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Tutor profile: Enjui K.

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Enjui K.
Tutor for 3 years
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Questions

Subject: Linear Algebra

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Question:

Given any constants $$a,b,c$$ where $$a≠0$$, find all values of $$x$$ such that the matrix $$A$$ is invertible if $$ A= \begin{bmatrix} 1 & 0 & c \\ 0 & a & -b \\ -1/a & x & x^{2} \end{bmatrix} $$ .

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Enjui K.
Answer:

We know the $$A$$ is invertible if and only if $$\det(A)\neq 0$$. $$\begin{align*} \det(A) &= 1 \begin{vmatrix} a & -b \\ x & x^{2} \end{vmatrix} +c \begin{vmatrix} 0 & a \\ -1/a & x \end{vmatrix} = 1(ax^{2}+bx) +c(0+1) \\ &= ax^{2}+bx+c . \end{align*}$$ We can solve $$x$$. Therefore, $$A$$ is invertible so long as $$x$$ satisfy both of them: $$ x\neq \dfrac{-b+\sqrt{b^{2}-4ac}}{2} ,\quad x\neq \dfrac{-b-\sqrt{b^{2}-4ac}}{2} . $$

Subject: Pre-Calculus

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Question:

Find a set of polar coordinates for the point with rectangular coordinates $$(5,5\sqrt{3})$$.

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Enjui K.
Answer:

We use this transformation: $$ r={\sqrt {y^{2}+x^{2}}} , \theta =arctan^{-1}\frac{y}{x}.$$ We need to be careful about $$\theta$$. So the good way is to plot the point and know the region $$\theta$$ to be. So $$r=10, \theta=arctan^{-1} \sqrt{3}=\frac{\pi}{3}$$

Subject: Calculus

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Question:

Consider the function $$f:\mathbb{R}^2 \to \mathbb{R}$$ defined as $$f(x,y)=2x^2 +3y^2 +2xy+2x+3y+2.$$ Find the minimizer(s) of the function if any. Explain your work

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Enjui K.
Answer:

$$ \frac{\partial f}{\partial x}=4x+2y+2, \frac{\partial f}{\partial x}=6x+2y+3 $$. In order to solve the minimal or maximum, we need to solve $$ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$$. We get $$x=-\frac{3}{10}, y=-\frac{2}{5}$$. We get $$f(-\frac{3}{10}, -\frac{2}{5})=\frac{11}{10}$$. This is a quadratic form. So the only minimum and maximum must be global. It is obvious that there is no global maximum. So the result must be a global minimum.

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