{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-determinant-01", "mko_id": "mko-determinant", "title": { "zh-Hant": "二階行列式：乘積公式、縮放公式與實數域可逆判準", "en": "2×2 determinants: the product rule, the scaling rule and the invertibility criterion over ℝ" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-determinant-01", "source": { "mko_id": "mko-determinant", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "d49050d70b58e0d9d2cab2f5ddc64133813f5a99418865b5941af4f1ec240093", "english_sha256": "c20e33f6fef226f483add30aa030f5fa68b5878d5ea16a6e3e75b0b360299311" }, "mode": "theorem_proof", "audience_prerequisites": [ "distributive", "mul_comm", "mul_assoc", "matrix_product_def", "mul_inverse_nonzero", "numeric_evaluation" ], "goal": { "zh-Hant": "對所有 2×2 實矩陣證明 det(AB)=det(A)det(B) 與 det(tM)=t²det(M)，並證明 M 在實數域可逆當且僅當 det(M)≠0。以 A=[[2,1],[−1,3]]、B=[[1,−2],[4,0]] 核對乘積公式並顯示 AB≠BA；以 t=−3 核對縮放公式；求 A⁻¹，並說明 A 雖是整數矩陣且 det(A)=7≠0，卻沒有整數反矩陣；以 C=[[1,2],[3,4]]（det=−2）說明非零行列式不必為正。", "en": "For all 2×2 real matrices prove det(AB)=det(A)det(B) and det(tM)=t²det(M), and prove that M is invertible over ℝ exactly when det(M)≠0. With A=[[2,1],[−1,3]] and B=[[1,−2],[4,0]], check the product rule and show AB≠BA; check the scaling rule with t=−3; find A⁻¹ and explain why A, although an integer matrix with det(A)=7≠0, has no integer inverse; use C=[[1,2],[3,4]] (det=−2) to show that a nonzero determinant need not be positive." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "所有矩陣是 2×2 實矩陣，數域是 ℝ；AB 是通常的行乘列乘積，不是逐項乘積（MKO 假設 a1、a3）。", "en": "All matrices are 2×2 real matrices over the field ℝ; AB is the usual row-by-column product, not the entrywise product (MKO assumptions a1, a3)." } }, { "id": "G2", "statement": { "zh-Hant": "例子：A=[[2,1],[−1,3]]，B=[[1,−2],[4,0]]，t=−3，C=[[1,2],[3,4]]，D=[[2,4],[1,2]]（每個內層括號是一行）。", "en": "Examples: A=[[2,1],[−1,3]], B=[[1,−2],[4,0]], t=−3, C=[[1,2],[3,4]], D=[[2,4],[1,2]] (each inner bracket is a row)." } }, { "id": "G3", "statement": { "zh-Hant": "M 在 ℝ 上可逆，指存在實矩陣 X 使 MX=XM=I；「整數反矩陣」指這樣的 X 的元素全為整數。", "en": "M is invertible over ℝ if there is a real matrix X with MX=XM=I; an 'integer inverse' is such an X whose entries are all integers." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "M=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}", "after_tex": "\\det(M)=ad-bc", "relation": "definition", "rule": "mko_det_def", "conditions": [], "explanation": { "zh-Hant": "寫出 MKO 的定義：主對角線乘積 ad 減去另一條對角線乘積 bc。符號與位置都重要：不能改成 ad+bc，也不能取絕對值（MKO 第一段）。", "en": "State the MKO's definition: the product ad of the main diagonal minus the product bc of the other diagonal. Signs and positions matter: it may not be changed to ad+bc, nor may an absolute value be taken (MKO first paragraph)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_det_def" ], "explanation": { "zh-Hant": "此步只寫出定義，不推出新事實。", "en": "This step only states the definition; it derives no new fact." } }, "leaf_basis": "mko_det_def + def_unfold" }, { "id": "S02", "uses": [ "S01", "G2" ], "before_tex": "\\det(A),\\ \\det(B)", "after_tex": "2\\cdot 3-1\\cdot(-1)=7,\\quad 1\\cdot 0-(-2)\\cdot 4=8", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "det(A)=2·3−1·(−1)=6−(−1)=7：減去一個負的乘積等於加上它的相反數。det(B)=1·0−(−2)·4=0−(−8)=8。", "en": "det(A)=2·3−1·(−1)=6−(−1)=7: subtracting a negative product adds its opposite. det(B)=1·0−(−2)·4=0−(−8)=8." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation", "neg_mul" ], "explanation": { "zh-Hant": "精確整數計算（N1、N2），雙向成立。", "en": "Exact integer computation (N1, N2), two-way." } }, "leaf_basis": "numeric_evaluation + neg_mul + mko_det_def" }, { "id": "S03", "uses": [ "G1", "matrix_product_def" ], "before_tex": "A=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix},\\ B=\\begin{pmatrix}e&f\\\\g&h\\end{pmatrix}", "after_tex": "AB=\\begin{pmatrix}ae+bg&af+bh\\\\ce+dg&cf+dh\\end{pmatrix}", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "依矩陣乘法定義，(AB)(i,j)=Σ_{k=1}^{2}A(i,k)B(k,j)：第 i 行與第 j 列的乘積和。