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

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 (B). (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., (A) = (A^T). Step 5: Apply the determinant property. Since we found (B) = -124, then (B^T) must also be -124. |B^T| = |B| = -124 The determinant of B^T is -124. Send me the next one 📸