1. Find the determinant of matrices: i. $$A = \begin{bmatrix} 8 & 2 & 5 \\ 1 & 2 & 2 \\ 0 & 4 & 3 \end{bmatrix}$$ Step 1: Use the formula for determinant of 3x3: $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ where matrix entries are: $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} 8 & 2 & 5 \\ 1 & 2 & 2 \\ 0 & 4 & 3 \end{bmatrix}$$ Step 2: Substitute values: $$= 8(2*3 - 2*4) - 2(1*3 - 2*0) + 5(1*4 - 2*0)$$ $$= 8(6 - 8) - 2(3 - 0) + 5(4 - 0)$$ $$= 8(-2) - 2(3) + 5(4)$$ $$= -16 - 6 + 20$$ $$= (-16 - 6) + 20 = -22 + 20 = -2$$ ii. $$B = \begin{bmatrix} 1 & 6 & 2 \\ 7 & 3 & 2 \\ 2 & 4 & 9 \end{bmatrix}$$ Step 1: Apply determinant formula: $$= 1(3*9 - 2*4) - 6(7*9 - 2*2) + 2(7*4 - 3*2)$$ $$= 1(27 - 8) - 6(63 - 4) + 2(28 - 6)$$ $$= 1(19) - 6(59) + 2(22)$$ $$= 19 - 354 + 44$$ $$= (19 + 44) - 354 = 63 - 354 = -291$$ 2. Prove: $$\det\begin{bmatrix} 1+x & y & z \\ x & 1+y & z \\ x & y & 1+z \end{bmatrix} = 1 + x + y + z$$ Step 1: Expand determinant by first row: $$= (1+x)((1+y)(1+z) - yz) - y(x(1+z) - xz) + z(xy - x(1+y))$$ Step 2: Simplify each part: $$= (1+x)(1 + y + z + yz - yz) - y(x + xz - xz) + z(xy - x - xy)$$ $$= (1+x)(1 + y + z) - y(x) + z(-x)$$ Step 3: Distribute: $$= (1)(1 + y + z) + x(1 + y + z) - xy - xz$$ Step 4: Simplify: $$= 1 + y + z + x + xy + xz - xy - xz$$ $$= 1 + x + y + z$$ 3. Prove: $$\det\begin{bmatrix} -x^2 & xy & xz \\ xy & -y^2 & yz \\ xz & yz & -z^2 \end{bmatrix} = 4 x^2 y^2 z^2$$ Step 1: Calculate determinant using cofactor expansion (first row): $$= (-x^2)(-y^2 * -z^2 - y z * y z) - (xy)(xy * -z^2 - yz * xz) + (xz)(xy * yz - (-y^2)(xz))$$ Step 2: Simplify each minor: $$= (-x^2)(y^2 z^2 - y^2 z^2) - xy(-x y z^2 - x y z^2) + x z(x y^2 z + y^2 x z)$$ Note first minor is 0. Step 3: Simplify second term: $$- xy(-2 x y z^2) = 2 x^2 y^2 z^2$$ Step 4: Simplify third term: $$x z (2 x y^2 z) = 2 x^2 y^2 z^2$$ Step 5: Sum terms: $$0 + 2 x^2 y^2 z^2 + 2 x^2 y^2 z^2 = 4 x^2 y^2 z^2$$ 4. Prove: $$\det \begin{bmatrix} x - y & y - z & z - x \\ y - z & z - x & x - y \\ z - x & x - y & y - z \end{bmatrix} = 0$$ Step 1: Note that sum of each row: $$(x-y) + (y-z) + (z-x) = 0$$ Step 2: This implies rows are linearly dependent, so determinant is 0. 5. Find $x$ if matrix $$A = \begin{bmatrix} x & 6 & 8 \\ 1 & x & 2 \\ 3 & 4 & x \end{bmatrix}$$ is singular (determinant zero). Step 1: Calculate determinant: $$= x(x*x - 2*4) - 6(1*x - 2*3) + 8(1*4 - x*3)$$ $$= x(x^2 - 8) - 6(x - 6) + 8(4 - 3x)$$ $$= x^3 - 8x - 6x + 36 + 32 - 24x$$ $$= x^3 - 38x + 68$$ Step 2: Set equal to 0: $$x^3 - 38x + 68 = 0$$ Step 3: Try rational roots; $x=2$: $$2^3 - 38(2) + 68 = 8 - 76 + 68 = 0$$ So $x=2$ is a root. 6. Solve for $x$: i. $$\det \begin{bmatrix} x & 2 & 3 \\ 2 & 7 & 2 \\ 8 & 8 & 1 \end{bmatrix} = 8$$ Step 1: Calculate determinant: $$= x(7*1 - 2*8) - 2(2*1 - 2*8) + 3(2*8 - 7*8)$$ $$= x(7 - 16) - 2(2 - 16) + 3(16 - 56)$$ $$= x(-9) - 2(-14) + 3(-40)$$ $$= -9x + 28 - 120$$ $$= -9x - 92$$ Step 2: Set equal to 8: $$-9x - 92 = 8$$ $$-9x = 100$$ $$x = -\frac{100}{9}$$ ii. $$\det \begin{bmatrix} x + 1 & 4 & 2 \\ 3 & 5 & 1 \\ 9 & 1 & 3 \end{bmatrix} = 10$$ Step 1: Calculate determinant: $$= (x+1)(5*3 - 1*1) - 4(3*3 - 1*9) + 2(3*1 - 5*9)$$ $$= (x+1)(15 - 1) - 4(9 - 9) + 2(3 - 45)$$ $$= (x+1)(14) - 4(0) + 2(-42)$$ $$= 14x + 14 - 84$$ $$= 14x - 70$$ Step 2: Set equal to 10: $$14x - 70 = 10$$ $$14x = 80$$ $$x = \frac{80}{14} = \frac{40}{7}$$