Advanced Mathematical Formatting Test
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Matrix Operations
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Problem: Matrix Multiplication
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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
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Logarithmic Differentiation
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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
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Integration by Parts
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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
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De Moivre’s Theorem Application
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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
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First Order Linear ODE
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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.