Use the square's rule to find the determinant of B = [[2, 5, 1], [6, 0, 4], [−2, −3, 3]]

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Part 1: Use Sarrus' rule to find the determinant of B Step 1: Write down the matrix B and extend it with the first two columns. B = 2 & 5 & 1 6 & 0 & 4 −2 & −3 & 3 To apply Sarrus' rule, we repeat the first two columns next to the matrix: 2 & 5 & 1 6 & 0 & 4 −2 & −3 & 3 2 & 5 6 & 0 −2 & −3 Step 2: Calculate the sum of the products of the main diagonals (top-left to bottom-right). (2 × 0 × 3) + (5 × 4 × −2) + (1 × 6 × −3) 0 + (−40) + (−18) = −58 Step 3: Calculate the sum of the products of the anti-diagonals (top-right to bottom-left). (1 × 0 × −2) + (2 × 4 × −3) + (5 × 6 × 3) 0 + (−24) + (90) = 66 Step 4: Subtract the sum of the anti-diagonal products from the sum of the main diagonal products to find det(B). det(B) = −58 − 66 = −124 The determinant of B is −124. Part 2: Find |B^T| The matrix given for this part is: B^T = 2 & 6 & −2 5 & 0 & −3 1 & 4 & 3 This matrix is the transpose of matrix B from Part 1. A fundamental property of determinants states that the determinant of a matrix is equal to the determinant of its transpose, i.e., det(A) = det(A^T). Step 5: Apply the determinant property. Since we found det(B) = −124, then det(B^T) must also be −124. |B^T| = |B| = −124 The determinant of B^T is −124.

Use the square's rule to find the determinant of B = [[2, 5, 1], [6, 0, 4], [-2, -3, 3]]

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Step-by-step solution

Part 1· 4 steps

Write down the matrix B and extend it with the first two columns.

Calculate the sum of the products of the main diagonals (top-left to bottom-right).

Calculate the sum of the products of the anti-diagonals (top-right to bottom-left).

Subtract the sum of the anti-diagonal products from the sum of the main diagonal products to find $\det(B)$ .

Part 2· 1 step-124

Apply the determinant property.

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