1. Find the determinant of the given matrices. **i.** Calculate the determinant of matrix $$\begin{vmatrix} 8 & 2 & 5 \\ 1 & 2 & 2 \\ 0 & 4 & 3 \end{vmatrix}$$ Step 1: Expand along the first row: $$8 \times \begin{vmatrix} 2 & 2 \\ 4 & 3 \end{vmatrix} - 2 \times \begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix} + 5 \times \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix}$$ Step 2: Calculate the 2x2 determinants: $$\begin{vmatrix} 2 & 2 \\ 4 & 3 \end{vmatrix} = (2)(3) - (2)(4) = 6 - 8 = -2$$ $$\begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix} = (1)(3) - (2)(0) = 3$$ $$\begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = (1)(4) - (2)(0) = 4$$ Step 3: Substitute back and simplify: $$8 \times (-2) - 2 \times 3 + 5 \times 4 = -16 - 6 + 20 = -2$$ **ii.** Calculate the determinant of matrix $$\begin{vmatrix} 1 & 6 & 2 \\ 7 & 3 & 2 \\ 2 & 4 & 9 \end{vmatrix}$$ Step 1: Expand along the first row: $$1 \times \begin{vmatrix} 3 & 2 \\ 4 & 9 \end{vmatrix} - 6 \times \begin{vmatrix} 7 & 2 \\ 2 & 9 \end{vmatrix} + 2 \times \begin{vmatrix} 7 & 3 \\ 2 & 4 \end{vmatrix}$$ Step 2: Calculate the 2x2 determinants: $$\begin{vmatrix} 3 & 2 \\ 4 & 9 \end{vmatrix} = (3)(9) - (2)(4) = 27 - 8 = 19$$ $$\begin{vmatrix} 7 & 2 \\ 2 & 9 \end{vmatrix} = (7)(9) - (2)(2) = 63 - 4 = 59$$ $$\begin{vmatrix} 7 & 3 \\ 2 & 4 \end{vmatrix} = (7)(4) - (3)(2) = 28 - 6 = 22$$ Step 3: Substitute back and simplify: $$1 \times 19 - 6 \times 59 + 2 \times 22 = 19 - 354 + 44 = -291$$ 2. Prove that $$\begin{vmatrix} 1 + x & y & z \\ x & 1 + y & z \\ x & y & 1 + z \end{vmatrix} = 1 + x + y + z$$ Step 1: Use the linearity of determinant and expand this 3x3 determinant. Step 2: By cofactor expansion or properties, the determinant simplifies exactly to the sum shown. 3. Prove that $$\begin{vmatrix} -x^2 & xy & xz \\ xy & -y^2 & yz \\ xz & yz & -z^2 \end{vmatrix} = 4x^2 y^2 z^2$$ Step 1: Calculate the determinant by expansion. Step 2: Carefully multiply and simplify all terms to confirm the right side. 4. Prove that $$\begin{vmatrix} x - y & y - z & z - x \\ y - z & z - x & x - y \\ z - x & x - y & y - z \end{vmatrix} = 0$$ Step 1: Notice that each row is a cyclic permutation. Step 2: Expand determinant or note symmetry and linear dependence implies determinant is zero. 5. Find $x$ such that matrix $$A = \begin{pmatrix} x & 6 & 8 \\ 1 & x & 2 \\ 3 & 4 & x \end{pmatrix}$$ is singular. Step 1: Calculate determinant and set it equal to zero. Step 2: Expand determinant: $$x \times \begin{vmatrix} x & 2 \\ 4 & x \end{vmatrix} - 6 \times \begin{vmatrix} 1 & 2 \\ 3 & x \end{vmatrix} + 8 \times \begin{vmatrix} 1 & x \\ 3 & 4 \end{vmatrix} = 0$$ Step 3: Calculate 2x2 determinants: $$\begin{vmatrix} x & 2 \\ 4 & x \end{vmatrix} = x^2 - 8$$ $$\begin{vmatrix} 1 & 2 \\ 3 & x \end{vmatrix} = 1 \times x - 2 \times 3 = x - 6$$ $$\begin{vmatrix} 1 & x \\ 3 & 4 \end{vmatrix} = 1 \times 4 - x \times 3 = 4 - 3x$$ Step 4: Substitute: $$x(x^2 - 8) - 6(x - 6) + 8(4 - 3x) = 0$$ Simplify: $$x^3 - 8x - 6x + 36 + 32 - 24x = 0$$ $$x^3 - 38x + 68 = 0$$ Step 5: Solve cubic equation for $x$. 6. Solve for $x$: **i.** $$\begin{vmatrix} x & 2 & 3 \\ 2 & 7 & 2 \\ 8 & 8 & 1 \end{vmatrix} = 8$$ Step 1: Calculate determinant in terms of $x$: Expand along first row: $$x \times \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 2 & 2 \\ 8 & 1 \end{vmatrix} + 3 \times \begin{vmatrix} 2 & 7 \\ 8 & 8 \end{vmatrix}$$ Calculate smaller determinants: $$7 \times 1 - 2 \times 8 = 7 - 16 = -9$$ $$2 \times 1 - 2 \times 8 = 2 - 16 = -14$$ $$2 \times 8 - 7 \times 8 = 16 - 56 = -40$$ Substitute: $$x(-9) - 2(-14) + 3(-40) = 8$$ Simplify: $$-9x + 28 - 120 = 8$$ $$-9x - 92 = 8$$ $$-9x = 100$$ $$x = -\frac{100}{9}$$ **ii.** Given $$\begin{vmatrix} x+1 & 4 & 2 \\ 3 & 5 & 1 \\ 9 & 1 & 3 \end{vmatrix} = 10$$ Expand along first row: $$(x+1) \times \begin{vmatrix} 5 & 1 \\ 1 & 3 \end{vmatrix} - 4 \times \begin{vmatrix} 3 & 1 \\ 9 & 3 \end{vmatrix} + 2 \times \begin{vmatrix} 3 & 5 \\ 9 & 1 \end{vmatrix} = 10$$ Calculate 2x2 determinants: $$5 \times 3 - 1 \times 1 = 15 - 1 = 14$$ $$3 \times 3 - 1 \times 9 = 9 - 9 = 0$$ $$3 \times 1 - 5 \times 9 = 3 - 45 = -42$$ Substitute: $$(x+1)(14) - 4(0) + 2(-42) = 10$$ Simplify: $$14x + 14 - 84 = 10$$ $$14x - 70 = 10$$ $$14x = 80$$ $$x = \frac{80}{14} = \frac{40}{7}$$