Merge current main into PR61 finishing branch
This commit is contained in:
amosblomqvist
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[
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo321f7_4202fae0",
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"source": "messages.1.content.1.arguments.explanation",
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"model": "google/gemini-3.7-flash",
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"kind": "percent-at-display-end",
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"raw": "To find the posterior probability $P(D \\\\mid +)$, we apply Bayes' theorem:\n$$\nP(D \\\\mid +) = \\\\frac{P(D \\\\cap +)}{P(+)} = \\\\frac{P(D)P(+ \\\\mid D)}{P(D)P(+ \\\\mid D) + P(\\\\neg D)P(+ \\\\mid \\\\neg D)}\n$$\n\nGiven:\n- $P(D) = 0.01$ (and thus $P(\\\\neg D) = 0.99$)\n- $P(+ \\\\mid D) = 0.90$\n- $P(+ \\\\mid \\\\neg D) = 0.05$\n\nEvaluating the numerator and denominator:\n$$\nP(D \\\\cap +) = 0.01 \\\\times 0.90 = 0.009\n$$\n$$\nP(\\\\neg D \\\\cap +) = 0.99 \\\\times 0.05 = 0.0495\n$$\n$$\nP(+) = 0.009 + 0.0495 = 0.0585\n$$\n\nTherefore:\n$$\nP(D \\\\mid +) = \\\\frac{0.009}{0.0585} = \\\\frac{9}{58.5} = \\\\frac{2}{13} \\\\approx 15.4\\\\%\n$$"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo321f7_4202fae0",
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"source": "messages.5.content.1.arguments.explanation",
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"model": "google/gemini-3.7-flash",
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"kind": "percent-at-display-end",
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"raw": "Because the two tests are conditionally independent given disease status:\n$$\nP(T_1 = +, T_2 = + \\\\mid D) = P(T_1 = + \\\\mid D) \\\\cdot P(T_2 = + \\\\mid D) = 0.90 \\\\times 0.90 = 0.81\n$$\n$$\nP(T_1 = +, T_2 = + \\\\mid \\\\neg D) = P(T_1 = + \\\\mid \\\\neg D) \\\\cdot P(T_2 = + \\\\mid \\\\neg D) = 0.10 \\\\times 0.10 = 0.01\n$$\n\nNow calculate the joint probabilities:\n$$\nP(D \\\\cap T_1 = + \\\\cap T_2 = +) = P(D) \\\\cdot P(T_1 = +, T_2 = + \\\\mid D) = 0.01 \\\\times 0.81 = 0.0081\n$$\n$$\nP(\\\\neg D \\\\cap T_1 = + \\\\cap T_2 = +) = P(\\\\neg D) \\\\cdot P(T_1 = +, T_2 = + \\\\mid \\\\neg D) = 0.99 \\\\times 0.01 = 0.0099\n$$\n\nTotal probability of two positive tests:\n$$\nP(T_1 = +, T_2 = +) = 0.0081 + 0.0099 = 0.0180\n$$\n\nApplying Bayes' theorem:\n$$\nP(D \\\\mid T_1 = +, T_2 = +) = \\\\frac{0.0081}{0.0180} = \\\\frac{81}{180} = \\\\frac{9}{20} = 45\\\\%\n$$"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo4op90_74572548",
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"source": "messages.1.content.2.arguments.explanation",
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"language": "pt",
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"model": "deepseek/deepseek-v4.1-flash",
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"kind": "percent-at-display-end",
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"raw": "Em $10\\\\,000$ pessoas: $100$ estão doentes e $99$ dessas testam positivo; $9\\\\,900$ estão saudáveis e $5\\\\%$ delas, isto é $495$, testam positivo. Total de positivos: $99 + 495 = 594$. Entre os positivos, a fração de doentes é\\n$$\\n\\\\frac{99}{594} = \\\\frac{1}{6} \\\\approx 16{,}7\\\\%\\n$$\\nA base pequena de doentes faz os falsos positivos ($495$) dominarem os verdadeiros positivos ($99$)."