例如 (AB)(1,1)=ae+bg，(AB)(2,2)=cf+dh。", "en": "By the definition of the matrix product, (AB)(i,j)=Σ_{k=1}^{2}A(i,k)B(k,j): the sum of products of row i with column j. For example (AB)(1,1)=ae+bg and (AB)(2,2)=cf+dh." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def" ], "explanation": { "zh-Hant": "這是定義展開，四個位置逐一相等，雙向成立。", "en": "This unfolds the definition at all four positions, two-way." } }, "leaf_basis": "matrix_product_def + def_unfold" }, { "id": "S04", "uses": [ "S01", "S03", "L_det_mul" ], "before_tex": "\\det(AB)=(ae+bg)(cf+dh)-(af+bh)(ce+dg)", "after_tex": "\\det(AB)=(ad-bc)(eh-fg)=\\det(A)\\det(B)", "relation": "equivalent", "rule": "L_det_mul", "conditions": [], "explanation": { "zh-Hant": "把 S01 的定義套到 S03 的 AB 上得到左式。依引理 L_det_mul，它等於 det(A)det(B)。引理以展開與比較單項式在 S04.1 到 S04.4 證明，對所有實數 a 到 h 成立，所以對所有 2×2 實矩陣成立。", "en": "Applying the definition of S01 to AB from S03 gives the left side. By Lemma L_det_mul it equals det(A)det(B). The lemma is proved in S04.1 to S04.4 by expanding and comparing monomials, for all reals a through h, so it holds for all 2×2 real matrices." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_det_mul", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "兩邊是對所有實數成立的多項式恆等式（N14），等號雙向。", "en": "The two sides form a polynomial identity valid for all reals (N14), so the equality is two-way." } }, "leaf_basis": "L_det_mul + mko_det_def" }, { "id": "S04.1", "uses": [ "distributive", "mul_comm", "mul_assoc" ], "before_tex": "(ae+bg)(cf+dh)", "after_tex": "acef+adeh+bcfg+bdgh", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "用分配律展開成四個乘積：ae·cf、ae·dh、bg·cf、bg·dh；再用乘法交換律與結合律把每個單項式的字母依字母序排列，得 acef、adeh、bcfg、bdgh。", "en": "Expand by distributivity into four products: ae·cf, ae·dh, bg·cf, bg·dh; then use commutativity and associativity of multiplication to put the letters of each monomial in alphabetical order: acef, adeh, bcfg, bdgh." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_comm", "mul_assoc" ], "explanation": { "zh-Hant": "所用的都是恆等式，雙向成立。", "en": "Only identities are used, so it is two-way." } }, "leaf_basis": "distributive + mul_comm + mul_assoc" }, { "id": "S04.2", "uses": [ "distributive", "mul_comm", "mul_assoc" ], "before_tex": "(af+bh)(ce+dg)", "after_tex": "acef+adfg+bceh+bdgh", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "同樣展開：af·ce=acef、af·dg=adfg、bh·ce=bceh、bh·dg=bdgh。", "en": "Expand the same way: af·ce=acef, af·dg=adfg, bh·ce=bceh, bh·dg=bdgh." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_comm", "mul_assoc" ], "explanation": { "zh-Hant": "恆等式，雙向成立。", "en": "Identities, two-way." } }, "leaf_basis": "distributive + mul_comm + mul_assoc" }, { "id": "S04.3", "uses": [ "S04.1", "S04.2" ], "before_tex": "(acef+adeh+bcfg+bdgh)-(acef+adfg+bceh+bdgh)", "after_tex": "adeh+bcfg-adfg-bceh", "relation": "equivalent", "rule": "add_inverse", "conditions": [], "explanation": { "zh-Hant": "相減時 acef 與 bdgh 兩項各出現在兩個展開式中，互相抵消（u−u=0）；剩下 adeh+bcfg−adfg−bceh。", "en": "On subtraction, acef and bdgh appear in both expansions and cancel (u−u=0); what remains is adeh+bcfg−adfg−bceh." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "add_inverse", "add_comm", "add_assoc" ], "explanation": { "zh-Hant": "加法反元素與重排都是恆等式，雙向成立。", "en": "Additive inverses and rearrangement are identities, so it is two-way." } }, "leaf_basis": "add_inverse + add_comm + add_assoc + distributive" }, { "id": "S04.4", "uses": [ "S04.3", "distributive" ], "before_tex": "(ad-bc)(eh-fg)", "after_tex": "adeh-adfg-bceh+bcfg", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開右邊：ad·eh=adeh，ad·(−fg)=−adfg，(−bc)·eh=−bceh，(−bc)(−fg)=bcfg（neg_mul）。與 S04.3 的四項逐項相同，所以 det(AB)=det(A)det(B)。