Enable contrast version

Tutor profile: Christopher B.

Inactive
Christopher B.
passionate about the field of mathematics
Tutor Satisfaction Guarantee

Questions

Subject: Linear Algebra

TutorMe
Question:

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

Inactive
Christopher B.
Answer:

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

TutorMe
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) $)

Inactive
Christopher B.
Answer:

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

TutorMe
Question:

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

Inactive
Christopher B.
Answer:

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} $)

Contact tutor

Send a message explaining your
needs and Christopher will reply soon.
Contact Christopher

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.