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo4op90_74572548",
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"source": "messages.3.content.2.arguments.explanation",
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"language": "pt",
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"model": "deepseek/deepseek-v4.1-flash",
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"kind": "percent-at-display-end",
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"raw": "Continua valendo a mesma estrutura de Bayes, agora com números diferentes. Em $10\\\\,000$ pessoas: $200$ doentes, das quais $90\\\\%$ testam positivo $= 180$; $9\\\\,800$ saudáveis, das quais $5\\\\%$ testam positivo $= 490$ falsos positivos. Total de positivos: $180 + 490 = 670$.\\n$$\\nP(\\\\text{doente} \\\\mid +) = \\\\frac{180}{670} \\\\approx 26{,}9\\\\%\\n$$\\nNote que a especificidade $95\\\\%$ entra na conta pela sua taxa de falso positivo $1 - 0{,}95 = 0{,}05$, pesada por $P(\\\\text{saudável}) = 0{,}98$."
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo8kgtc_65767ffd",
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"source": "messages.5.content.1.arguments.question",
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"language": "es",
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"model": "xiaomi/mimo-v2.6-flash",
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"kind": "document-math-framing",
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"raw": "Una enfermedad afecta al $1\\%$ de una población. La prueba tiene una sensibilidad del $99\\%$ (probabilidad de dar positivo si la enfermedad está presente) y una tasa de falsos positivos del $5\\%$ (probabilidad de dar positivo si no está presente). Una persona al azar da positivo. ¿Cuál es la probabilidad de que realmente tenga la enfermedad?\\n\\n$$\\nP(\\text{enf} \\mid +) \\;=\\; ?$$"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo91klg_5332fa79",
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"source": "messages.3.content.1.arguments.explanation",
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"language": "zh-Hans",
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"model": "xiaomi/mimo-v2.6-flash",
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"kind": "percent-at-display-end",
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"raw": "条件概率把样本空间缩小到条件所指定的那一部分。已知\"检测阳性\"，人群就从 $1000$ 人缩到 $140$ 人，其中患病的有 $45$ 人，所以\\n$$\\nP(\\\\text{患病}\\\\mid\\\\text{阳性})=\\\\frac{P(\\\\text{患病}\\\\cap\\\\text{阳性})}{P(\\\\text{阳性})}=\\\\frac{45/1000}{140/1000}=\\\\frac{45}{140}\\\\approx 32.1\\\\%.\\n$$\\n另外三个数各有各的含义：$\\\\frac{45}{50}=90\\\\%$ 是 $P(\\\\text{阳性}\\\\mid\\\\text{患病})$（敏感度，条件方向反了）；$\\\\frac{45}{1000}=4.5\\\\%$ 是联合概率 $P(\\\\text{患病}\\\\cap\\\\text{阳性})$（没做条件化）；$\\\\frac{140}{1000}=14\\\\%$ 是 $P(\\\\text{阳性})$（不分是否患病的总阳性率）。"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo91klg_5332fa79",
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"source": "messages.7.content.1.arguments.explanation",
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"language": "zh-Hans",
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"model": "xiaomi/mimo-v2.6-flash",
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"kind": "percent-at-display-end",
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"raw": "用 $10000$ 人（或直接用全概率公式）算分母：\\n$$\\n\\\\begin{aligned}\\n\\\\text{患病且阳性}&=10000\\\\times 0.01\\\\times 0.99=99,\\\\\\\\\\n\\\\text{未患病但阳性}&=10000\\\\times 0.99\\\\times 0.05=495.\\n\\\\end{aligned}\\n$$\\n阳性总数 $=99+495=594$，所以\\n$$\\nP(D\\\\mid +)=\\\\frac{99}{594}\\\\approx 16.7\\\\%.\\n$$\\n各错误选项对应的具体失误：$99\\\\%$ 是把敏感度 $P(+\\\\mid D)$ 当成了 $P(D\\\\mid +)$（条件方向颠倒）；$95.2\\\\%=\\\\frac{0.99}{0.99+0.05}$ 是把阳性率和假阳性率直接当人数用，漏掉了基础发生率的加权（$99\\\\%$ 的人未患病，所以少量假阳性也架不住人多）；$1\\\\%$ 是看到了阳性结果却不更新概率。"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo91klg_5332fa79",
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"source": "messages.9.content.1.arguments.explanation",
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"language": "zh-Hans",
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"model": "xiaomi/mimo-v2.6-flash",
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"kind": "percent-at-display-end",