S04.1 到 S04.4 合起來證得引理 L_det_mul。", "en": "Expand the right side: ad·eh=adeh, ad·(−fg)=−adfg, (−bc)·eh=−bceh and (−bc)(−fg)=bcfg (neg_mul). These are term for term the four terms of S04.3, so det(AB)=det(A)det(B). S04.1 to S04.4 together prove Lemma L_det_mul." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "neg_mul", "mul_comm" ], "explanation": { "zh-Hant": "恆等式，雙向成立；N14 在格點上核對同一多項式恆等式。", "en": "Identities, two-way; N14 checks the same polynomial identity on a grid." } }, "leaf_basis": "distributive + neg_mul + mul_comm + mul_assoc" }, { "id": "S05", "uses": [ "S02", "S03", "S04", "G2" ], "before_tex": "AB", "after_tex": "AB=\\begin{pmatrix}6&-4\\\\11&2\\end{pmatrix},\\ \\det(AB)=12+44=56=7\\cdot 8", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "依 S03 直接計算：(AB)(1,1)=2·1+1·4=6，(AB)(1,2)=2·(−2)+1·0=−4，(AB)(2,1)=(−1)·1+3·4=11，(AB)(2,2)=(−1)(−2)+3·0=2。det(AB)=6·2−(−4)·11=12+44=56，與 det(A)det(B)=7·8=56 相同。這是以實際乘積重算的核對（MKO 第三段）。", "en": "Compute directly by S03: (AB)(1,1)=2·1+1·4=6, (AB)(1,2)=2·(−2)+1·0=−4, (AB)(2,1)=(−1)·1+3·4=11, (AB)(2,2)=(−1)(−2)+3·0=2. det(AB)=6·2−(−4)·11=12+44=56, the same as det(A)det(B)=7·8=56. This is a check by recomputing from the actual product (MKO third paragraph)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "numeric_evaluation" ], "explanation": { "zh-Hant": "有限計算只核對本例（N3 到 N8），不取代 S04 的一般證明。", "en": "The finite computation checks this example only (N3 to N8) and does not replace the general proof of S04." } }, "leaf_basis": "numeric_evaluation + matrix_product_def + mko_det_def" }, { "id": "S06", "uses": [ "S03", "S05", "G2" ], "before_tex": "BA", "after_tex": "BA=\\begin{pmatrix}4&-5\\\\8&4\\end{pmatrix}\\neq AB,\\ \\det(BA)=16+40=56", "relation": "implies", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "BA：(1,1)=1·2+(−2)(−1)=4，(1,2)=1·1+(−2)·3=−5，(2,1)=4·2+0·(−1)=8，(2,2)=4·1+0·3=4。左上角 4≠6，所以 BA≠AB；但 det(BA)=4·4−(−5)·8=56=det(AB)。乘積公式只說行列式相等，並不表示 A、B 可交換（MKO 假設 a3）。", "en": "BA: (1,1)=1·2+(−2)(−1)=4, (1,2)=1·1+(−2)·3=−5, (2,1)=4·2+0·(−1)=8, (2,2)=4·1+0·3=4. The top-left entries 4≠6, so BA≠AB; yet det(BA)=4·4−(−5)·8=56=det(AB). The product rule says only that the determinants agree; it does not say that A and B commute (MKO assumption a3)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "matrix_product_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個不相等的元素就足以推出 BA≠AB。", "en": "One unequal entry suffices for BA≠AB." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S07", "uses": [ "S01", "matrix_alg" ], "before_tex": "\\det(tM),\\ tM=\\begin{pmatrix}ta&tb\\\\tc&td\\end{pmatrix}", "after_tex": "(ta)(td)-(tb)(tc)=t^2(ad-bc)=t^2\\det(M)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "tM 的每個元素乘 t。依定義 det(tM)=(ta)(td)−(tb)(tc)；用乘法交換律與結合律，(ta)(td)=t²ad、(tb)(tc)=t²bc；再用分配律提出 t²。每個乘積含兩個元素，所以出現 t²，而不是 t。", "en": "Every entry of tM is multiplied by t. By definition det(tM)=(ta)(td)−(tb)(tc); by commutativity and associativity of multiplication, (ta)(td)=t²ad and (tb)(tc)=t²bc; distributivity then factors out t². Each product contains two entries, which is why t² appears and not t." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mul_comm", "mul_assoc", "distributive" ], "explanation": { "zh-Hant": "對所有實數 t、a、b、c、d 成立的恆等式（N15），雙向成立。", "en": "An identity for all reals t, a, b, c, d (N15), two-way." } }, "leaf_basis": "mul_comm + mul_assoc + distributive + power_laws" }, { "id": "S08", "uses": [ "S02", "S07", "G2" ], "before_tex": "-3A=\\begin{pmatrix}-6&-3\\\\3&-9\\end{pmatrix}", "after_tex": "\\det(-3A)=54+9=63=(-3)^2\\cdot 7;\\ \\det(0A)=0", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "直接算：det(−3A)=(−6)(−9)−(−3)·3=54+9=63；依 S07 應為 (−3)²·7=9·7=63，一致。負的 t 經由 t² 作用，行列式不變號（MKO 假設 a2）。