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# Tutor profile: Maheshsinh S.

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Maheshsinh S.
Mathematics tutor with 6 Years of experience
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## Questions

### Subject:Linear Algebra

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Question:

Find the eigenvalue and eigenvector of the matrix $$A=\begin{bmatrix} 5 &2 \\ -8 &-3 \end{bmatrix}$$.

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Maheshsinh S.

We have $$A=\begin{bmatrix} 5 &2 \\ -8 &-3 \end{bmatrix}$$ Now (i) Subtract $$\lambda$$ from the diagonal entries of the given square matrix. That is $$A-\lambda=\begin{bmatrix} 5-\lambda & 2 \\ -8 & -3-\lambda \end{bmatrix}$$. (ii) Find out the determinant of the matrix generated in step(i). That is $$|A-\lambda|=\begin{bmatrix} 5-\lambda & 2 \\ -8 & -3-\lambda \end{bmatrix}$$ By expanding with respect to the last column we get: $$=(5-\lambda)(-3-\lambda)+16\\ =1-2\lambda+{\lambda}^2\\$$ (iii) Set the determinant equal to zero. That is $$|A-\lambda|=0 \implies 1-2\lambda+{\lambda}^2=0$$ Solve the equation generated from step (iii). $$1-2\lambda+{\lambda}^2=0\\ \implies (\lambda-1)(\lambda-1)=0\\ \implies \lambda=1 \, or \, 1$$ The eigenvalues will be $$\lambda_1=1,\lambda_2=1$$. (iv) Using the $$\lambda=1$$ we get $$\begin{bmatrix} 4 &2 \\ -8 &-4 \end{bmatrix}$$ Using RREF we get $$\begin{bmatrix} 1 & \frac{1}{2} \\ 0 &0 \end{bmatrix}$$ Now solve the matrix equation $$\begin{bmatrix} 1 & \frac{1}{2} \\ 0 &0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$$ Thus $$v_2=t;v_1=\frac{t}{2}$$ Therefore eigen vector is $$\begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix}$$.

### Subject:Applied Mathematics

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Question:

In 1986, the Chernobyl nuclear disaster in the Soviet Union contaminated the atmosphere. The buildup of radioactive material in the atmosphere satisfies the differential equation $$\frac{dM}{dt} = r((\frac{k}{r})-M)$$, where $$M$$ is the mass of radioactive material in the atmosphere after time $$t$$ (in years); $$k$$ is the rate at which the radioactive material is introduced into the atmosphere; $$r$$ is the annual decay rate of the radioactive material. Find the solution of this differential equation in terms of $$k \enspace and \enspace r$$.

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Maheshsinh S.

We know that the buildup of radioactive material in the atmosphere satisfies the differential equation $$\frac{dM}{dt} = r((\frac{k}{r})-M)$$. where $$M$$ is the mass of radioactive material in the atmosphere after time $$t$$ (in years); $$k$$ is the rate at which the radioactive material is introduced into the atmosphere; $$r$$ is the annual decay rate of the radioactive material. Now using the separable variable we get $$\frac{dM}{((\frac{k}{r})-M)} = r dt$$ Integrating both the sides we get $$\implies \int \frac{dM}{((\frac{k}{r})-M)} =\int r dt$$ $$\implies -\int \frac{dM}{(M-(\frac{k}{r}))} =r \int 1 dt$$ $$\implies -\ln {(M-(\frac{k}{r}))} =r t+C$$ $$\implies \ln {(M-(\frac{k}{r}))} =-r t+C$$ $$\implies (M-(\frac{k}{r})) =e^{-r t+C}$$ $$\implies M(t)=(\frac{k}{r})) + e^{-r t+C}$$ is our required solution of the differential equation.

### Subject:Basic Math

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Question:

Find the slope, if it exists, of the line containing the pair of points. $$(9, 2)$$ and $$(10, -1)$$.

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Maheshsinh S.

We are given two points $$(a,b)=(9,2)$$ and $$(c,d)=(10,-1)$$. The $$y-$$ difference of the points is $$b-d=2-(-1)=3$$ and $$x-$$ difference of the points is $$a-c=9-10=-1$$. Thus the slope of the line passing through two points (9,2) and (10,-1) is $$\frac{y-difference}{x-difference}=\frac{3}{-1}=-3$$.

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