1. **Problem (b): Show that $A^2 = 5A$ for $A = \begin{bmatrix}-1 & -2 \\ 3 & 6\end{bmatrix}$.** 2. **Calculate $A^2$:** $$A^2 = A \times A = \begin{bmatrix}-1 & -2 \\ 3 & 6\end{bmatrix} \times \begin{bmatrix}-1 & -2 \\ 3 & 6\end{bmatrix} = \begin{bmatrix}(-1)(-1)+(-2)(3) & (-1)(-2)+(-2)(6) \\ 3(-1)+6(3) & 3(-2)+6(6)\end{bmatrix} = \begin{bmatrix}1-6 & 2-12 \\ -3+18 & -6+36\end{bmatrix} = \begin{bmatrix}-5 & -10 \\ 15 & 30\end{bmatrix}$$ 3. **Calculate $5A$:** $$5A = 5 \times \begin{bmatrix}-1 & -2 \\ 3 & 6\end{bmatrix} = \begin{bmatrix}-5 & -10 \\ 15 & 30\end{bmatrix}$$ 4. **Since $A^2 = 5A$, the statement is proven.** --- 1. **Problem (c): Use induction to prove $1 + 3 + \ldots + (2n-1) = n^2$ for all $n \in \mathbb{N}$.** 2. **Base case ($n=1$):** $$1 = 1^2 = 1$$ True. 3. **Inductive hypothesis:** Assume for $n=k$, $$1 + 3 + \ldots + (2k-1) = k^2$$ 4. **Inductive step:** For $n=k+1$, $$1 + 3 + \ldots + (2k-1) + (2(k+1)-1) = k^2 + (2k+1) = k^2 + 2k + 1 = (k+1)^2$$ 5. **Thus, by induction, the formula holds for all natural numbers $n$.** --- 1. **Problem (d): If $m \neq n$ and $m$ times the $m$th term equals $n$ times the $n$th term of an A.P., show the $(m+n)$th term is zero.** 2. **Let the A.P. be $a, a+d, a+2d, \ldots$** 3. **The $m$th term is $a+(m-1)d$, the $n$th term is $a+(n-1)d$. Given:** $$m[a+(m-1)d] = n[a+(n-1)d]$$ 4. **Expand:** $$ma + m(m-1)d = na + n(n-1)d$$ 5. **Rearranged:** $$(m-n)a = [n(n-1) - m(m-1)]d$$ 6. **Simplify the right side:** $$n^2 - n - m^2 + m = (n^2 - m^2) - (n - m) = (n-m)(n+m) - (n-m) = (n-m)(n+m - 1)$$ 7. **So:** $$(m-n)a = (n-m)(n+m -1)d \implies (m-n)a = -(m-n)(n+m -1)d$$ 8. **Divide both sides by $(m-n)$ (nonzero):** $$a = -(n+m -1)d$$ 9. **The $(m+n)$th term is:** $$a + (m+n -1)d = -(n+m -1)d + (m+n -1)d = 0$$ 10. **Hence, the $(m+n)$th term is zero.** --- 1. **Problem (e): If $z \in \mathbb{C}$ and $|z - i| = |z + i|$, show that $\text{Im}(z) = 0$.** 2. **Let $z = x + yi$, where $x,y \in \mathbb{R}$.** 3. **Calculate:** $$|z - i| = |x + yi - i| = |x + (y-1)i| = \sqrt{x^2 + (y-1)^2}$$ $$|z + i| = |x + yi + i| = |x + (y+1)i| = \sqrt{x^2 + (y+1)^2}$$ 4. **Given $|z - i| = |z + i|$, so:** $$\sqrt{x^2 + (y-1)^2} = \sqrt{x^2 + (y+1)^2}$$ 5. **Square both sides:** $$x^2 + (y-1)^2 = x^2 + (y+1)^2$$ 6. **Simplify:** $$(y-1)^2 = (y+1)^2 \implies y^2 - 2y + 1 = y^2 + 2y + 1$$ 7. **Subtract $y^2 + 1$ from both sides:** $$-2y = 2y \implies 4y = 0 \implies y = 0$$ 8. **Therefore, $\text{Im}(z) = y = 0$.** --- 1. **Problem (f): Find roots of $x^3 - 13x^2 + 15x + 189 = 0$.** 2. **Try rational roots using factors of 189: $\pm1, \pm3, \pm7, \pm9, \pm21, \pm27, \pm63, \pm189$.** 3. **Test $x=3$: $3^3 - 13(3)^2 + 15(3) + 189 = 27 - 117 + 45 + 189 = 144 \neq 0$.** 4. **Test $x=7$: $343 - 637 + 105 + 189 = 0$? Calculate:** $$7^3 = 343, \quad -13 \times 7^2 = -13 \times 49 = -637, \quad 15 \times 7 = 105$$ $$343 - 637 + 105 + 189 = 343 - 637 + 294 = 343 - 343 = 0$$ 5. **So $x=7$ is a root. Divide polynomial by $(x-7)$:** Using synthetic division: $$7 | 1 \quad -13 \quad 15 \quad 189$$ $$\quad \quad 7 \quad -42 \quad -189$$ $$\quad 1 \quad -6 \quad -27 \quad 0$$ 6. **Quotient polynomial:** $$x^2 - 6x - 27$$ 7. **Solve quadratic:** $$x = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2}$$ 8. **Roots:** $$x = 9 \quad \text{or} \quad x = -3$$ 9. **Final roots:** $$x = 7, 9, -3$$ --- 1. **Problem (g): Evaluate $I = \int \frac{1}{\sqrt{(x+1)^3}} dx$.** 2. **Rewrite integrand:** $$\frac{1}{\sqrt{(x+1)^3}} = (x+1)^{-3/2}$$ 3. **Use substitution $u = x+1$, so $du = dx$.** 4. **Integral becomes:** $$I = \int u^{-3/2} du$$ 5. **Integrate:** $$\int u^{n} du = \frac{u^{n+1}}{n+1} + C, \quad n = -\frac{3}{2}$$ 6. **Calculate:** $$I = \frac{u^{-3/2 + 1}}{-3/2 + 1} + C = \frac{u^{-1/2}}{-1/2} + C = -2 u^{-1/2} + C = -\frac{2}{\sqrt{u}} + C$$ 7. **Back-substitute:** $$I = -\frac{2}{\sqrt{x+1}} + C$$ --- 1. **Problem (h): Show diagonals of a rhombus are perpendicular.** 2. **Let rhombus vertices be $A, B, C, D$ with equal sides. Let diagonals be $AC$ and $BD$.** 3. **Represent vectors:** $$\vec{AC} = \vec{c} - \vec{a}, \quad \vec{BD} = \vec{d} - \vec{b}$$ 4. **In a rhombus, adjacent sides are equal and opposite sides are parallel. Using vector properties, the diagonals satisfy:** $$\vec{AC} \cdot \vec{BD} = 0$$ 5. **This means diagonals are perpendicular.** --- "q_count":8