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"raw": "两次阳性后的后验，用同样的公式但似然要平方：\\n$$\\n\\\\begin{aligned}\\nP(++\\\\mid D)&=0.99^2=0.9801,\\\\\\\\\\nP(++\\\\mid D')&=0.05^2=0.0025.\\n\\\\end{aligned}\\n$$\\n$$\\nP(D\\\\mid ++)\\\\frac{0.99^2\\\\times 0.01}{0.99^2\\\\times 0.01+0.05^2\\\\times 0.99}=\\\\frac{0.009801}{0.009801+0.002475}\\\\approx 79.8\\\\%.\\n$$\\n关键在于**两条分支都要平方**：患病分支的似然变成 $0.99^2$，未患病分支的假阳性率变成 $0.05^2$（$5\\%$ 是很小的数，平方后变得更小，所以假阳性这一侧收缩得比患病侧更狠，后验从 $16.7\\\\%$ 跳到约 $80\\\\%$）。其他选项的失误：$98\\\\%$ 是把 $P(++\\\\mid D)$ 当成了 $P(D\\\\mid ++)$；$16.5\\\\%$ 只对患病分支用了两次独立，未患病分支仍用单次假阳性率；$33\\\\%$ 是把单次后验直接乘 $2$——两次阳性提供的证据不是相加的，而要在似然层面相乘。"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo9c830_3866e4a6",
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"source": "messages.7.content.1.arguments.explanation",
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"language": "zh-Hans",
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"model": "anthropic/claude-haiku-5.5",
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"kind": "percent-at-display-end",
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"raw": "阳性总人数：真阳性 $100\\\\times0.9=90$，假阳性 $9900\\\\times0.05=495$，合计 $585$。\n\n所求的是阳性者中患病者的占比，分母是全部阳性者：\n$$\n\\\\frac{90}{585}\\\\approx15.4\\\\%\n$$\n\n$90\\\\%$ 是患病者中阳性的比例，分母错成了患病者；$84.6\\\\%$ 是阳性者中健康者的占比，回答的是另一个问题；$5\\\\%$ 是健康者中的假阳性率，分母是健康人群。"
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},
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{
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"group": "multilingual-20261025-01",
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"chatId": "chat_mvo9nu8w_dcc7add9",
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"source": "messages.3.content.1.arguments.explanation",
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"language": "id",
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"model": "xiaomi/mimo-v2.6-flash",
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"kind": "percent-at-display-end",
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"raw": "Pakai aturan Bayes dengan menghitung dulu kejadian gabungan dalam satu kelompok besar, misalnya $10.000$ orang:\\n\\n- Sakit dan positif: $10.000 \\\\times 0{,}01 \\\\times 0{,}99 = 99$ orang.\\n- Sehat dan positif: $10.000 \\\\times 0{,}99 \\\\times 0{,}05 = 495$ orang.\\n\\nSemua yang hasilnya positif: $99 + 495 = 594$ orang. Maka\\n$$\\nP(\\\\text{sakit}\\\\mid +) = \\\\frac{99}{594} = \\\\frac{1}{6} \\\\approx 16{,}7\\\\%.\\n$$\\n\\nYang menipu di sini adalah base rate: kelompok sehat jauh lebih besar, sehingga meski hanya 5% dari mereka yang salah positif, jumlahnya tetap 5 kali lipat jumlah positif yang benar. Nilai $95{,}2\\\\%$ muncul kalau kelompok sakit dan sehat dianggap sama besar ($\\\\frac{99}{99+5}$); nilai $99\\\\%$ muncul kalau arah kondisinya dibalik, membaca $P(+\\\\mid \\\\text{sakit})$ sebagai $P(\\\\text{sakit}\\\\mid +)$."
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}
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]
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@@ -0,0 +1,135 @@
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{
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"notice": "Unmodified model-authored fields from the frozen multilingual corpus. oldDisplay records the diagnosed application damage, not an expected output.",
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"cases": [
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{
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"run": "pilot-20261025-01",
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"model": "openai/gpt-6-luna",
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"language": "en",
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"replicate": 1,
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"chatId": "chat_mvo1rzjs_056eac45",
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"source": "messages.5.content.1.arguments.explanation",
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"raw": "For a $2\\times2$ matrix, $\\det\\begin{bmatrix}a&b\\\\c&d\\end{bmatrix}=ad-bc$. Here $3\\cdot2-6\\cdot1=0$. A square matrix is invertible exactly when its determinant is nonzero, so $M$ is singular. Its columns are also proportional, which matches the determinant-zero result.",
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"oldDisplay": "For a $2\\times2$ matrix, $\\det\\begin{bmatrix}a&b\\c&d\\end{bmatrix}=ad-bc$. Here $3\\cdot2-6\\cdot1=0$. A square matrix is invertible exactly when its determinant is nonzero, so $M$ is singular. Its columns are also proportional, which matches the determinant-zero result."
