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Tutor profile: Kirsi M.

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Kirsi M.
Student Teacher (Math specialization)
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Questions

Subject: Linear Algebra

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Question:

Find the eigenvalues of the matrix $$A = \begin{bmatrix} 0&1\\ -2&-3 \end{bmatrix}$$.

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Kirsi M.
Answer:

First, we want to find the characteristic equation of the matrix $$A$$. To find this, we can use the equation: $$|A - \lambda \cdot I| = 0$$, where $$A$$ is our matrix, $$\lambda$$ is our variable, and $$I$$ is the Identity matrix. First, lets find our new matrix: $$A - \lambda \cdot I = \begin{bmatrix} 0&1\\ -2&-3 \end{bmatrix} - \lambda \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&1\\ -2&-3 \end{bmatrix} - \begin{bmatrix} \lambda&0\\ 0&\lambda \end{bmatrix} = \begin{bmatrix} -\lambda&1\\ -2&-3- \lambda \end{bmatrix}$$ Now we take the determinate of this matrix. $$\begin{vmatrix} -\lambda&1\\ -2&-3- \lambda \end{vmatrix} = (- \lambda)(-3-\lambda) - (-2)(1) = \lambda ^2 + 3\lambda + 2$$. Now we factor this characteristic polynomial and set it to $$0$$ to solve for $$\lambda$$. Two numbers which add to 3 and multiply to 2 are $$\mathbf{1}$$ and $$\mathbf{2}$$. So we get $$(\lambda + 1)(\lambda + 2) = 0$$, therefore $$\lambda = -1 $$ and $$\lambda = -2$$.

Subject: Pre-Calculus

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Question:

Multiply and simplify the following complex numbers: $$(-1 + 4i)*(4 - 3i)$$.

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Kirsi M.
Answer:

We know that when multiplying two binomials, we can use FOIL. First: $$-1 * 4 = \mathbf{-4}$$ Outside: $$-1 * -3i = \mathbf{3i}$$ Inside: $$4i * 4 = \mathbf{16i}$$ Last: $$4i * -3i = -12i^2 = -12 * -1 = \mathbf{12}$$. (Don't forget, $$i^2 = -1$$ ! ) Now we can add our four terms together and combine like terms. $$-4 + 3i + 16i + 12 = 8 + 19i$$ Thus, $$(-1 + 4i)*(4 - 3i) = (8 + 19i)$$.

Subject: Basic Math

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Question:

A rectangular garden has a total area of $$x^2 - 4x - 12$$ m$$^2$$. The width of the garden is $$x +2$$ m. What is the length of the garden?

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Kirsi M.
Answer:

We know that the formula for the area of a rectangle is $$A = l * w$$. Since we know what the area (A) and width (w) are, we can write our formula like so: $$x^2 - 4x - 12 = l * (x + 2)$$. If we divide each side of the equality by $$x +2$$, we get: $$\frac{x^2 - 4x - 12}{x+2} = l$$. Now, we can check to see if our numerator on the left side is factorable. We are looking for two integers (whole numbers) so that added together, their sum is -4, and when multiplied, their product is -12. These numbers come from the coefficients in the polynomial $$x^2 \mathbf{- 4}x \mathbf{- 12}$$. With some trial and error, we find that these integers are 2 and -6. Now we can write the polynomial in its factored form: $$x^2 - 4x - 12 = (x \mathbf{+2} )(x \mathbf{-6})$$, and substitute this back into our area equation. $$\frac{(x+2)(x-6)}{(x + 2)} = l$$ Now our $$(x+2)$$ terms divide into 1, so we are left with $$(x-6) = l$$. Thus, the length of the garden is $$(x-6)$$m.

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