Canonical URI: https://w3id.org/kmath/textbook_solution/chunjae_hong_solution_p155_p129_problem_02
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 천재홍 p129 문제 2 풀이 |
| rdfs:comment | 천재홍 공통수학1 정답 및 풀이 p155의 p129 문제 2 풀이. |
| math:answerText | \(A+B=B+A\), \((A+B)+C=A+(B+C)\) |
| math:explanationText | (1) \[ A+B=\begin{pmatrix}0&11\\13&4\end{pmatrix},\quad B+A=\begin{pmatrix}0&11\\13&4\end{pmatrix} \] 이므로 \(A+B=B+A\)이다. (2) \[ A+B=\begin{pmatrix}0&11\\13&4\end{pmatrix},\quad B+C=\begin{pmatrix}-5&4\\9&0\end{pmatrix} \] 이고 \[ (A+B)+C = \begin{pmatrix}0&11\\13&4\end{pmatrix} + \begin{pmatrix}-2&-1\\5&-1\end{pmatrix} = \begin{pmatrix}-2&10\\18&3\end{pmatrix} \] \[ A+(B+C) = \begin{pmatrix}3&6\\9&3\end{pmatrix} + \begin{pmatrix}-5&4\\9&0\end{pmatrix} = \begin{pmatrix}-2&10\\18&3\end{pmatrix} \] 따라서 \((A+B)+C=A+(B+C)\)이다. |
| math:mappingConfidence | 1.0 |
| math:pageStart | 155 |
| math:problem | textbook_problem:chunjae_hong_vision_p129_problem_02 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:matrix_linear_combination_entries |