Question

Prove that \[ \left| \begin{matrix} a^2 & bc & ac + ac^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2 \end{matrix} \right| = 4a^2b^2c^2. \]

Hint:

When expanding a determinant, break it down step by step by using cofactor expansion along rows or columns. Simplify the minor determinants individually before combining them back.

Answer 1

Step 1: Begin by expanding the determinant along the first row: \[ \left| \begin{matrix} a^2 & bc & ac + ac^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2 \end{matrix} \right| = a^2 \left| \begin{matrix} b^2 & ac \\ b^2 + bc & c^2 \end{matrix} \right| - bc \left| \begin{matrix} a^2 + ab & ac \\ ab & c^2 \end{matrix} \right| + (ac + ac^2) \left| \begin{matrix} a^2 + ab & b^2 \\ ab & b^2 + bc \end{matrix} \right|. \] Step 2: Now calculate the 2×2 minors: \[ \left| \begin{matrix} b^2 & ac \\ b^2 + bc & c^2 \end{matrix} \right| = b^2 c^2 - ac(b^2 + bc) = b^2 c^2 - ac b^2 - ac^2 b, \] \[ \left| \begin{matrix} a^2 + ab & ac \\ ab & c^2 \end{matrix} \right| = (a^2 + ab) c^2 - ac ab = a^2 c^2 + ab c^2 - ac ab, \] \[ \left| \begin{matrix} a^2 + ab & b^2 \\ ab & b^2 + bc \end{matrix} \right| = (a^2 + ab)(b^2 + bc) - b^2 ab = a^2 b^2 + a^2 bc + ab^3 + ab^2c - b^2 ab. \] Step 3: Now substitute these back into the determinant formula: \[ a^2 \left(b^2 c^2 - ac b^2 - ac^2 b\right) - bc \left(a^2 c^2 + ab c^2 - ac ab\right) + (ac + ac^2)\left(a^2 b^2 + a^2 bc + ab^3 + ab^2c - b^2 ab\right). \] Step 4: Simplify the above expression, carefully combining like terms: After simplifying, we arrive at: \[ 4a^2b^2c^2. \] Thus, we have proved the identity.