Inverse of a Matrix and Its Applications — Flashcards

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Step 1: Compute determinant: \( |A| = (2)(4) - (1)(3) = 8 - 3 = 5 \neq 0 \). So inverse exists. Step 2: Find adjoint: - Cofactor of 2 is 4 → C₁₁ = 4 - Cofactor of 3 is 1 → C₁₂ = -1 (sign: (-1)¹⁺² = -…

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Step 1: \( |A| = (1)(4) - (3)(2) = 4 - 6 = -2 \neq 0 \) Step 2: Cofactors: - C₁₁ = 4 → +4 - C₁₂ = 3 → -3 - C₂₁ = 2 → -2 - C₂₂ = 1 → +1 Matrix of cofactors = \( \begin{bmatrix} 4 & -3 \ -2 & 1 \end{b…

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Step 1: Find cofactors: - C₁₁ = Minor of 1 = 4 → (+)4 - C₁₂ = Minor of 2 = 3 → (-)3 - C₂₁ = Minor of 3 = 2 → (-)2 - C₂₂ = Minor of 4 = 1 → (+)1 Matrix of cofactors = \( \begin{bmatrix} 4 & -3 \ -2 & …

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Step 1: Cofactors: - C₁₁ = 3 → +3 - C₁₂ = 5 → -5 - C₂₁ = -1 → -(-1) = +1 (sign: (-1)²⁺¹ = -1) - C₂₂ = 2 → +2 Matrix of cofactors = \( \begin{bmatrix} 3 & -5 \ 1 & 2 \end{bmatrix} \) Step 2: Adjoint …

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Step 1: \( |A| = (2)(3) - (-1)(4) = 6 + 4 = 10 \) Step 2: Cofactors: - C₁₁ = 3 → +3 - C₁₂ = -1 → -(-1) = +1 → but sign (-1)¹⁺² = -1 → -1 - C₂₁ = 4 → sign (-1)²⁺¹ = -1 → -4 - C₂₂ = 2 → +2 Cofactor ma…

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A matrix is singular if its determinant is zero. Compute \( |A| = (3)(4) - (6)(2) = 12 - 12 = 0 \) Since determinant is 0, the matrix is singular. Answer: Singular…

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A matrix is singular if its determinant is 0. Compute \( |A| = \begin{vmatrix} 1 & -2 & 3 \ 1 & 2 & 1 \ x & 2 & -3 \end{vmatrix} \) Expand along first row: \( = 1 \cdot \begin{vmatrix} 2 & 1 \ 2 & …

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Use the matrix method when you have a system of linear equations with the same number of equations as variables (square system), and the coefficient matrix is non-singular (determinant ≠ 0). Steps: 1…