1. Define Rank of a matrix. The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. It represents the dimension of the vector space spanned by its rows or columns. 2. Solve the system of equations: $$\begin{cases} x - y + 2z = 0 \\ 4x + y + 5z = 0 \\ 3x + 2y + z = 0 \end{cases}$$ Step 1: Write the augmented matrix: $$\left[\begin{array}{ccc|c}1 & -1 & 2 & 0 \\ 4 & 1 & 5 & 0 \\ 3 & 2 & 1 & 0 \end{array}\right]$$ Step 2: Use Gaussian elimination. Multiply row 1 by 4 and subtract from row 2: Row2 = Row2 - 4*Row1: $$4 - 4(1) = 0, \quad 1 - 4(-1) = 5, \quad 5 - 4(2) = -3, \quad 0 - 0 = 0$$ Row2 becomes: $[0, 5, -3, 0]$ Multiply row 1 by 3 and subtract from row 3: Row3 = Row3 - 3*Row1: $$3 - 3(1) = 0, \quad 2 - 3(-1) = 5, \quad 1 - 3(2) = -5, \quad 0 - 0 = 0$$ Row3 becomes: $[0, 5, -5, 0]$ Step 3: Subtract row 2 from row 3: Row3 = Row3 - Row2: $$0, 5 - 5 = 0, -5 - (-3) = -2, 0 - 0 = 0$$ Row3 becomes: $[0, 0, -2, 0]$ Step 4: Back substitution: From row 3: $-2z = 0 \Rightarrow z = 0$ From row 2: $5y - 3z = 0 \Rightarrow 5y = 0 \Rightarrow y = 0$ From row 1: $x - y + 2z = 0 \Rightarrow x = 0$ Solution: $x = 0, y = 0, z = 0$ 3. State the Cayley-Hamilton theorem. Every square matrix satisfies its own characteristic equation. If $A$ is an $n \times n$ matrix and $p(\lambda) = \det(\lambda I - A)$ is its characteristic polynomial, then $p(A) = 0$ (the zero matrix). 4. Define group. A group is a set $G$ with a binary operation $\cdot$ satisfying: 1. Closure: For all $a,b \in G$, $a \cdot b \in G$. 2. Associativity: For all $a,b,c \in G$, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$. 3. Identity: There exists $e \in G$ such that $e \cdot a = a \cdot e = a$ for all $a \in G$. 4. Inverse: For each $a \in G$, there exists $a^{-1} \in G$ such that $a \cdot a^{-1} = a^{-1} \cdot a = e$. 5. Define sub-group. A subgroup $H$ of a group $G$ is a subset of $G$ that is itself a group under the operation of $G$. 6. If $y = a^{mx}$, find $y_n$ (the $n^{th}$ derivative). Step 1: Write $y = a^{mx} = e^{mx \ln a}$. Step 2: The $n^{th}$ derivative of $y$ is: $$y_n = \frac{d^n}{dx^n} e^{mx \ln a} = (m \ln a)^n e^{mx \ln a} = (m \ln a)^n a^{mx}$$ 7. Find the $n^{th}$ derivative of $\log(5x+1)$. Step 1: Let $f(x) = \log(5x+1)$ (natural log). Step 2: First derivative: $$f'(x) = \frac{5}{5x+1}$$ Step 3: Use the formula for $n^{th}$ derivative of $\frac{1}{g(x)}$: $$f^{(n)}(x) = (-1)^{n-1} (n-1)! \frac{5^n}{(5x+1)^n}$$ 8. Find the $n^{th}$ derivative of $\cosh 2x \sin^2(2x)$. This is a product of functions; the $n^{th}$ derivative can be found using Leibniz theorem: $$\frac{d^n}{dx^n} [u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)$$ where $u(x) = \cosh 2x$, $v(x) = \sin^2(2x)$. 9. State the Leibnitz theorem. The $n^{th}$ derivative of a product $uv$ is: $$\frac{d^n}{dx^n} (uv) = \sum_{k=0}^n \binom{n}{k} u^{(k)} v^{(n-k)}$$ 10. Evaluate $\lim_{x \to 5} \frac{x^3 - 125}{x - 5}$. Step 1: Factor numerator: $$x^3 - 125 = (x - 5)(x^2 + 5x + 25)$$ Step 2: Simplify limit: $$\lim_{x \to 5} \frac{(x - 5)(x^2 + 5x + 25)}{x - 5} = \lim_{x \to 5} (x^2 + 5x + 25)$$ Step 3: Substitute $x=5$: $$5^2 + 5 \times 5 + 25 = 25 + 25 + 25 = 75$$ 11. Discuss differentiability of $$f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2 - x, & x > 1 \end{cases}$$ at $x=1$. Step 1: Check continuity at $x=1$: $$f(1^-) = 1, \quad f(1^+) = 2 - 1 = 1$$ So, $f$ is continuous at $x=1$. Step 2: Check left derivative: $$f'(1^-) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{1+h - 1}{h} = 1$$ Step 3: Check right derivative: $$f'(1^+) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{2 - (1+h) - 1}{h} = \lim_{h \to 0^+} \frac{1 - h - 1}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1$$ Since left and right derivatives are not equal, $f$ is not differentiable at $x=1$. 12. State Rolle’s theorem. If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$.