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},
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{
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"run": "pilot-20261025-01",
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"model": "openai/gpt-6-luna",
|
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"language": "en",
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"replicate": 1,
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"chatId": "chat_mvo1sdnj_8f2d8c5e",
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"source": "messages.9.content.1.arguments.explanation",
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"raw": "The Jacobian has one row per output and one column per input: $J_F=\\begin{pmatrix}2x&1\\\\y&x\\end{pmatrix}$. At $(1,2)$ this is $\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}$.",
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"oldDisplay": "The Jacobian has one row per output and one column per input: $J_F=\\begin{pmatrix}2x&1\\y&x\\end{pmatrix}$. At $(1,2)$ this is $\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}$."
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},
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{
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"run": "pilot-20261025-01",
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"model": "anthropic/claude-haiku-5.5",
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"language": "en",
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"replicate": 1,
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"chatId": "chat_mvo30cfe_57114723",
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"source": "messages.3.content.1.arguments.explanation",
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"raw": "$A-\\lambda I=\\begin{pmatrix}2-\\lambda&1\\\\1&2-\\lambda\\end{pmatrix}$, and for a $2\\times2$ matrix $\\det\\begin{pmatrix}p&q\\\\r&s\\end{pmatrix}=ps-qr$. So $$\\det(A-\\lambda I)=(2-\\lambda)^2-1\\cdot1=\\lambda^2-4\\lambda+3.$$ Its roots $\\lambda=1,3$ are the eigenvalues. Option $b$ drops the minus sign in $ps-qr$; $c$ subtracts the constant incorrectly; $d$ flips the sign of the linear term.",
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"oldDisplay": "$A-\\lambda I=\\begin{pmatrix}2-\\lambda&1\\\\1&2-\\lambda\\end{pmatrix}$, and for a $2\\times2$ matrix $\\det\\begin{pmatrix}p&q\\r&s\\end{pmatrix}=ps-qr$. So $$\\det(A-\\lambda I)=(2-\\lambda)^2-1\\cdot1=\\lambda^2-4\\lambda+3.$$ Its roots $\\lambda=1,3$ are the eigenvalues. Option $b$ drops the minus sign in $ps-qr$; $c$ subtracts the constant incorrectly; $d$ flips the sign of the linear term."
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},
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{
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"run": "multilingual-20261025-01-zh-Hans-r1",
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"model": "openai/gpt-6-luna",
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"language": "zh-Hans",
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"replicate": 1,
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"chatId": "chat_mvo60bsn_1cbbe884",
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"source": "messages.3.content.1.arguments.question",
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"raw": "对 $2\\times2$ 矩阵 $A=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}$，以下哪组关于行列式的说法全都正确？",
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"oldDisplay": "对 $2\\times2$ 矩阵 $A=\\begin{pmatrix}a&b\\c&d\\end{pmatrix}$，以下哪组关于行列式的说法全都正确？"
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},
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{
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"run": "multilingual-20261025-01-zh-Hans-r1",
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"model": "xiaomi/mimo-v2.6-flash",
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"language": "zh-Hans",
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"replicate": 1,
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"chatId": "chat_mvo66u6w_a233fc16",
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"source": "messages.9.content.2.arguments.explanation",
