Advanced Mathematical Formatting Test

Matrix Operations

Problem: Matrix Multiplication

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix}$ then find AB & BA.

Solution for AB: $AB = \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix}$

Calculating each element:

$(AB)_{11} = (1)(1) + (-1)(4) + (1)(1) = 1 - 4 + 1 = -2$
$(AB)_{12} = (1)(2) + (-1)(2) + (1)(7) = 2 - 2 + 7 = 7$
$(AB)_{21} = (3)(1) + (2)(4) + (1)(1) = 3 + 8 + 1 = 12$
$(AB)_{22} = (3)(2) + (2)(2) + (1)(7) = 6 + 4 + 7 = 17$

Therefore: $AB = \begin{bmatrix} -2 & 7 \\ 12 & 17 \end{bmatrix}$

Solution for BA: $BA = \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix}$

Calculating each element:

$(BA)_{11} = (1)(1) + (2)(3) = 1 + 6 = 7$
$(BA)_{12} = (1)(-1) + (2)(2) = -1 + 4 = 3$
$(BA)_{13} = (1)(1) + (2)(1) = 1 + 2 = 3$
$(BA)_{21} = (4)(1) + (2)(3) = 4 + 6 = 10$
$(BA)_{22} = (4)(-1) + (2)(2) = -4 + 4 = 0$
$(BA)_{23} = (4)(1) + (2)(1) = 4 + 2 = 6$
$(BA)_{31} = (1)(1) + (7)(3) = 1 + 21 = 22$
$(BA)_{32} = (1)(-1) + (7)(2) = -1 + 14 = 13$
$(BA)_{33} = (1)(1) + (7)(1) = 1 + 7 = 8$

Therefore: $BA = \begin{bmatrix} 7 & 3 & 3 \\ 10 & 0 & 6 \\ 22 & 13 & 8 \end{bmatrix}$

Calculus - Advanced Differentiation

Logarithmic Differentiation

If $y = \log\left(\frac{\cos x}{1+\sin x}\right)$ then find $\frac{dy}{dx}$

Solution: $y = \log\left(\frac{\cos x}{1+\sin x}\right)$

Using logarithm properties: $y = \log(\cos x) - \log(1+\sin x)$

Differentiating both sides: $\frac{dy}{dx} = \frac{d}{dx}[\log(\cos x)] - \frac{d}{dx}[\log(1+\sin x)]$

$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) - \frac{1}{1+\sin x} \cdot \cos x$

$\frac{dy}{dx} = -\frac{\sin x}{\cos x} - \frac{\cos x}{1+\sin x}$

$\frac{dy}{dx} = -\tan x - \frac{\cos x}{1+\sin x}$

Simplifying further: $\frac{dy}{dx} = -\frac{\sin x(1+\sin x) + \cos^2 x}{\cos x(1+\sin x)}$

$= -\frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1+\sin x)}$

Since $\sin^2 x + \cos^2 x = 1$ : $\frac{dy}{dx} = -\frac{\sin x + 1}{\cos x(1+\sin x)} = -\frac{1}{\cos x} = -\sec x$

Integration Techniques

Integration by Parts

Evaluate: $\int x \log x \, dx$

Solution: Using integration by parts: $\int u \, dv = uv - \int v \, du$

Let:

$u = \log x \Rightarrow du = \frac{1}{x} dx$
$dv = x \, dx \Rightarrow v = \frac{x^2}{2}$

$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$

$= \frac{x^2 \log x}{2} - \int \frac{x}{2} dx$

$= \frac{x^2 \log x}{2} - \frac{x^2}{4} + C$

$= \frac{x^2}{2}\left(\log x - \frac{1}{2}\right) + C$

Complex Numbers - Polar Form

De Moivre's Theorem Application

Simplify: $\frac{(\cos 2\theta + i\sin 2\theta)^3 \cdot (\cos 3\theta - i\sin 3\theta)^2}{(\cos 2\theta + i\sin 2\theta)^7 \cdot (\cos 5\theta - i\sin 5\theta)^3}$

Solution: Using Euler's formula: $\cos\theta + i\sin\theta = e^{i\theta}$ and $\cos\theta - i\sin\theta = e^{-i\theta}$

Numerator: $(\cos 2\theta + i\sin 2\theta)^3 \cdot (\cos 3\theta - i\sin 3\theta)^2 = (e^{i2\theta})^3 \cdot (e^{-i3\theta})^2$ $= e^{i6\theta} \cdot e^{-i6\theta} = e^{i(6\theta-6\theta)} = e^{i0} = 1$

Denominator: $(\cos 2\theta + i\sin 2\theta)^7 \cdot (\cos 5\theta - i\sin 5\theta)^3 = (e^{i2\theta})^7 \cdot (e^{-i5\theta})^3$ $= e^{i14\theta} \cdot e^{-i15\theta} = e^{i(14\theta-15\theta)} = e^{-i\theta}$

Therefore: $\frac{1}{e^{-i\theta}} = e^{i\theta} = \cos\theta + i\sin\theta$

Differential Equations

First Order Linear ODE

Solve: $\frac{dy}{dx} + \frac{y}{x} = e^x$ , $y(0) = 2$

Solution: This is a first-order linear ODE of the form $\frac{dy}{dx} + P(x)y = Q(x)$

Where $P(x) = \frac{1}{x}$ and $Q(x) = e^x$

Finding Integrating Factor: $\mu(x) = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x| = x \text{ (for } x > 0\text{)}$

Multiplying the equation by $\mu(x) = x$ : $x\frac{dy}{dx} + y = xe^x$

The left side is the derivative of $xy$ : $\frac{d}{dx}(xy) = xe^x$

Integrating both sides: $xy = \int xe^x dx$

Using integration by parts with $u = x$ , $dv = e^x dx$ : $\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C$

Therefore: $xy = e^x(x-1) + C$ $y = \frac{e^x(x-1) + C}{x}$

Applying initial condition $y(0) = 2$ : This creates an issue since we have $x$ in the denominator. Let's reconsider the problem setup or assume $y$ approaches a limit as $x \to 0^+$ .

For a proper solution, we typically need $x \neq 0$ in the domain.

Formulas Reference

Trigonometric Identities

$\sin^2\theta + \cos^2\theta = 1$
$\tan^2\theta + 1 = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$
$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

Derivatives

$\frac{d}{dx}(x^n) = nx^{n-1}$
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(\log x) = \frac{1}{x}$
$\frac{d}{dx}(e^x) = e^x$

Integrals

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$ )
$\int \frac{1}{x} dx = \log|x| + C$
$\int e^x dx = e^x + C$
$\int \sin x dx = -\cos x + C$
$\int \cos x dx = \sin x + C$