Canonical URI: https://w3id.org/kmath/textbook_solution/chunjae_hong_solution_p155_p135_problem_03
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 천재홍 p135 문제 3 풀이 |
| rdfs:comment | 천재홍 공통수학1 정답 및 풀이 p155의 p135 문제 3 풀이. |
| math:answerText | \(AE=EA=A\) |
| math:explanationText | \[ AE=\begin{pmatrix}2&1\\-1&3\end{pmatrix} \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}2&1\\-1&3\end{pmatrix}=A \] \[ EA=\begin{pmatrix}1&0\\0&1\end{pmatrix} \begin{pmatrix}2&1\\-1&3\end{pmatrix} = \begin{pmatrix}2&1\\-1&3\end{pmatrix}=A \] 따라서 \(AE=EA=A\)이다. |
| math:mappingConfidence | 1.0 |
| math:pageStart | 155 |
| math:problem | textbook_problem:chunjae_hong_vision_p135_problem_03 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:identity_matrix_product_property |