t=0 時 0A 是零矩陣，行列式 0=0²·7。", "en": "Directly: det(−3A)=(−6)(−9)−(−3)·3=54+9=63; by S07 it should be (−3)²·7=9·7=63, which agrees. A negative t acts through t², so the determinant does not change sign (MKO assumption a2). For t=0, 0A is the zero matrix, with determinant 0=0²·7." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07", "numeric_evaluation" ], "explanation": { "zh-Hant": "兩種算法（直接與公式）一致（N16、N17），只核對本例。", "en": "The two methods (direct and by the formula) agree (N16, N17); this checks the example only." } }, "leaf_basis": "numeric_evaluation + mko_det_def" }, { "id": "S09", "uses": [ "S01", "matrix_product_def" ], "before_tex": "M\\,\\operatorname{adj}(M),\\ \\operatorname{adj}(M)\\,M,\\ \\operatorname{adj}(M)=\\begin{pmatrix}d&-b\\\\-c&a\\end{pmatrix}", "after_tex": "\\begin{pmatrix}ad-bc&0\\\\0&ad-bc\\end{pmatrix}=\\det(M)\\,I", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "M·adj(M) 的四個元素：(1,1)=ad+b(−c)=ad−bc，(1,2)=a(−b)+ba=0，(2,1)=cd+d(−c)=0，(2,2)=c(−b)+da=ad−bc。adj(M)·M：(1,1)=da+(−b)c=ad−bc，(1,2)=db+(−b)d=0，(2,1)=(−c)a+ac=0，(2,2)=(−c)b+ad=ad−bc。兩個乘積都等於 det(M)·I。", "en": "The four entries of M·adj(M): (1,1)=ad+b(−c)=ad−bc, (1,2)=a(−b)+ba=0, (2,1)=cd+d(−c)=0, (2,2)=c(−b)+da=ad−bc. For adj(M)·M: (1,1)=da+(−b)c=ad−bc, (1,2)=db+(−b)d=0, (2,1)=(−c)a+ac=0, (2,2)=(−c)b+ad=ad−bc. Both products equal det(M)·I." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "neg_mul", "mul_comm" ], "explanation": { "zh-Hant": "每個元素都是恆等式（N18、N19），雙向成立。", "en": "Every entry is an identity (N18, N19), so it is two-way." } }, "leaf_basis": "matrix_product_def + neg_mul + mul_comm + add_inverse" }, { "id": "S10", "uses": [ "S09", "matrix_alg", "mul_inverse_nonzero", "G3" ], "before_tex": "\\det(M)\\neq 0", "after_tex": "X=\\tfrac{1}{\\det(M)}\\operatorname{adj}(M):\\ MX=XM=I", "relation": "implies", "rule": "mul_inverse_nonzero", "conditions": [ "\\det(M)\\neq 0" ], "explanation": { "zh-Hant": "若 det(M)≠0，則 1/det(M) 存在。令 X=(1/det(M))·adj(M)。依 matrix_alg 純量可提出：MX=(1/det(M))·(M·adj(M))=(1/det(M))·det(M)·I=I；同理 XM=I。所以 M 在 ℝ 上可逆。", "en": "If det(M)≠0, then 1/det(M) exists. Let X=(1/det(M))·adj(M). Since scalars factor out (matrix_alg), MX=(1/det(M))·(M·adj(M))=(1/det(M))·det(M)·I=I, and likewise XM=I. So M is invertible over ℝ." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "mul_inverse_nonzero", "matrix_alg" ], "explanation": { "zh-Hant": "關鍵是除以 det(M)，需要 det(M)≠0 且在可做除法的 ℝ 中；這是「det≠0 ⇒ 可逆」方向。", "en": "The key is dividing by det(M), which needs det(M)≠0 and a number system with division, such as ℝ; this is the direction 'det≠0 ⇒ invertible'." } }, "leaf_basis": "mul_inverse_nonzero + matrix_alg + frac_rules" }, { "id": "S11", "uses": [ "S04", "matrix_alg", "G3" ], "before_tex": "MX=XM=I", "after_tex": "\\det(M)\\det(X)=\\det(I)=1\\ \\Rightarrow\\ \\det(M)\\neq 0", "relation": "implies", "rule": "contradiction", "conditions": [], "explanation": { "zh-Hant": "設 M 在 ℝ 上可逆，MX=I。依 L_det_mul，det(M)det(X)=det(MX)=det(I)=1·1−0·0=1。若 det(M)=0，左邊是 0·det(X)=0，得 0=1，矛盾。所以 det(M)≠0。", "en": "Suppose M is invertible over ℝ with MX=I. By L_det_mul, det(M)det(X)=det(MX)=det(I)=1·1−0·0=1. If det(M)=0, the left side would be 0·det(X)=0, giving 0=1, a contradiction. So det(M)≠0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_det_mul", "mul_zero", "contradiction" ], "explanation": { "zh-Hant": "這是「可逆 ⇒ det≠0」方向，只用到乘積公式與 det(I)=1。", "en": "This is the direction 'invertible ⇒ det≠0', using only the product rule and det(I)=1." } }, "leaf_basis": "L_det_mul + mul_zero + contradiction + numeric_evaluation" }, { "id": "S12", "uses": [ "S10", "S11", "G2" ], "before_tex": "\\det(M)\\neq 0\\Rightarrow\\text{invertible},\\ \\text{invertible}\\Rightarrow\\det(M)\\neq 0", "after_tex": "M\\ \\text{invertible over }\\mathbb{R}\\iff\\det(M)\\neq 0;\\ \\det(D)=2\\cdot 2-4\\cdot 1=0", "relation": "implies", "rule": "iff_intro", "conditions": [], "explanation": { "zh-Hant": "S10、S11 兩個方向合起來得到實數域可逆判準。