Canonical URI: https://w3id.org/kmath/textbook_solution/visang_solution_body_p122_01
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 비상 p122 예제 1 풀이 |
| rdfs:comment | 비상 공통수학1 교과서 p122의 예제 1 본문 풀이. |
| math:answerText | 답 (1) \[ \begin{pmatrix}1&6\\-16&-4\end{pmatrix} \] (2) \[ \begin{pmatrix}6&-9\\9&6\end{pmatrix} \] |
| math:explanationText | (1) \[ \begin{aligned} 4A-3(A+B)&=4A-3A-3B=A-3B\\ &= \begin{pmatrix}4&-3\\-1&2\end{pmatrix} -3\begin{pmatrix}1&-3\\5&2\end{pmatrix}\\ &= \begin{pmatrix}4&-3\\-1&2\end{pmatrix} - \begin{pmatrix}3&-9\\15&6\end{pmatrix}\\ &= \begin{pmatrix}1&6\\-16&-4\end{pmatrix} \end{aligned} \] (2) \(3X-2A=X+4B\)에서 \(2X=2A+4B\)이므로 \[ \begin{aligned} X&=A+2B\\ &= \begin{pmatrix}4&-3\\-1&2\end{pmatrix} +2\begin{pmatrix}1&-3\\5&2\end{pmatrix}\\ &= \begin{pmatrix}4&-3\\-1&2\end{pmatrix} + \begin{pmatrix}2&-6\\10&4\end{pmatrix}\\ &= \begin{pmatrix}6&-9\\9&6\end{pmatrix} \end{aligned} \] |
| math:mappingConfidence | 1.0 |
| math:pageStart | 122 |
| math:problem | textbook_problem:visang_vision_p122_01 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:matrix_linear_combination_entries |