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# Tutor profile: Ryan G.

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Ryan G.
University Math Tutor
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## Questions

### Subject:Linear Algebra

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Question:

Find a basis for the row space and a basis for the column space of the matrix $(A=\begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ -1 & \phantom{-} 1 & -1\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix}.$)

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Ryan G.

Begin by applying elementary row operations in order to reduce $$A$$ to echelon form. Designate the rows of $$A$$ by $$r_i$$, $$i\in\{1,2,3\}$$. First, row replace the second row by adding the first row to the second row, $$r_1+r_2$$, obtaining $(A=\begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ -1 & \phantom{-} 1 & -1\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix} \rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix}.$) Next, row replace the third row by adding $$-2$$ times the first row to the third row, $$-2r_1+r_3$$, obtaining $(A=\begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ -1 & \phantom{-} 1 & -1\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix} \rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-}0 & -2 & \phantom{-} 0 \end{bmatrix}.$) Finally, row replace the third row by adding the second row to the third row, $$r_2+r_3$$, obtaining $(A=\begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ -1 & \phantom{-} 1 & -1\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix} \rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-} 2 & \phantom{-} 0 & \phantom{-} 2 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-}0 & -2 & \phantom{-} 0 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & \phantom{-} 1 & \phantom{-} 1\\ \phantom{-}0 & \phantom{-} 2 & \phantom{-}0\\ \phantom{-}0 & \phantom{-}0 & \phantom{-} 0 \end{bmatrix}.$) The last matrix listed above is the echelon form of $$A$$, and its nonzero rows form a basis for the row space of $$A$$. We can conclude that a basis for the row space of $$A$$ is the set $(\{(1,1,1),(0,2,0)\}.$) To find a basis for the column space, we will apply the same method on the transpose of the given matrix, $$A^T$$. The matrix $$A^T$$ is produced by listing the columns of $$A$$ as rows (or by listing the rows as columns). The result is $(A^T=\begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}1 & \phantom{-} 1 & \phantom{-}0\\ \phantom{-} 1 & -1 & \phantom{-} 2 \end{bmatrix}.$) Again applying elementary row operations, we row replace the second row by adding $$-1$$ times the first row to the second row, obtaining $(A^T=\begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}1 & \phantom{-} 1 & \phantom{-}0\\ \phantom{-} 1 & -1 & \phantom{-} 2 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}0 & \phantom{-} 2 & -2\\ \phantom{-} 1 & -1 & \phantom{-} 2 \end{bmatrix}.$) Next, row replace the third row by adding $$-1$$ times the first row to the third row, obtaining $(A^T=\begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}1 & \phantom{-} 1 & \phantom{-}0\\ \phantom{-} 1 & -1 & \phantom{-} 2 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}0 & \phantom{-} 2 & -2\\ \phantom{-} 1 & -1 & \phantom{-} 2 \end{bmatrix}\rightarrow \begin{bmatrix}\phantom{-} 1 & -1 & \phantom{-} 2\\ \phantom{-}0 & \phantom{-} 2 & -2\\ \phantom{-} 0 & \phantom{-}0 & \phantom{-} 0 \end{bmatrix}.$) This last matrix is the echelon form of $$A^T$$, and its nonzero rows form a basis for the row space of $$A^T$$, which is also the column space of $$A$$. We conclude that a basis for the column space of $$A$$ is the set $(\{(1,-1,2)^T,(0,2,-2)^T\}.$)

### Subject:Pre-Calculus

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Question:

As an application of logarithms, solve the following equation for $$x$$, $(4^x=2^x+3.$)

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Ryan G.

By the properties of exponents, express the left side of the equation as $$4^x=(2^2)^x=(2^x)^2$$. The equation becomes $((2^x)^2=2^x+3,$) or $((2^x)^2-2^x-3=0,$) where the latter is a quadratic equation in $$2^x$$. By the quadratic formula, the solution of this equation is $(2^x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-3)}}{2(1)}=\frac{1\pm\sqrt{13}}{2}.$) Recognizing that $$(1-\sqrt{13})/2$$ is negative but $$2^x$$ is positive for all $$x$$, we disregard the negative solution so that $(2^x=\frac{1+\sqrt{13}}{2}.$) Take the base $$2$$ logarithm of both sides to obtain $(x=\log_2\left(\frac{1+\sqrt{13}}{2}\right).$) Using the properties of logarithms this simplifies to $(x=\log_2(1+\sqrt{13})-\log_22=\log_2(1+\sqrt{13})-1.$)

### Subject:Calculus

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Question:

Evaluate the indefinite integral $(\int \sin^3x\cos^2x\ dx.$)

Inactive
Ryan G.

This problem demonstrates the simplifying power of an appropriate substitution. Since I have no trigonometric identities for $$\sin^3x$$, I'll begin by factoring that term $(\int \sin^3x\cos^2x\ dx=\int \sin x\sin^2x\cos^2x\ dx.$) The trigonometric identity $$\sin^2x+\cos^2x=1$$ allows us to write $$\sin^2x=1-\cos^2x$$. Making this replacement, the integral becomes $(\int \sin^3x\cos^2x\ dx=\int \sin x(1-\cos^2x)\cos^2x\ dx.$) Now consolidate the cosine terms by multiplying $$(1-\cos^2x)$$ and $$\cos^2x$$. It follows that $(\int \sin^3x\cos^2x\ dx=\int \sin x(\cos^2x-\cos^4x)\ dx.$) The cosine makes for a natural substitution, since it is related to sine by differentiation. We substitute $$u=\cos x$$ and $$du=-\sin x\ dx$$ so that, after some manipulation, the integral becomes $(\int \sin^3x\cos^2x\ dx=\int u^4-u^2\ du.$) From here we can evaluate the integral, remembering the constant of integration, $(\int \sin^3x\cos^2x\ dx=\frac{1}{5}u^5-\frac{1}{3}u^3+C.$) As a final step, we resubstitute $$\cos x=u$$, giving the solution $(\int \sin^3x\cos^2x\ dx=\frac{1}{5}\cos^5x-\frac{1}{3}\cos^3x+C.$).

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