例如 D=[[2,4],[1,2]] 的行列式 2·2−4·1=0，所以 D 在 ℝ 上不可逆；確實 D(2,−1)ᵀ=(0,0)ᵀ，非零向量被送到零，不能還原。", "en": "The two directions S10 and S11 give the invertibility criterion over ℝ. For example D=[[2,4],[1,2]] has determinant 2·2−4·1=0, so D is not invertible over ℝ; indeed D(2,−1)ᵀ=(0,0)ᵀ sends a nonzero vector to zero, which cannot be undone." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "S11", "iff_intro" ], "explanation": { "zh-Hant": "兩個方向各有證明；D 的例子只是說明，不參與證明。", "en": "Each direction has its own proof; the example D only illustrates and is not part of the proof." } }, "leaf_basis": "iff_intro + numeric_evaluation" }, { "id": "S13", "uses": [ "S02", "S09", "S10", "G2" ], "before_tex": "A^{-1}", "after_tex": "A^{-1}=\\tfrac{1}{7}\\begin{pmatrix}3&-1\\\\1&2\\end{pmatrix},\\ A\\cdot\\tfrac{1}{7}\\begin{pmatrix}3&-1\\\\1&2\\end{pmatrix}=\\tfrac{1}{7}\\begin{pmatrix}7&0\\\\0&7\\end{pmatrix}=I", "relation": "implies", "rule": "mul_inverse_nonzero", "conditions": [ "\\det(A)=7\\neq 0" ], "explanation": { "zh-Hant": "det(A)=7≠0，依 S10，A⁻¹=(1/7)adj(A)，adj(A)=[[3,−1],[1,2]]。直接核對 A·adj(A)：(1,1)=2·3+1·1=7，(1,2)=2·(−1)+1·2=0，(2,1)=(−1)·3+3·1=0，(2,2)=(−1)(−1)+3·2=7，所以乘以 1/7 後是 I。A⁻¹ 的元素 3/7、−1/7、1/7、2/7 都不是整數。", "en": "det(A)=7≠0, so by S10 A⁻¹=(1/7)adj(A) with adj(A)=[[3,−1],[1,2]]. Check A·adj(A) directly: (1,1)=2·3+1·1=7, (1,2)=2·(−1)+1·2=0, (2,1)=(−1)·3+3·1=0, (2,2)=(−1)(−1)+3·2=7, so after multiplying by 1/7 it is I. The entries 3/7, −1/7, 1/7, 2/7 of A⁻¹ are not integers." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "numeric_evaluation" ], "explanation": { "zh-Hant": "數值核對（N20 到 N23）確認 A·A⁻¹=I；A⁻¹·A=I 由 S09、S10 的一般論證保證。", "en": "The numeric check (N20 to N23) confirms A·A⁻¹=I; A⁻¹·A=I is guaranteed by the general argument of S09 and S10." } }, "leaf_basis": "mul_inverse_nonzero + matrix_product_def + numeric_evaluation" }, { "id": "S14", "uses": [ "S13", "matrix_alg", "G3" ], "before_tex": "AY=YA=I,\\ Y\\ \\text{with integer entries}\\ ?", "after_tex": "Y=A^{-1}\\ \\text{and}\\ \\tfrac{3}{7}\\notin\\mathbb{Z}\\ \\Rightarrow\\ \\text{no integer inverse}", "relation": "implies", "rule": "contradiction", "conditions": [], "explanation": { "zh-Hant": "反矩陣唯一：若 X=A⁻¹ 且 Y 也滿足 AY=YA=I，則依 matrix_alg 的結合律 X=XI=X(AY)=(XA)Y=IY=Y。所以任何反矩陣都等於 A⁻¹，其 (1,1) 元素是 3/7，不是整數（7 不整除 3）。因此 A 沒有整數反矩陣，雖然 A 的元素都是整數且 det(A)=7≠0。實數域的「非零即可逆」不能原樣搬到整數環（MKO 假設 a1）。", "en": "The inverse is unique: if X=A⁻¹ and Y also satisfies AY=YA=I, then by associativity (matrix_alg) X=XI=X(AY)=(XA)Y=IY=Y. So every inverse equals A⁻¹, whose (1,1) entry is 3/7, not an integer (7 does not divide 3). Hence A has no integer inverse, although its entries are integers and det(A)=7≠0. The real-field rule 'nonzero means invertible' does not carry over unchanged to the integer ring (MKO assumption a1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S13", "matrix_alg", "contradiction", "divides_def" ], "explanation": { "zh-Hant": "需要反矩陣唯一（結合律）與 3/7 不是整數；不宣稱整數環上的一般可逆判準。", "en": "It needs uniqueness of the inverse (associativity) and that 3/7 is not an integer; no general invertibility criterion over the integers is claimed." } }, "leaf_basis": "matrix_alg + contradiction + divides_def" }, { "id": "S15", "uses": [ "S10", "S12", "G2" ], "before_tex": "C=\\begin{pmatrix}1&2\\\\3&4\\end{pmatrix}", "after_tex": "\\det(C)=-2\\neq 0,\\ C^{-1}=\\begin{pmatrix}-2&1\\\\ \\tfrac{3}{2}&-\\tfrac{1}{2}\\end{pmatrix}", "relation": "implies", "rule": "iff_intro", "conditions": [], "explanation": { "zh-Hant": "det(C)=1·4−2·3=−2，是負數但非零，依 S12 C 可逆；依 S10，C⁻¹=(1/(−2))·[[4,−2],[−3,1]]=[[−2,1],[3/2,−1/2]]。核對 C·C⁻¹：(1,1)=1·(−2)+2·(3/2)=1，(1,2)=1·1+2·(−1/2)=0，(2,1)=3·(−2)+4·(3/2)=0，(2,2)=3·1+4·(−1/2)=1。