Determinant Evaluation

A three by three determinant is evaluated by cofactor expansion along any row or column, multiplying entries by signed minors, a basic JEE Advanced computation. Expanding along the first row of rows (2,-1,3),(1,0,4),(3,2,-1): the first term is 2 times the minor det[(0,4),(2,-1)] = 2(0·(-1) - 4·2) = 2(-8) = -16. The second term is -(-1) times det[(1,4),(3,-1)] = +1·(1·(-1) - 4·3) = (-1 - 12) = -13. The third term is 3 times det[(1,0),(3,2)] = 3(1·2 - 0·3) = 3·2 = 6. The cofactor of position (1,2) carries sign (-1)^{1+2} = -1, and with entry -1 its contribution is -1·(-1)·(minor) = (minor) = -13. Summing the three terms gives -16 + (-13) + 6 = -23. Option 43 ignores signs. Option -7 drops a term. Option 17 mishandles a minor. Hence the determinant is -23. Plausibility check: expanding along a different row should reproduce the same value, a built-in consistency guarantee of determinants. Cofactor expansion may be carried out along whichever row or column contains the most zeros, since the determinant is independent of that choice, and exploiting this freedom dramatically reduces arithmetic. Tracking the alternating sign pattern of the cofactors is where most computational errors arise, so writing the checkerboard of plus and minus signs before expanding is a reliable discipline under exam conditions.