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  "body": "\n## Proof\n\nWe factor the block matrix into a lower and an upper block triangular matrix.\n\n1. Let $M = \\begin{pmatrix} A \u0026 B \\\\ C \u0026 D \\end{pmatrix}$ be the [[block matrix|01AXG9KZQF]] of the statement, with $p, q \\ge 1$. Let $S = D - C A^{-1} B$.\n2. Let $L = \\begin{pmatrix} I_p \u0026 0 \\\\ C A^{-1} \u0026 I_q \\end{pmatrix}$ and $U = \\begin{pmatrix} A \u0026 B \\\\ 0 \u0026 S \\end{pmatrix}$, where $I_p, I_q$ are [[identity matrices|06DW9BAWB1]] and $0$ are zero blocks.\n3. Block products: by the definition of the [[matrix product|076JGP6R3T]], entry $(r, s)$ of a product is $\\sum_k X_{rk} Y_{ks}$.\n4. Split this sum into the indices $k \\le p$ and $k \u003e p$. This shows that block matrices with the same splitting multiply block by block, like $2 \\times 2$ matrices.\n5. By step 4, the top blocks of $LU$ are $I_p A + 0 \\cdot 0 = A$ and $I_p B + 0 \\cdot S = B$.\n6. The bottom left block of $LU$ is $C A^{-1} A + I_q \\cdot 0 = C (A^{-1} A) = C$, by [[associativity|073E4XC016]] and the definition of the [[inverse|0TWTS88TWN]].\n7. The bottom right block of $LU$ is $C A^{-1} B + S = C A^{-1} B + D - C A^{-1} B = D$. So $LU = M$.\n8. By the rule for [[block triangular matrices|0J9TVTNB4K]], $\\det U = \\det A \\cdot \\det S$.\n9. The [[transpose|0W809K6AW0]] of $L$ is $\\begin{pmatrix} I_p \u0026 (C A^{-1})^T \\\\ 0 \u0026 I_q \\end{pmatrix}$, which is upper block triangular.\n10. By the rule for the [[determinant of a transpose|0ZZQ2E1MTE]], the rule for block triangular matrices and $\\det I_p = \\det I_q = 1$ by the [[determinant of the identity|0Z4BWARC38]], $\\det L = \\det L^T = 1$.\n11. By the [[product rule|04CPJ6AF00]] and steps 7, 8 and 10, $\\det M = \\det L \\cdot \\det U = \\det A \\cdot \\det(D - C A^{-1} B)$.\n\n## Idea\n\nSubtracting $C A^{-1}$ times the top block row from the bottom block row clears the block $C$. This row operation is multiplication by a block triangular matrix with identity diagonal blocks, so it does not change the determinant. What remains is block triangular, with diagonal blocks $A$ and the Schur complement.\n",
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