非零行列式包含正與負，不必等於一（MKO 第一段）。", "en": "det(C)=1·4−2·3=−2, negative but nonzero, so by S12 C is invertible; by S10, C⁻¹=(1/(−2))·[[4,−2],[−3,1]]=[[−2,1],[3/2,−1/2]]. Check C·C⁻¹: (1,1)=1·(−2)+2·(3/2)=1, (1,2)=1·1+2·(−1/2)=0, (2,1)=3·(−2)+4·(3/2)=0, (2,2)=3·1+4·(−1/2)=1. Nonzero determinants include positive and negative values and need not equal one (MKO first paragraph)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "numeric_evaluation" ], "explanation": { "zh-Hant": "依賴判準與數值核對（N24 到 N28）。", "en": "It relies on the criterion and the numeric checks (N24 to N28)." } }, "leaf_basis": "iff_intro + mul_inverse_nonzero + numeric_evaluation" }, { "id": "S16", "uses": [ "S04", "S05", "S06", "S07", "S08", "S12", "S13", "S14", "S15" ], "before_tex": "\\text{tasks of the goal}", "after_tex": "\\det(AB)=\\det A\\det B,\\ \\det(tM)=t^2\\det M,\\ \\text{invertible over }\\mathbb{R}\\iff\\det\\neq 0", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併：乘積公式與縮放公式對所有 2×2 實矩陣成立（S04、S07）；例中 det(AB)=det(BA)=56 但 AB≠BA（S05、S06），det(−3A)=63（S08）；實數域可逆 ⇔ det≠0（S12）；A⁻¹=(1/7)[[3,−1],[1,2]] 不是整數矩陣（S13、S14）；det(C)=−2 的 C 可逆（S15）。", "en": "Combine: the product and scaling rules hold for all 2×2 real matrices (S04, S07); in the example det(AB)=det(BA)=56 but AB≠BA (S05, S06), and det(−3A)=63 (S08); invertible over ℝ ⇔ det≠0 (S12); A⁻¹=(1/7)[[3,−1],[1,2]] is not an integer matrix (S13, S14); C with det(C)=−2 is invertible (S15)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S04", "S05", "S06", "S07", "S08", "S12", "S13", "S14", "S15" ], "explanation": { "zh-Hant": "目標 G：題目要求的各項。足夠條件 H：S04 到 S15 的結論；H⇒G 只用合取。H 由 G1 到 G3 正向推出，沒有把 G 當作前提。", "en": "Goal G: the items asked for. Sufficient condition H: the conclusions of S04 to S15; H⇒G uses only conjunction. H is derived forwards from G1 to G3; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "對所有 2×2 實矩陣，det(AB)=det(A)det(B)、det(tM)=t²det(M)，且 M 在 ℝ 上可逆當且僅當 det(M)≠0。例中 det(A)=7、det(B)=8、det(AB)=det(BA)=56 而 AB≠BA；det(−3A)=63；A⁻¹=(1/7)[[3,−1],[1,2]]，A 沒有整數反矩陣；det(C)=−2，C 可逆。", "en": "For all 2×2 real matrices, det(AB)=det(A)det(B) and det(tM)=t²det(M), and M is invertible over ℝ exactly when det(M)≠0. In the example det(A)=7, det(B)=8, det(AB)=det(BA)=56 while AB≠BA; det(−3A)=63; A⁻¹=(1/7)[[3,−1],[1,2]], and A has no integer inverse; det(C)=−2 and C is invertible." }, "negative_controls": [ { "claim": { "zh-Hant": "det(C)=−2<0，所以 C 不可逆", "en": "det(C)=−2<0, so C is not invertible" }, "expected": "reject", "reason": { "zh-Hant": "判準是 det≠0；C⁻¹=[[−2,1],[3/2,−1/2]] 存在。", "en": "The criterion is det≠0; C⁻¹=[[−2,1],[3/2,−1/2]] exists." } }, { "claim": { "zh-Hant": "逐項乘積 [[2,−2],[−4,0]] 的行列式等於 det(A)det(B)", "en": "The entrywise product [[2,−2],[−4,0]] has determinant det(A)det(B)" }, "expected": "reject", "reason": { "zh-Hant": "它的行列式是 −8，而 det(A)det(B)=56。", "en": "Its determinant is −8, whereas det(A)det(B)=56." } }, { "claim": { "zh-Hant": "det(AB)=det(BA)，所以 AB=BA", "en": "det(AB)=det(BA), so AB=BA" }, "expected": "reject", "reason": { "zh-Hant": "AB 的左上角是 6，BA 的左上角是 4。", "en": "The top-left entry of AB is 6 and that of BA is 4." } }, { "claim": { "zh-Hant": "det(−3A)=−3·det(A)=−21", "en": "det(−3A)=−3·det(A)=−21" }, "expected": "reject", "reason": { "zh-Hant": "2×2 時 det(tM)=t²det(M)，所以是 9·7=63。", "en": "For 2×2 matrices det(tM)=t²det(M), so it is 9·7=63." } }, { "claim": { "zh-Hant": "A 的元素是整數且 det(A)=7≠0，所以 A 有整數反矩陣", "en": "A has integer entries and det(A)=7≠0, so A has an integer inverse" }, "expected": "reject", "reason": { "zh-Hant": "唯一的反矩陣 (1/7)[[3,−1],[1,2]] 含非整數 3/7。", "en": "The unique inverse (1/7)[[3,−1],[1,2]] contains the non-integer 3/7." