldsymbol{v_1}+\boldsymbol{v_2})$$ Hence, since we are getting that $$A(\boldsymbol{v_1}+\boldsymbol{v_2})=1(\boldsymbol{v_1}+\boldsymbol{v_2}$$ this means by definition of eigenvalue that 1 is an eigenvalue of $$A$$. Let's also notice the following $$A(\boldsymbol{v_1}-\boldsymbol{v_2})=A\boldsymbol{v_1}-A\boldsymbol{v_2}$$ $$=2\boldsymbol{v_1}-\boldsymbol{v_2}-(-\boldsymbol{v_1}+2\boldsymbol{v_2})$$ $$=2\boldsymbol{v_1}-\boldsymbol{v_2}+\boldsymbol{v_1}-2\boldsymbol{v_2}$$ $$=3\boldsymbol{v_1}-3\boldsymbol{v_2}$$ $$=3(\boldsymbol{v_1}-\boldsymbol{v_2})$$ Since $$A(\boldsymbol{v_1}-\boldsymbol{v_2})=3(\boldsymbol{v_1}-\boldsymbol{v_2})$$ this means by definition of eigenvalue that 3 is an eigenvalue of $$A$$. Finally, since $$A$$ is a $$2\times 2$$ matrix, then the maximum number of different eigenvalues it could have is 2, and since we already found two different eigenvalues, we can conclude that the only eigenvalues of $$A$$ are: $$1$$, and $$3$$.

Subject: Basic Math

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Question:

Solve the following equation for the variable $$x$$. $$8\log(5x+2)=24$$

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Carlos V.
Answer:

First, let's divide by 8 on both sides of the equation $$8\log(5x+2)=24$$, this way we get $$8\log(5x+2)=24$$ $$\frac{8\log(5x+2)}{8}=\frac{24}{8}$$ $$\log(5x+2)=3$$ Since the inverse function of $$\log x$$ is $$10^x$$, this means that $$10^{\log x}=x$$; therefore, by taking $$10^x$$ on both sides of the equation we get $$10^{\log(5x+2)}=10^3$$ $$5x+2=10^3$$ since $$10^{\log(5x+2)}=5x+2$$ $$5x+2=1000$$ since $$10^3=1000$$ Subtracting 2 on both sides we get $$5x+2-2=1000-2$$ $$5x=998$$ Dividing by 5 on both sides, $$\frac{5x}{5}=\frac{998}{5}$$ $$x=\frac{998}{5}$$ Finally, the solution of the equation $$8\log(5x+2)=24$$ is $$x=\frac{998}{5}$$

Subject: Calculus

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Question:

Evaluate the improper integral: $$\int_{0}^{\infty}xe^{x/2}dx$$

Inactive
Carlos V.
Answer:

Let's notice that one of the limits for this integral is $$\infty$$; therefore, we need to start by rewriting this integral by using limits as follows: $$\int_{0}^{\infty}xe^{-x/2}dx=\lim\limits_{t\rightarrow\infty}\int_{0}^{t}xe^{-x/2}dx$$ Now in order to solve this integral let's use integration by parts as follows $$\begin{array}{cc} u=x & du=dx\\ dv=e^{-x/2}dx & v=-2e^{-x/2} \end{array}$$ This way we get $$\int_{0}^{\infty}xe^{-x/2}dx=\lim\limits_{t\rightarrow\infty}\int_{0}^{t}xe^{-x/2}dx$$ $$=\lim\limits_{t\rightarrow\infty}\left( \left[ -2xe^{-x/2} \right]_0^t-\int_{0}^{t}-2e^{-x/2}dx \right)$$ $$=\lim\limits_{t\rightarrow\infty}\left( -2\left[ \frac{x}{e^{x/2}} \right]_0^t+2\int_{0}^{t}e^{-x/2}dx \right)$$ $$=\lim\limits_{t\rightarrow\infty}\left( -2\left( \frac{t}{e^{t/2}}-\frac{0}{e^{0/2}} \right)-4\left[ e^{-x/2} \right]_0^t \right)$$ $$=\lim\limits_{t\rightarrow\infty}\left( -2\frac{t}{e^{t/2}}-4\left[ \frac{1}{e^{x/2}} \right]_0^t \right)$$ $$=-2\frac{\infty}{e^{\infty/2}}-4\lim\limits_{t\rightarrow\infty}\left( \frac{1}{e^{t/2}}-\frac{1}{e^{0/2}} \right)$$ $$=-2(0)-4\lim\limits_{t\rightarrow\infty}\left( \frac{1}{e^{t/2}}-1 \right)$$ $$=-4\left( \frac{1}{e^{\infty/2}}-1 \right)$$ $$=-4(0-1)$$ $$4$$ Finally, since we are getting just a constant number as result, this means that the improper integral is convergent, and besides $$\int_{0}^{\infty}xe^{-x/2}dx=4$$

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