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# Tutor profile: Christopher B.

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Christopher B.
passionate about the field of mathematics
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## Questions

### Subject:Linear Algebra

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Question:

Does the given $$2\times2$$ matrix have an inverse? $( \begin{bmatrix} 3 &6 \\ 2 &4 \end{bmatrix}$)

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Christopher B.

Recall that a matrix has an inverse if and only if it is nonsingular, i.e., the determinant of the matrix is not equal to zero. To determine if this matrix has an inverse, we can find the determinant. If the determinant is zero, the matrix does not have an inverse, otherwise, it does. The determinant of a $$2\times 2$$ of the form \begin{bmatrix} a &b \\ c&d \end{bmatrix} is given by $$(a)(d)-(c)(b)$$. Thus, the determinant of the given matrix is ( \begin{aligned} &(3)(4) - (2)(6) \\ \\ &= 12 - 12 \\ \\ &= 0 \end{aligned}) Since the determinant of the given matrix is zero, it is singular, and thus does not have an inverse.

### Subject:Trigonometry

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Question:

Use one of the sum or difference identities for sine and cosine to find the exact value of $( \csc \left(\frac{3\pi}{2} \right)$)

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Christopher B.

One way to solve this problem is to rewrite the cosecant function in terms of sine and use the sine of sum identity. As \begin{aligned} \csc x = 1/ \sin x \end{aligned}, The given function becomes $( \frac{1}{\sin \left( \frac{3\pi}{2}\right)}$) Rewriting \begin{aligned} \frac{3\pi} {2} \text{ as } \frac{2\pi}{2} + \frac{\pi}{2} \end{aligned}, we get ( \begin{aligned} \frac{1}{\sin \left( \frac{2\pi}{2} + \frac{\pi}{2}\right)}\end{aligned}) Recalling the sine of sum identity, \begin{aligned} \sin \left( a+b \right) = \sin (a) \cos (b) + \cos (a) \sin (b) \end{aligned}, we have ( \begin{aligned} & \frac{1}{\sin \left(\frac{2\pi}{2}\right) \cos \left(\frac{\pi}{2}\right) + \cos \left(\frac{2\pi}{2}\right) \sin \left(\frac{\pi}{2}\right)} \\ \\ = & \, \, \frac{1}{\sin \left(\pi\right) \cos \left(\frac{\pi}{2}\right) + \cos \left(\pi\right) \sin \left(\frac{\pi}{2}\right)} \\ \\ = & \, \, \frac{1}{\left(0\right) \left(0\right) + \left(-1\right) \left(1\right)} \\ \\ =& \, \, \frac{1}{0-1} \\ \\ =& \, \, \frac{1}{-1} \\ \\ =&\, \, \boxed{ -1} \\ \\ \end{aligned})

### Subject:Calculus

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Question:

Evaluate the integral $( \int_0^e \ln x^2 \, dx$)

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Christopher B.

Using the power rule for logarithms, we can re-write $$\ln x^2$$ as $$2 \ln x$$. Now the given integral becomes $(\int _0^e 2 \ln x\, dx$) Since $$2$$ is a constant, we can pull it out of the integral. This gives us $( 2\int_0^e \ln x \, dx$) We can now solve the integral using integration by parts. Recalling that the formula for integration by parts is \begin{aligned} \int_a^b u\, dv = \left. uv \right \vert_a^b - \int_a^b v\, du \end{aligned}, we can let $$u= \ln x$$ and $$dv = dx$$. This gives $$du = \frac{1}{x} dx$$ and $$v = x$$. Thus, ( \begin{aligned} &2\int_0^e \ln x \, dx \\ \\ &= 2 \left[ \left. x\ln x \right \vert_0^e - \int_0^e \frac{x}{x}\, dx \right] \\ \\ &= 2 \left[ \left. x\ln x \right \vert_0^e - \int_0^e dx \right] \\ \\ & = 2 \left[ \left. x\ln x \right \vert_0^e - \left. x \right \vert_0^e \right] \\ \\ & = 2 \left[ \left(e \ln e - 0\right) - \left( e-0 \right) \right]\\ \\ & = 2 \left(e \ln e - e \right) \\ \\ & = 2 \left(e - e \right) && \text{as } \ln e = 1 \\ \\ & = 2(0) \\ \\ & = \boxed{0} \end{aligned})

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