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "id": "mul_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "乘法結合律：(uv)w=u(vw)。", "en": "Associativity of multiplication: (uv)w=u(vw)." } }, { "id": "matrix_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。", "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative." } }, { "id": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "neg_mul", "kind": "prerequisite", "statement": { "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。", "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv." } }, { "id": "eq_refl_symm_trans", "kind": "prerequisite", "statement": { "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。", "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "id": "frac_rules", "kind": "prerequisite", "statement": { "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。", "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0." } }, { "id": "contradiction", "kind": "prerequisite", "statement": { "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。", "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds." } }, { "id": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "id": "iff_intro", "kind": "prerequisite", "statement": { "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。", "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold." } }, { "id": "divides_def", "kind": "prerequisite", "statement": { "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。", "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "mko_det_def", "kind": "cited", "source": { "mko_id": "mko-determinant", "sha256": "d49050d70b58e0d9d2cab2f5ddc64133813f5a99418865b5941af4f1ec240093", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對實矩陣 M=[[a,b],[c,d]]，本輪 D(M)=det(M)=ad−bc。M 在實數域可逆當且僅當 det(M)≠0；對同為2×2的實矩陣 A、B，det(AB)=det(A)det(B)，且 det(tM)=t²det(M)。", "en": "For M=[[a,b],[c,d]] over R, define D(M)=det(M)=ad−bc. M is invertible over R exactly when det(M)≠0. For 2×2 real matrices A,B, det(AB)=det(A)det(B), and det(tM)=t²det(M)." } }, { "id": "matrix_alg", "kind": "prerequisite", "statement": { "zh-Hant": "2×2 矩陣乘法結合：(XY)Z=X(YZ)；I=[[1,0],[0,1]] 滿足 IX=XI=X；純量可提出：X(kY)=(kX)Y=k(XY)；tM 是每個元素乘 t。", "en": "Multiplication of 2×2 matrices is associative: (XY)Z=X(YZ); I=[[1,0],[0,1]] satisfies IX=XI=X; scalars factor out: X(kY)=(kX)Y=k(XY); tM multiplies every entry by t." } }, { "id": "L_det_mul", "kind": "lemma", "proved_in": [ "S04.1", "S04.2", "S04.3", "S04.4" ], "statement": { "zh-Hant": "對所有 2×2 實矩陣 A、B：det(AB)=det(A)det(B)。", "en": "For all 2×2 real matrices A and B: det(AB)=det(A)det(B)." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S02", "lhs": "2*3-1*(-1)", "rhs": "7" }, { "id": "N2", "step": "S02", "lhs": "1*0-(-2)*4", "rhs": "8" }, { "id": "N3", "step": "S05", "lhs": "2*1+1*4", "rhs": "6" }, { "id": "N4", "step": "S05", "lhs": "2*(-2)+1*0", "rhs": "-4" }, { "id": "N5", "step": "S05", "lhs": "(-1)*1+3*4", "rhs": "11" }, { "id": "N6", "step": "S05", "lhs": "(-1)*(-2)+3*0", "rhs": "2" }, { "id": "N7", "step": "S05", "lhs": "6*2-(-4)*11", "rhs": "56" }, { "id": "N8", "step": "S05", "lhs": "7*8", "rhs": "56" }, { "id": "N9", "step": "S06", "lhs": "1*2+(-2)*(-1)", "rhs": "4" }, { "id": "N10", "step": "S06", "lhs": "1*1+(-2)*3", "rhs": "-5" }, { "id": "N11", "step": "S06", "lhs": "4*2+0*(-1)", "rhs": "8" }, { "id": "N12", "step": "S06", "lhs": "4*1+0*3", "rhs": "4" }, { "id": "N13", "step": "S06", "lhs": "4*4-(-5)*8", "rhs": "56" }, { "id": "N14", "step": "S04", "vars": [ "a", "b", "c", "d", "e", "f", "g", "h" ], "degree": 2, "lhs": "(a*e+b*g)*(c*f+d*h)-(a*f+b*h)*(c*e+d*g)", "rhs": "(a*d-b*c)*(e*h-f*g)" }, { "id": "N15", "step": "S07", "vars": [ "t", "a", "b", "c", "d" ], "degree": 2, "lhs": "(t*a)*(t*d)-(t*b)*(t*c)", "rhs": "t**2*(a*d-b*c)" }, { "id": "N16", "step": "S08", "lhs": "(-6)*(-9)-(-3)*3", "rhs": "63" }, { "id": "N17", "step": "S08", "lhs": "(-3)**2*7", "rhs": "63" }, { "id": "N18", "step": "S09", "vars": [ "a", "b" ], "degree": 2, "lhs": "a*(-b)+b*a", "rhs": "0" }, { "id": "N19", "step": "S09", "vars": [ "c", "d" ], "degree": 