The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 500. The cost of 1 kg onion, 2 kg wheat and 3 kg rice is ₹ 300 . The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹ 575 . The above situation can be represented in matrix form as AX = B . Then |5 A⁻¹ |=

Step-by-step solution

Let x, y, z be the cost per kg of onion, wheat, and rice respectively. The given system of linear equations is: 4x + 3y + 2z = 500 x + 2y + 3z = 300 6x + 2y + 3z = 575 The matrix A is given by: A = pmatrix 4 & 3 & 2 1 & 2 & 3 6 & 2 & 3 pmatrix Calculate the determinant |A| : |A| = 4(2 3 - 3 2) - 3(1 3 - 3 6) + 2(1 2 - 2 6) |A| = 4(6 - 6) - 3(3 - 18) + 2(2 - 12) |A| = 4(0) - 3(-15) + 2(-10) |A| = 0 + 45 - 20 = 25 We need to find |5A⁻¹| . Using the property |kA| = k^n |A| where n is the order of the matrix: |5A⁻¹| =