1. Simplify (a) $\log_a b^m$ and (b) $\log_a b^n$. Using the logarithm power rule: $\log_a b^m = m \log_a b$ and $\log_a b^n = n \log_a b$. 2. Given matrices: $$A = \begin{bmatrix}1 & 2\\ \end{bmatrix}, \quad B = \begin{bmatrix}3 & 4 & 5\\ \end{bmatrix}$$ Since $4A - 3B + C = 0$, solve for $C$: $$C = 3B - 4A$$ Calculate $4A$ and $3B$: $$4A = 4 \times \begin{bmatrix}1 & 2\\ \end{bmatrix} = \begin{bmatrix}4 & 8\\ \end{bmatrix}$$ $$3B = 3 \times \begin{bmatrix}3 & 4 & 5\\ \end{bmatrix} = \begin{bmatrix}9 & 12 & 15\\ \end{bmatrix}$$ Note: Dimensions of $A$ and $B$ differ, so $C$ cannot be computed as stated unless matrices are clarified. 3. Prove $f(x) = -2x$ is one-one. A function is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Assume $f(x_1) = f(x_2)$: $$-2x_1 = -2x_2 \implies x_1 = x_2$$ Hence, $f$ is one-one. 4. Find distance between $P(2, -5)$ and $Q(-4, 7)$. Distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Calculate: $$d = \sqrt{(-4 - 2)^2 + (7 - (-5))^2} = \sqrt{(-6)^2 + (12)^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5}$$ 5. Solve $\frac{d}{dx} \left( \frac{7x+2}{5x+2} \right)$. Use quotient rule: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$$ Let $u=7x+2$, $v=5x+2$. Calculate derivatives: $$u' = 7, \quad v' = 5$$ Apply: $$\frac{d}{dx} = \frac{(5x+2)(7) - (7x+2)(5)}{(5x+2)^2} = \frac{35x + 14 - 35x - 10}{(5x+2)^2} = \frac{4}{(5x+2)^2}$$ 6. Resolve into partial fractions: $$\frac{2x+3}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}$$ Multiply both sides by $(x+1)(x+3)$: $$2x + 3 = A(x+3) + B(x+1)$$ Expand: $$2x + 3 = A x + 3A + B x + B$$ Group terms: $$2x + 3 = (A + B) x + (3A + B)$$ Equate coefficients: $$A + B = 2$$ $$3A + B = 3$$ Solve: Subtract first from second: $$(3A + B) - (A + B) = 3 - 2 \implies 2A = 1 \implies A = \frac{1}{2}$$ Then: $$B = 2 - A = 2 - \frac{1}{2} = \frac{3}{2}$$ So: $$\frac{2x+3}{(x+1)(x+3)} = \frac{1/2}{x+1} + \frac{3/2}{x+3}$$ 7. Solve: $$\int \frac{(x-1)^3}{x} dx$$ Expand numerator: $$(x-1)^3 = x^3 - 3x^2 + 3x -1$$ Rewrite integral: $$\int \frac{x^3 - 3x^2 + 3x -1}{x} dx = \int (x^2 - 3x + 3 - \frac{1}{x}) dx$$ Integrate term-wise: $$\int x^2 dx = \frac{x^3}{3}, \quad \int (-3x) dx = -\frac{3x^2}{2}, \quad \int 3 dx = 3x, \quad \int -\frac{1}{x} dx = -\ln|x|$$ Final answer: $$\frac{x^3}{3} - \frac{3x^2}{2} + 3x - \ln|x| + C$$ 8. Prove for matrix $$A = \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$ Show: $$A^2 - 7A + 10I = 0$$ Calculate $A^2$: $$A^2 = A \times A = \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix}$$ Calculate $-7A$: $$-7A = -7 \times \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix}-21 & -14 & 0 \\ -7 & -28 & 0 \\ 0 & 0 & -35 \end{bmatrix}$$ Calculate $10I$: $$10I = 10 \times \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}$$ Sum: $$A^2 - 7A + 10I = \begin{bmatrix}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} + \begin{bmatrix}-21 & -14 & 0 \\ -7 & -28 & 0 \\ 0 & 0 & -35 \end{bmatrix} + \begin{bmatrix}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Hence proved. 9. Verify Cayley-Hamilton theorem for $$M = \begin{bmatrix}2 & 1 & 1 \\ 6 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}$$ Characteristic polynomial $p(\lambda) = \det(M - \lambda I)$: Calculate: $$\det \begin{bmatrix}2-\lambda & 1 & 1 \\ 6 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \end{bmatrix}$$ Expanding determinant: $$= (2-\lambda) \left[(1-\lambda)(2-\lambda) - 0 \times 1\right] - 1 \left[6(2-\lambda) - 0 \times 1\right] + 1 \left[6 \times 1 - (1-\lambda) \times 1\right]$$ Simplify: $$= (2-\lambda)((1-\lambda)(2-\lambda)) - 6(2-\lambda) + (6 - (1-\lambda))$$ Calculate $(1-\lambda)(2-\lambda) = 2 - 3\lambda + \lambda^2$. So: $$= (2-\lambda)(2 - 3\lambda + \lambda^2) - 6(2-\lambda) + (6 - 1 + \lambda)$$ Expand: $$= (2)(2 - 3\lambda + \lambda^2) - \lambda(2 - 3\lambda + \lambda^2) - 12 + 6\lambda + 5 + \lambda$$ Calculate terms: $$= 4 - 6\lambda + 2\lambda^2 - 2\lambda + 3\lambda^2 - \lambda^3 - 12 + 6\lambda + 5 + \lambda$$ Combine like terms: $$= (4 - 12 + 5) + (-6\lambda - 2\lambda + 6\lambda + \lambda) + (2\lambda^2 + 3\lambda^2) - \lambda^3$$ Simplify: $$= (-3) + (-1\lambda) + 5\lambda^2 - \lambda^3 = -\lambda^3 + 5\lambda^2 - \lambda - 3$$ Characteristic polynomial: $$p(\lambda) = -\lambda^3 + 5\lambda^2 - \lambda - 3$$ Rewrite as: $$p(\lambda) = - (\lambda^3 - 5\lambda^2 + \lambda + 3)$$ By Cayley-Hamilton theorem, matrix $M$ satisfies: $$p(M) = 0 \implies M^3 - 5M^2 + M + 3I = 0$$ This can be verified by direct matrix multiplication (omitted here for brevity). Final answers are provided for each question.