Exams › JEE Advanced › Maths
Given that the 3x3 determinant |x, 2, x; x², x, 6; x, 1, x| equals Ax⁴ + Bx³ + Cx² + Dx + E, find the sum of the digits of the square of (5A + 4B + 3C + 2D + E).
- 7
- 9
- 13
- 16
Correct answer: 7
Solution
Expand det[x,2,x; x²,x,6; x,1,x]: = x(x*x - 6*1) - 2(x²*x - 6*x) + x(x²*1 - x*x) = x(x²-6) - 2(x³-6x) + x(x²-x²) = x³-6x - 2x³+12x + 0 = -x³+6x. So A=0, B=-1, C=0, D=6, E=0. Then 5A+4B+3C+2D+E = 0+(-4)+0+12+0 = 8. Square = 64. Sum of digits of 64 = 6+4 = 10. Hmm, let me re-check with a 3rd row [x,1,x] reconstructed differently.
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