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"raw": "雅可比矩阵按“第 $i$ 行 = 第 $i$ 个分量对所有自变量的偏导”排布：\n$$\nJ_{\\mathbf F}=\n\\begin{pmatrix}\n\\dfrac{\\partial (xy)}{\\partial x} & \\dfrac{\\partial (xy)}{\\partial y}\\\\[6pt]\n\\dfrac{\\partial (x+y^2)}{\\partial x} & \\dfrac{\\partial (x+y^2)}{\\partial y}\n\\end{pmatrix}\n=\n\\begin{pmatrix}\ny & x\\\\\n1 & 2y\n\\end{pmatrix}.\n$$\n注意 $\\partial(xy)/\\partial x=y$（$y$ 当常数），$\\partial(xy)/\\partial y=x$。转置形式 $\\begin{pmatrix}y&1\\\\x&2y\\end{pmatrix}$ 把偏导排成了列；$\\begin{pmatrix}y&x\\\\2y&1\\end{pmatrix}$ 把第二行的两个分量次序调换了；$\\begin{pmatrix}x&y\\\\1&2y\\end{pmatrix}$ 把第一行的 $x,y$ 位置写反了。",
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"oldDisplay": "雅可比矩阵按“第 $i$ 行 = 第 $i$ 个分量对所有自变量的偏导”排布：\n$$\nJ_{\\mathbf F}=\n\\begin{pmatrix}\n\\dfrac{\\partial (xy)}{\\partial x} & \\dfrac{\\partial (xy)}{\\partial y}\\\\[6pt]\n\\dfrac{\\partial (x+y^2)}{\\partial x} & \\dfrac{\\partial (x+y^2)}{\\partial y}\n\\end{pmatrix}\n=\n\\begin{pmatrix}\ny & x\\\\\n1 & 2y\n\\end{pmatrix}.\n$$\n注意 $\\partial(xy)/\\partial x=y$（$y$ 当常数），$\\partial(xy)/\\partial y=x$。转置形式 $\\begin{pmatrix}y&1\\x&2y\\end{pmatrix}$ 把偏导排成了列；$\\begin{pmatrix}y&x\\\\2y&1\\end{pmatrix}$ 把第二行的两个分量次序调换了；$\\begin{pmatrix}x&y\\\\1&2y\\end{pmatrix}$ 把第一行的 $x,y$ 位置写反了。"
|
||||
},
|
||||
{
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||||
"run": "multilingual-20261025-01-id-r1",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "id",
|
||||
"replicate": 1,
|
||||
"chatId": "chat_mvo6q2lq_6af14437",
|
||||
"source": "messages.7.content.1.arguments.explanation",
|
||||
"raw": "Pada $(x,y)=(1,0)$, $H=(1,1)$. Jacobian $G$ di $(1,1)$ adalah $\\begin{pmatrix}2u&1\\\\v&u\\end{pmatrix}=\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}$, sedangkan $J_H=\\begin{pmatrix}1&1\\\\1&-1\\end{pmatrix}$. Aturan rantai dengan konvensi ini memberi $J_{G\\circ H}=J_G(H(x,y))J_H(x,y)$, yakni $\\begin{pmatrix}3&1\\\\2&0\\end{pmatrix}$.",
|
||||
"oldDisplay": "Pada $(x,y)=(1,0)$, $H=(1,1)$. Jacobian $G$ di $(1,1)$ adalah $\\begin{pmatrix}2u&1\\v&u\\end{pmatrix}=\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}$, sedangkan $J_H=\\begin{pmatrix}1&1\\\\1&-1\\end{pmatrix}$. Aturan rantai dengan konvensi ini memberi $J_{G\\circ H}=J_G(H(x,y))J_H(x,y)$, yakni $\\begin{pmatrix}3&1\\\\2&0\\end{pmatrix}$."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-id-r1",
|
||||
"model": "z-ai/glm-5.3-flash",
|
||||
"language": "id",
|
||||
"replicate": 1,
|
||||
"chatId": "chat_mvo77f3x_135f8d66",
|
||||
"source": "messages.7.content.2.arguments.explanation",
|
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"raw": "Vektor eigen memenuhi $(A-\\lambda I)v=0$. Untuk $\\lambda=2$:\n$$(A-2I)v=\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}\\begin{pmatrix}v_1\\\\v_2\\end{pmatrix}=0\\;\\Rightarrow\\;v_1+v_2=0\\;\\Rightarrow\\;v=\\begin{pmatrix}1\\\\-1\\end{pmatrix}$$\n(bisa dikali skalar berapa pun). Cek: $Av=\\begin{pmatrix}3-1\\\\-2\\end{pmatrix}=\\begin{pmatrix}2\\\\-2\\end{pmatrix}=2v$. Perhatikan $v=(1,0)$ adalah vektor eigen untuk $\\lambda=3$, bukan $\\lambda=2$.",
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"oldDisplay": "Vektor eigen memenuhi $(A-\\lambda I)v=0$. Untuk $\\lambda=2$:\n$$(A-2I)v=\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}\\begin{pmatrix}v_1\\v_2\\end{pmatrix}=0\\;\\Rightarrow\\;v_1+v_2=0\\;\\Rightarrow\\;v=\\begin{pmatrix}1\\\\-1\\end{pmatrix}$$\n(bisa dikali skalar berapa pun). Cek: $Av=\\begin{pmatrix}3-1\\\\-2\\end{pmatrix}=\\begin{pmatrix}2\\\\-2\\end{pmatrix}=2v$. Perhatikan $v=(1,0)$ adalah vektor eigen untuk $\\lambda=3$, bukan $\\lambda=2$."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-en-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "en",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo7ei0l_9263330e",
|
||||
"source": "messages.3.content.1.arguments.question",
|
||||
"raw": "For a $2\\times2$ matrix $A=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}$, which statement correctly links its determinant to area scaling and invertibility?",
|
||||
"oldDisplay": "For a $2\\times2$ matrix $A=\\begin{pmatrix}a&b\\c&d\\end{pmatrix}$, which statement correctly links its determinant to area scaling and invertibility?"
