question 37 points 3

find the determinant of the 3 by 3 matrix: \begin{bmatrix} 2 & 4 & 6 \\ 1 & 0 & 1 \\ 5 & 1 & 3 \end{bmatrix}.

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Question

question 37 points 3

find the determinant of the 3 by 3 matrix: \begin{bmatrix} 2 & 4 & 6 \\ 1 & 0 & 1 \\ 5 & 1 & 3 \end{bmatrix}.

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Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Matrix Determinant", "Determinant Expansion" ], "new_concepts": [], "current_concepts": [ "Matrix Determinant", "Determinant Expansion" ] } ## Define the matrix and target Using the *Matrix Determinant* knowledge point $$ A = \begin{bmatrix} 2 & 4 & 6 \ 1 & 0 & 1 \ 5 & 1 & 3 \end{bmatrix} $$ We need to find $\det(A)$. ## Expand along the second row Using the *Determinant Expansion* knowledge point $$ \begin{aligned} \det(A) &= -1 \cdot \det\begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} + 0 \cdot \det\begin{bmatrix} 2 & 6 \ 5 & 3 \end{bmatrix} - 1 \cdot \det\begin{bmatrix} 2 & 4 \ 5 & 1 \end{bmatrix} \end{aligned} $$ ## Calculate the 2 by 2 determinants Using the *Matrix Determinant* knowledge point $$ \begin{aligned} \det\begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} &= 4(3) - 6(1) = 12 - 6 = 6\ \det\begin{bmatrix} 2 & 4 \ 5 & 1 \end{bmatrix} &= 2(1) - 4(5) = 2 - 20 = -18 \end{aligned} $$ ## Compute final determinant value Using the *Matrix Determinant* knowledge point $$ \begin{aligned} \det(A) &= -1(6) + 0 - 1(-18)\ &= -6 + 18\ &= 12 \end{aligned} $$

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{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Matrix Determinant" ] }

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QUESTION IMAGE

Question

Explanation:

Define the matrix and target

Expand along the second row

Calculate the 2 by 2 determinants

Compute final determinant value

Answer: