YBM p149 해설 - p133 마무리 15

Canonical URI: https://w3id.org/kmath/textbook_solution/ybm_solution_p149_p133_03

rdf:typemath:TextbookSolution
rdfs:labelYBM p149 해설 - p133 마무리 15
rdfs:commentYBM 공통수학1 정답과 해설 p149의 행렬 곱셈 등식 풀이 원문.
math:answerText\(-4\)
math:explanationText15 \[ AB=\begin{pmatrix}a&b\\3&-1\end{pmatrix} \begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}b&a\\-1&3\end{pmatrix}. \quad \text{▶ 30 \%} \] \[ BA=\begin{pmatrix}0&1\\1&0\end{pmatrix} \begin{pmatrix}a&b\\3&-1\end{pmatrix} = \begin{pmatrix}3&-1\\a&b\end{pmatrix}. \quad \text{▶ 30 \%} \] \[ \begin{pmatrix}b&a\\-1&3\end{pmatrix} = \begin{pmatrix}3&-1\\a&b\end{pmatrix} \] 에서 \(a=-1,\ b=3\)이므로 \[ a-b=-4. \quad \text{▶ 40 \%} \]
math:mappingConfidence0.92
math:pageStart149
math:problemtextbook_problem:ybm_vision_p133_03
math:reviewStatusreviewed
math:solutionKindworked_solution
math:usesSolutionPatternsolution_pattern:matrix_equality_component_equations
math:usesSolutionPatternsolution_pattern:matrix_product_row_column