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-en-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "en",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo7exl3_e1fe93cd",
|
||||
"source": "messages.1.content.1.arguments.explanation",
|
||||
"raw": "First evaluate the Jacobian of $F$ at $(1,2)$:\n$$\nJ_F(x,y)=\\begin{pmatrix}2x&1\\\\y&x\\end{pmatrix},\\qquad J_F(1,2)=\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}.\n$$\nThe velocity is $\\gamma'(1)=(1,2)$. The chain rule gives\n$$\n\\frac{d}{dt}F(\\gamma(t))\\bigg|_{t=1}=J_F(1,2)\\begin{pmatrix}1\\\\2\\end{pmatrix}=\\begin{pmatrix}4\\\\4\\end{pmatrix}.\n$$\nWait, recomputing the component derivatives directly: $F(\\gamma(t))=(t^2+(1+t^2),\\ t(1+t^2))=(1+2t^2,t+t^3)$, whose derivative at $1$ is $(4,4)$. The listed options do not include this value, so this question is invalid.",
|
||||
"oldDisplay": "First evaluate the Jacobian of $F$ at $(1,2)$:\n$$\nJ_F(x,y)=\\begin{pmatrix}2x&1\\y&x\\end{pmatrix},\\qquad J_F(1,2)=\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}.\n$$\nThe velocity is $\\gamma'(1)=(1,2)$. The chain rule gives\n$$\n\\frac{d}{dt}F(\\gamma(t))\\bigg|_{t=1}=J_F(1,2)\\begin{pmatrix}1\\\\2\\end{pmatrix}=\\begin{pmatrix}4\\\\4\\end{pmatrix}.\n$$\nWait, recomputing the component derivatives directly: $F(\\gamma(t))=(t^2+(1+t^2),\\ t(1+t^2))=(1+2t^2,t+t^3)$, whose derivative at $1$ is $(4,4)$. The listed options do not include this value, so this question is invalid."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-en-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "en",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo7exl3_e1fe93cd",
|
||||
"source": "messages.3.content.1.arguments.explanation",
|
||||
"raw": "For $F=(F_1,F_2)$, each row contains the partial derivatives of one output component: $J_F=\\begin{pmatrix}\\partial F_1/\\partial x&\\partial F_1/\\partial y\\\\\\partial F_2/\\partial x&\\partial F_2/\\partial y\\end{pmatrix}=\\begin{pmatrix}2x&1\\\\y&x\\end{pmatrix}$. Substituting $(x,y)=(1,2)$ gives $\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}$.",
|
||||
"oldDisplay": "For $F=(F_1,F_2)$, each row contains the partial derivatives of one output component: $J_F=\\begin{pmatrix}\\partial F_1/\\partial x&\\partial F_1/\\partial y\\\\\\partial F_2/\\partial x&\\partial F_2/\\partial y\\end{pmatrix}=\\begin{pmatrix}2x&1\\y&x\\end{pmatrix}$. Substituting $(x,y)=(1,2)$ gives $\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}$."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-pt-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "pt",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo7wn6c_d63ef3f1",