2, "lhs": "c*d+d*(-c)", "rhs": "0" }, { "id": "N20", "step": "S13", "lhs": "2*3+1*1", "rhs": "7" }, { "id": "N21", "step": "S13", "lhs": "2*(-1)+1*2", "rhs": "0" }, { "id": "N22", "step": "S13", "lhs": "(-1)*3+3*1", "rhs": "0" }, { "id": "N23", "step": "S13", "lhs": "(-1)*(-1)+3*2", "rhs": "7" }, { "id": "N24", "step": "S15", "lhs": "1*4-2*3", "rhs": "-2" }, { "id": "N25", "step": "S15", "lhs": "1*(-2)+2*(3/2)", "rhs": "1" }, { "id": "N26", "step": "S15", "lhs": "1*1+2*(-1/2)", "rhs": "0" }, { "id": "N27", "step": "S15", "lhs": "3*(-2)+4*(3/2)", "rhs": "0" }, { "id": "N28", "step": "S15", "lhs": "3*1+4*(-1/2)", "rhs": "1" }, { "id": "N29", "step": "S12", "lhs": "2*2-4*1", "rhs": "0" }, { "id": "N30", "step": "S12", "lhs": "2*2+4*(-1)", "rhs": "0" }, { "id": "N31", "step": "S14", "lhs": "3%7", "relation": "!=", "rhs": "0" }, { "id": "N32", "lhs": "2*0-(-2)*(-4)", "rhs": "-8" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 範圍內：2×2 方陣、實數域、通常的行乘列矩陣乘積。完整證明乘積公式、縮放公式與實數域可逆判準；不處理高階行列式、一般環上的可逆判準，也不做 Lean 的 det_fin_two 橋接；不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO's scope: 2×2 square matrices, the real field, the usual row-by-column matrix product. It fully proves the product rule, the scaling rule and the invertibility criterion over ℝ; it does not treat higher-order determinants or invertibility criteria over general rings, and does not build the Lean det_fin_two bridge; it does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "\\det(M)", "2×2 矩陣 M=[[a,b],[c,d]] 的行列式 ad−bc", "the determinant ad−bc of a 2×2 matrix M=[[a,b],[c,d]]" ], [ "AB", "行乘列的矩陣乘積", "the row-by-column matrix product" ], [ "tM", "純量 t 乘矩陣 M", "the scalar multiple of M by t" ], [ "\\operatorname{adj}(M)", "伴隨矩陣 [[d,−b],[−c,a]]", "the adjugate [[d,−b],[−c,a]]" ], [ "I", "2×2 單位矩陣", "the 2×2 identity matrix" ] ], "md_error_controls": [ { "title": "非零不必為正 / Nonzero need not be positive", "zh": "MKO 的誤解之一是以為可逆需要正的行列式。det(C)=−2<0，但 C 可逆，反矩陣在 S15 明確寫出。判準是 det≠0，不是 det>0，也不是 det=1。", "en": "One misconception the MKO names is that invertibility needs a positive determinant. det(C)=−2<0, yet C is invertible, with the inverse written out in S15. The criterion is det≠0, not det>0 and not det=1." }, { "title": "AB 不是逐項乘積 / AB is not the entrywise product", "zh": "A 與 B 的逐項乘積是 [[2,−2],[−4,0]]，行列式 2·0−(−2)(−4)=−8，不是 56。乘積公式只對行乘列的 AB 成立；而且它不表示 AB=BA（S06）。", "en": "The entrywise product of A and B is [[2,−2],[−4,0]], with determinant 2·0−(−2)(−4)=−8, not 56. The product rule holds for the row-by-column product AB only, and it does not mean AB=BA (S06)." }, { "title": "可逆判準依賴數域 / The criterion depends on the number system", "zh": "在 ℝ 上 det(A)=7≠0 足以可逆；但 A⁻¹ 的元素是 3/7 等分數，而反矩陣唯一，所以 A 沒有整數反矩陣（S14）。忽略數域是 MKO 列出的誤解。", "en": "Over ℝ, det(A)=7≠0 is enough for invertibility; but the entries of A⁻¹ are fractions such as 3/7, and the inverse is unique, so A has no integer inverse (S14). Ignoring the number system is a misconception the MKO lists." } ], "md_open_items": [ { "zh-Hant": "本稿只處理 2×2；高階行列式與整數環上的一般可逆判準不在範圍內。", "en": "This guide treats 2×2 matrices only; higher-order determinants and the general invertibility criterion over the integers are out of scope." }, { "zh-Hant": "matrix_alg（矩陣乘法結合、單位矩陣、純量可提出）作為前提使用，未在本稿逐項證明。", "en": "matrix_alg (associativity of matrix multiplication, the identity matrix, factoring out scalars) is used as a prerequisite and is not proved entry by entry here." }, { "zh-Hant": "MKO 的 formal next（det_fin_two、det_mul、det_smul 與實數域 inverse 的橋接）未做。", "en": "The MKO's formal next step (bridging det_fin_two, det_mul, det_smul and the real-field inverse) is not done." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "a1bb7c2c935079e026c6f8947b4a0da75f8f3294086082e985d4bf8fb3003bb9", "original_source_sha256": "a1bb7c2c935079e026c6f8947b4a0da75f8f3294086082e985d4bf8fb3003bb9", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }