Determinants and their Applications

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Formula: |a₁₁ a₁₂; a₂₁ a₂₂| = a₁₁a₂₂ - a₁₂a₂₁ Calculation: |3 2; 1 4| = (3×4) - (2×1) = 12 - 2 = 10 Method: Multiply diagonal elements (3×4), subtract product of other diagonal (2×1).

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Minor (Mᵢⱼ): Value of determinant obtained by deleting the iᵗʰ row and jᵗʰ column containing element aᵢⱼ. Cofactor (Cᵢⱼ): Cᵢⱼ = (-1)ⁱ⁺ʲ × Mᵢⱼ Relationship: Cofactor = Minor × (-1)ⁱ⁺ʲ - If (i+j) is e

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Step 1: Identify elements of first row: 1, 2, 3 Step 2: Apply cofactor expansion formula |1 2 3; 4 5 6; 7 8 9| = 1×C₁₁ + 2×C₁₂ + 3×C₁₃ Step 3: Calculate cofactors: C₁₁ = +|5 6; 8 9| = 45-48 = -3 C₁₂

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Sarrus Diagram Method: Step 1: Write the determinant and repeat first two columns Step 2: Draw downward arrows (positive products) Step 3: Draw upward arrows (negative products) Step 4: Sum positive p

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Property: If two rows (or columns) are interchanged, the determinant changes sign. Demonstration: Original: |2 3; 1 4| = (2×4) - (3×1) = 8 - 3 = 5 After row interchange: |1 4; 2 3| = (1×3) - (4×2) =

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Property: If two rows (or columns) are identical, the determinant equals zero. Proof with given example: |3 2 1; 5 4 6; 3 2 1| Expanding using first row: = 3|4 6; 2 1| - 2|5 6; 3 1| + 1|5 4; 3 2| =

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Property: If each element of a row (or column) is multiplied by k, the determinant is multiplied by k. Step-by-step verification: Original: |3 2; 2 8| = (3×8) - (2×2) = 24 - 4 = 20 After multiplying

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Step 1: Write in standard form 2x + 3y = 7 1x - 1y = 1 Step 2: Calculate determinants D = |2 3; 1 -1| = (2×(-1)) - (3×1) = -2 - 3 = -5 Dₓ = |7 3; 1 -1| = (7×(-1)) - (3×1) = -7 - 3 = -10 Dᵧ = |2