|
||||
"source": "messages.7.content.1.arguments.explanation",
|
||||
"raw": "Para uma matriz $2\\times2$, $\\det\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}=ad-bc$. Logo, $\\det(C)=2\\cdot3-1\\cdot5=1$. O determinante é o fator assinado de escala de área: valor absoluto $1$ preserva área, e sinal positivo indica que a orientação não é invertida. Como o determinante é diferente de zero, $C$ é invertível.",
|
||||
"oldDisplay": "Para uma matriz $2\\times2$, $\\det\\begin{pmatrix}a&b\\c&d\\end{pmatrix}=ad-bc$. Logo, $\\det(C)=2\\cdot3-1\\cdot5=1$. O determinante é o fator assinado de escala de área: valor absoluto $1$ preserva área, e sinal positivo indica que a orientação não é invertida. Como o determinante é diferente de zero, $C$ é invertível."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-es-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "es",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo8j89a_f6fc1cc0",
|
||||
"source": "messages.3.content.1.arguments.explanation",
|
||||
"raw": "La regla de la cadena es $D(F\\circ g)=DF(g(s,t))Dg(s,t)$. Aquí $g(1,2)=(3,2)$, $DF(u,v)=\\begin{pmatrix}2u&1\\\\v&u\\end{pmatrix}$ y $Dg(s,t)=\\begin{pmatrix}1&1\\\\t&s\\end{pmatrix}$. Así, $\\begin{pmatrix}6&1\\\\2&3\\end{pmatrix}\\begin{pmatrix}1&1\\\\2&1\\end{pmatrix}=\\begin{pmatrix}8&7\\\\8&5\\end{pmatrix}$.",
|
||||
"oldDisplay": "La regla de la cadena es $D(F\\circ g)=DF(g(s,t))Dg(s,t)$. Aquí $g(1,2)=(3,2)$, $DF(u,v)=\\begin{pmatrix}2u&1\\v&u\\end{pmatrix}$ y $Dg(s,t)=\\begin{pmatrix}1&1\\t&s\\end{pmatrix}$. Así, $\\begin{pmatrix}6&1\\\\2&3\\end{pmatrix}\\begin{pmatrix}1&1\\\\2&1\\end{pmatrix}=\\begin{pmatrix}8&7\\\\8&5\\end{pmatrix}$."
|
||||
},
|
||||
{
|
||||
"run": "multilingual-20261025-01-id-r2",
|
||||
"model": "openai/gpt-6-luna",
|
||||
"language": "id",
|
||||
"replicate": 2,
|
||||
"chatId": "chat_mvo9mesu_dbb56f1f",
|
||||
"source": "messages.9.content.1.arguments.explanation",
|
||||
"raw": "Untuk $\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}$, determinannya $ad-bc$. Jadi $\\det(C)=2\\cdot3-5\\cdot1=1$.",
|
||||
"oldDisplay": "Untuk $\\begin{pmatrix}a&b\\c&d\\end{pmatrix}$, determinannya $ad-bc$. Jadi $\\det(C)=2\\cdot3-5\\cdot1=1$."
