Matrix Determinant 185E09

1. **State the problem:** We want to verify the determinant of the 3x3 matrix: $$\begin{vmatrix} a & b & ax + by \\ b & c & bx + cy \\ ax + by & bx + cy & 0 \end{vmatrix} = (b^2 - ac)(ax^2 + 2bxy + cy^2)$$ 2. **Recall the determinant formula for a 3x3 matrix:** For matrix \(M = \begin{bmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{bmatrix}\), $$\det(M) = m_{11}(m_{22}m_{33} - m_{23}m_{32}) - m_{12}(m_{21}m_{33} - m_{23}m_{31}) + m_{13}(m_{21}m_{32} - m_{22}m_{31})$$ 3. **Apply the formula to our matrix:** Let $$m_{11} = a, m_{12} = b, m_{13} = ax + by,$$ $$m_{21} = b, m_{22} = c, m_{23} = bx + cy,$$ $$m_{31} = ax + by, m_{32} = bx + cy, m_{33} = 0.$$ Calculate each term: - First term: $$a(c \cdot 0 - (bx + cy)(bx + cy)) = -a(bx + cy)^2$$ - Second term: $$-b(b \cdot 0 - (bx + cy)(ax + by)) = b(bx + cy)(ax + by)$$ - Third term: $$(ax + by)(b(bx + cy) - c(ax + by))$$ 4. **Simplify the third term inside parentheses:** $$b(bx + cy) - c(ax + by) = b^2 x + b c y - a c x - b c y = b^2 x - a c x = x(b^2 - a c)$$ So the third term becomes: $$(ax + by) \cdot x (b^2 - a c) = x (b^2 - a c)(ax + by)$$ 5. **Rewrite the determinant:** $$\det = -a(bx + cy)^2 + b(bx + cy)(ax + by) + x(b^2 - a c)(ax + by)$$ 6. **Expand and group terms:** First, expand \((bx + cy)^2 = b^2 x^2 + 2 b c x y + c^2 y^2\). So, $$-a(bx + cy)^2 = -a(b^2 x^2 + 2 b c x y + c^2 y^2) = -a b^2 x^2 - 2 a b c x y - a c^2 y^2$$ Next, expand \(b(bx + cy)(ax + by) = b (b x + c y)(a x + b y)\). Calculate \((b x + c y)(a x + b y) = a b x^2 + b^2 x y + a c x y + b c y^2\). Multiply by \(b\): $$b (a b x^2 + b^2 x y + a c x y + b c y^2) = a b^2 x^2 + b^3 x y + a b c x y + b^2 c y^2$$ 7. **Sum the first two terms:** $$-a b^2 x^2 - 2 a b c x y - a c^2 y^2 + a b^2 x^2 + b^3 x y + a b c x y + b^2 c y^2$$ Simplify: - \(-a b^2 x^2 + a b^2 x^2 = 0\) - \(-2 a b c x y + a b c x y = -a b c x y\) So sum is: $$-a b c x y - a c^2 y^2 + b^3 x y + b^2 c y^2$$ 8. **Add the third term:** $$x (b^2 - a c)(a x + b y) = x (b^2 - a c) a x + x (b^2 - a c) b y = a x^2 (b^2 - a c) + b x y (b^2 - a c)$$ Expand: $$a b^2 x^2 - a^2 c x^2 + b^3 x y - a b c x y$$ 9. **Add all terms together:** $$(-a b c x y - a c^2 y^2 + b^3 x y + b^2 c y^2) + (a b^2 x^2 - a^2 c x^2 + b^3 x y - a b c x y)$$ Group like terms: - \(a b^2 x^2 - a^2 c x^2 = x^2 (a b^2 - a^2 c)\) - \(-a b c x y - a b c x y + b^3 x y + b^3 x y = x y (-2 a b c + 2 b^3) = 2 x y (b^3 - a b c)\) - \(-a c^2 y^2 + b^2 c y^2 = y^2 (-a c^2 + b^2 c) = y^2 c (b^2 - a c)\) 10. **Factor the expression:** $$x^2 a (b^2 - a c) + 2 x y b (b^2 - a c) + y^2 c (b^2 - a c) = (b^2 - a c)(a x^2 + 2 b x y + c y^2)$$ **Final answer:** $$\det = (b^2 - a c)(a x^2 + 2 b x y + c y^2)$$ This matches the right side of the original equation, confirming the identity.