# Tutor profile: Suzana M.

## Questions

### Subject: Linear Algebra

In $$\mathbb{R}^3$$, let $$x_1 = (1, 2, 2)$$, $$x_2 = (2, -2, 1).$$ Show that the vector $$x = (0, 6, 3)$$ is a linear combination of $$x_1, x_2.$$

By definition, $$x = (0, 6, 3)$$ is a linear combination of $$x_1, x_2$$ if there exists scalars $$c_1, c_2$$ such that $$(0, 6, 3)=c_1\cdot(1, 2, 2)+c_2\cdot (2, -2, 1).$$ Thus $$(0, 6, 3)= (c_1 + 2c_2, 2c_1 - 2c_2, 2c_1 + c_2).$$ This implies that if we can find solutions to the system of equations \begin{align*} c_1 + 2c_2 &= 0 \\ 2c_1 - 2c_2 &= 6 \\ 2c_1 + c_2 &= 3 \end{align*} then $$x$$ is a linear combination of $$x_1$$ and $$x_2.$$ The first equation gives us $$c_1 = -2c_2$$. We can solve for $$c_2$$ by substituting into the second equation: \begin{align*} 2(-2c_2) - 2c_2 = 6 \\ -4c_2 - 2c_2 = 6 \\ c_2 = -1 \end{align*} and hence $$c_1 = -2(-1) = 2$$. This is a solution to the third equation: $$2(2) + (-1) = 3$$, so $$x$$ is a linear combination of $$x_1$$ and $$x_2$$. We have $$x = 2x_1 - x_2.$$

### Subject: Pre-Calculus

The floor of a one-story building is $$14$$ feet longer than it is wide. The building has $$1632$$ square feet of floor space. (a) Write a quadratic equation for the area of the floor in terms of the width. (b) Find the length and width of the floor.

(a) Let $$l$$ denote the length and $$w$$ denote the width. Since the floor is 14 feet longer than it is wide, we have that $$l=w+14$$. Recall that the area of a rectangle of length $$l$$ and width $$w$$ is given by $$A=lw.$$ Plugging in $$l=w+14$$ into $$A=lw$$ we obtain the area of the floor in terms of the width. That is, $$A=w(w+14).$$ (b) Since the building has 1632 square feet of floor space, $$1632=(w+14)w.$$ $$\Rightarrow w^2+14w-1632=0.$$ We solve this quadratic equation using the quadratic formula. $$\Delta = 14^2-4\cdot 1\cdot (-1632)=196+6528=6742$$ $$w=\frac{-14\pm \sqrt{6742}}{2}=\frac{-14\pm 82}{2}$$ $$\Rightarrow w=\frac{-14+82}{2}=34 \text{ (the other value is negative)}$$ $$\Rightarrow l=34+14=48.$$

### Subject: Calculus

Evaluate the definite integral. $$\int_{0}^{\pi}(x+1) \sin(x)\ dx$$

We use the integration by parts formula $$\int_a^b u\ dv=uv\bigg|_a^b-\int_a^b v\ du$$ with $$u=x+1$$, $$dv=\sin(x)dx$$, $$a=0$$ and $$b=\pi.$$ Then $$du=dx$$ and $$v=-\cos(x).$$ Hence $$\int_{0}^{\pi}(x+1) \sin(x)\ dx=-(x+1)\cos(x)\bigg|_0^{\pi}+\int_0^{\pi}\cos(x)dx=\left[-(x+1)\cos(x)+\sin(x)\right]\bigg|_0^{\pi}$$ $$= \left[-(\pi+1)\cos(\pi)+\sin(\pi)\right]-\left[-(0+1)\cos(0)+\sin(0)\right]=\pi+2.$$

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