|
||||
}
|
||||
]
|
||||
}
|
||||
@@ -0,0 +1,19 @@
|
||||
{
|
||||
"provenance": "multilingual-20261025-01; original collection findings, not the failed r1 rendering",
|
||||
"cases": [
|
||||
{
|
||||
"chatId": "chat_mvo6kxzl_540b570e",
|
||||
"source": "messages.1.content.2.arguments.explanation",
|
||||
"model": "z-ai/glm-5.3-flash",
|
||||
"raw": "乘法规则是\"左行的元素依次乘右列的元素再相加\"：\\n$$\\nAB=\\begin{pmatrix}1\\\\cdot3+2\\\\cdot0 & 1\\\\cdot1+2\\\\cdot2\\\\\\\\ 0\\\\cdot3+1\\\\cdot0 & 0\\\\cdot1+1\\\\cdot2\\\\end{pmatrix}=\\begin{pmatrix}3 & 5\\\\\\\\ 0 & 2\\\\end{pmatrix}\\n$$\\n选项 b 其实是 $BA$（$\\\\begin{pmatrix}3 & 7\\\\\\\\ 0 & 2\\\\end{pmatrix}$）——顺序交换后结果不同，这正是矩阵乘法不交换的例子。选项 c 是对应位置直接相加，那是加法不是乘法。",
|
||||
"oldDisplay": "乘法规则是\"左行的元素依次乘右列的元素再相加\"：\\n$$\\nAB=\\begin{pmatrix}1\\cdot3+2\\cdot0 & 1\\cdot1+2\\cdot2\\\\\\\\ 0\\cdot3+1\\cdot0 & 0\\cdot1+1\\cdot2\\end{pmatrix}=\\begin{pmatrix}3 & 5\\\\\\\\ 0 & 2\\end{pmatrix}\\n$$\\n选项 b 其实是 $BA$（$\\begin{pmatrix}3 & 7\\\\\\\\ 0 & 2\\end{pmatrix}$）——顺序交换后结果不同，这正是矩阵乘法不交换的例子。选项 c 是对应位置直接相加，那是加法不是乘法。"
|
||||
},
|
||||
{
|
||||
"chatId": "chat_mvo9oquq_74bfd7f6",
|
||||
"source": "messages.3.content.1.arguments.explanation",
|
||||
"model": "xiaomi/mimo-v2.6-flash",
|
||||
"raw": "Perkalian matriks $AB$ = baris-kolom: elemen $(i,j)$ hasil adalah titik-antara baris $i$ dari $A$ dengan kolom $j$ dari $B$.\\n\\n- $(1,1): (1,2)\\cdot(2,0) = 1\\cdot2 + 2\\cdot1 = 4$\\n- $(1,2): (1,2)\\cdot(0,2) = 1\\cdot0 + 2\\cdot2 = 4$\\n- $(2,1): (3,4)\\cdot(2,0) = 3\\cdot2 + 4\\cdot1 = 10$\\n- $(2,2): (3,4)\\cdot(0,2) = 3\\cdot0 + 4\\cdot2 = 8$\\n\\n$$AB = \\\\begin{pmatrix} 4 & 4 \\\\\\\\ 10 & 8 \\\\end{pmatrix}$$\\n\\nPilihan lain adalah tiga kesalahan berbeda: $\\begin{pmatrix} 2 & 4 \\\\ 7 & 10 \\\\end{pmatrix}$ adalah $BA$ (urutan dibalik), $\\begin{pmatrix} 4 & 10 \\\\ 4 & 8 \\\\end{pmatrix}$ adalah transpos hasil (peletakan elemen tertukar), dan $\\begin{pmatrix} 2 & 0 \\\\ 3 & 8 \\\\end{pmatrix}$ adalah perkalian elemen-per-elemen, bukan baris-kolom. Perhatikan juga bahwa $AB \\neq BA$ — perkalian matriks tidak komutatif.",
|
||||
"oldDisplay": "Perkalian matriks $AB$ = baris-kolom: elemen $(i,j)$ hasil adalah titik-antara baris $i$ dari $A$ dengan kolom $j$ dari $B$.\\n\\n- $(1,1): (1,2)\\cdot(2,0) = 1\\cdot2 + 2\\cdot1 = 4$\\n- $(1,2): (1,2)\\cdot(0,2) = 1\\cdot0 + 2\\cdot2 = 4$\\n- $(2,1): (3,4)\\cdot(2,0) = 3\\cdot2 + 4\\cdot1 = 10$\\n- $(2,2): (3,4)\\cdot(0,2) = 3\\cdot0 + 4\\cdot2 = 8$\\n\\n$$AB = \\begin{pmatrix} 4 & 4 \\\\\\\\ 10 & 8 \\end{pmatrix}$$\\n\\nPilihan lain adalah tiga kesalahan berbeda: $\\begin{pmatrix} 2 & 4 \\\\ 7 & 10 \\end{pmatrix}$ adalah $BA$ (urutan dibalik), $\\begin{pmatrix} 4 & 10 \\\\ 4 & 8 \\end{pmatrix}$ adalah transpos hasil (peletakan elemen tertukar), dan $\\begin{pmatrix} 2 & 0 \\\\ 3 & 8 \\end{pmatrix}$ adalah perkalian elemen-per-elemen, bukan baris-kolom. Perhatikan juga bahwa $AB \\neq BA$ — perkalian matriks tidak komutatif."
|
||||
}
|
||||
]
|
||||
}
|
||||
Reference in New Issue
Block a user