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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.
Answer:

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.
Answer:

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.
Answer:

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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