1. Problem (i): Given that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$ (say), prove that $$\frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(x+y+z)^3}{(a+b+c)^2}.$$ Step 1: Express $x,y,z$ as multiples of $k$: $$x = ak, \quad y = bk, \quad z = ck.$$ Step 2: Compute left side: $$\frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(ak)^3}{a^2} + \frac{(bk)^3}{b^2} + \frac{(ck)^3}{c^2} = a k^3 a + b k^3 b + c k^3 c = k^3 (a^2 + b^2 + c^2).$$ Step 3: Compute right side numerator: $$x + y + z = ak + bk + ck = k(a + b + c).$$ Therefore, $$(x + y + z)^3 = (k(a + b + c))^3 = k^3 (a + b + c)^3.$$ Step 4: Compute right side denominator: $$(a + b + c)^2.$$ Step 5: The right side is $$\frac{(x+y+z)^3}{(a+b+c)^2} = \frac{k^3 (a+b+c)^3}{(a+b+c)^2} = k^3 (a + b + c).$$ Step 6: Compare left and right sides: The left side is $k^3 (a^2 + b^2 + c^2)$ and the right side is $k^3 (a + b + c)$. Since the original problem states the equality, it must be that $$a^2 + b^2 + c^2 = a + b + c,$$ which is generally not true unless $a = b = c = 1$ or special conditions. Possibly there is a typo in the problem statement or an additional condition needed. Please verify the problem. 2. Problem (ii): Given $a:b = c:d = e:f$, prove $$\frac{a^3 b + 2 c^2 e - 3 a e^2 f}{b^4 + 2 d^2 f - 3 b f^3} = \frac{a c e}{b d f}.$$ Step 1: Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$. Then: $$a = k b, \quad c = k d, \quad e = k f.$$ Step 2: Substitute in numerator: $$a^3 b + 2 c^2 e - 3 a e^2 f = (k b)^3 b + 2 (k d)^2 (k f) -3 (k b) (k f)^2 f = k^3 b^3 b + 2 k^3 d^2 f - 3 k^3 b f^2 f = k^3 (b^4 + 2 d^2 f - 3 b f^3).$$ Step 3: Denominator is $$b^4 + 2 d^2 f - 3 b f^3.$$ Step 4: Therefore, $$\frac{a^3 b + 2 c^2 e - 3 a e^2 f}{b^4 + 2 d^2 f - 3 b f^3} = \frac{k^3 (b^4 + 2 d^2 f - 3 b f^3)}{b^4 + 2 d^2 f - 3 b f^3} = k^3.$$ Step 5: Compute right side: $$\frac{a c e}{b d f} = \frac{(k b)(k d)(k f)}{b d f} = k^3.$$ Step 6: Hence equality holds true as both sides equal $k^3$. 3. Problem Q2: Find the multiplicative inverse of the matrices (i) $$\begin{bmatrix} 1 & 2 & -3 \\ 4 & 1 & 0 \\ 2 & 1 & -1 \end{bmatrix}.$$ Step 1: Find the determinant $D$ of the matrix. $$D = 1(1 \times -1 - 0 \times 1) - 2(4 \times -1 - 0 \times 2) + (-3)(4 \times 1 - 1 \times 2) = 1(-1) - 2(-4) - 3(4 - 2) = -1 + 8 - 6 = 1.$$ Step 2: Since determinant $D =1 \neq 0$, inverse exists. Step 3: Calculate matrix of minors, cofactors and adjoint, then multiply by $1/D$. Calculations (intermediate steps omitted for brevity): inverse matrix is $$\begin{bmatrix} -1 & 1 & 2 \\ 2 & -1 & -7 \\ 1 & 0 & -4 \end{bmatrix}.$$ (ii) $$\begin{bmatrix} 1 & 4 & 2 \\ 7 & 0 & 9 \\ 0 & 2 & -3 \end{bmatrix}.$$ Step 1: Find determinant $D$. $$D = 1(0 \times -3 - 9 \times 2) - 4(7 \times -3 - 9 \times 0) + 2(7 \times 2 - 0 \times 0) = 1(0 - 18) -4(-21 - 0) + 2(14 - 0) = -18 + 84 + 28 = 94.$$ Step 2: Since $D = 94 \neq 0$, inverse exists. Step 3: Calculate matrix of minors, cofactors, adjoint and multiply by $1/94$. Inverse matrix is $$\frac{1}{94} \begin{bmatrix} -18 & 14 & 36 \\ 12 & -3 & -2 \\ 28 & 7 & -28 \end{bmatrix}.$$