Canonical URI: https://w3id.org/kmath/textbook_solution/chunjae_hong_solution_p132_p132_example_02
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 천재홍 p132 예제 2 풀이 |
| rdfs:comment | 천재홍 공통수학1 교과서 p132 본문에서 이미지 판독으로 추출한 예제 2 풀이. |
| math:answerText | \(\begin{pmatrix}3&-3\\4&14\end{pmatrix}\) |
| math:explanationText | \[ 3(A+2B)+A-5B=3A+6B+A-5B=4A+B \] 이므로 \[ 4A+B= 4\begin{pmatrix}1&-2\\0&3\end{pmatrix} + \begin{pmatrix}-1&5\\4&2\end{pmatrix} = \begin{pmatrix}4&-8\\0&12\end{pmatrix} + \begin{pmatrix}-1&5\\4&2\end{pmatrix} = \begin{pmatrix}3&-3\\4&14\end{pmatrix}. \] |
| math:mappingConfidence | 1.0 |
| math:pageStart | 132 |
| math:problem | textbook_problem:chunjae_hong_vision_p132_example_02 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:matrix_linear_combination_entries |