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# Tutor profile: Seth C.

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Seth C.
Math Tutor for 2.5 years
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## Questions

### Subject:Linear Algebra

TutorMe
Question:

How do I find the inverse matrix of $$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 3 & 1 \\ 0 & 1 & 3 \end{bmatrix}$$?

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Seth C.

The way I like to find the inverse of a matrix is to set your given matrix equal to the identity matrix, then solve the left side for the identity matrix. That may sound a bit confusing, so let's go through it. $$\begin{bmatrix} 2 & 2 & 2 & : & 1 & 0 & 0 \\ 2 & 3 & 1 & : & 0 & 1 & 0 \\ 0 & 1 & 3 & : & 0 & 0 & 1 \end{bmatrix}$$. Now we have this augmented matrix. Let us row reduce the left side until we get the identity matrix on the left. This gives us $$\begin{bmatrix} 1 & 0 & 0 & : \frac{3}{2} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 & : -1 & \frac{1}{4} & \frac{1}{4}\\ 0 & 0 & 1 & : 0 & -\frac{1}{4} & \frac{1}{4}\end{bmatrix}$$

### Subject:Calculus

TutorMe
Question:

How do I determine if a critical point is a maximum or a minimum?

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Seth C.

There are several different ways to determine this. My personal favorite is to check concavity. We do this by finding the double derivative of the function in question. We then plug in a critical point into that function. If the resulting value is negative, then we have a maximum, because it is concave down. If the result is positive, then we have a minimum.

### Subject:Differential Equations

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Question:

How do you solve the differential equation $$\frac{dy}{dx} = 5xy$$?

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Seth C.

We can solve this differential equation using a method called separation of variables. As the name implies, we first want to separate the variables, so everything with an $$x$$ is on one side, and everything with a $$y$$ is on the other side. This would give us $$\frac{1}{y}dy=5xdx$$. Now we can integrate both sides, which leads us to $$\int \frac{1}{y}dy=\int 5xdx$$. Evaluating this integral gets us $$\ln(y)+c_1=\frac{5}{2}x^2+c_2$$. Be sure to remember the constants of integration - they are important. The next step is to solve for $$y$$. We will define $$c_3$$ as $$c_2-c_1$$ to make things easier. $$y = e^{\frac{5}{2}x^2+c_3}$$. This can be simplified to $$y = c_4e^{\frac{5}{2}x^2}$$, where $$c_4=e^{c_3}.$$

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