Question

Arrange the following determinants in decreasing order. \[ A = \begin{vmatrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix} \quad B = \begin{vmatrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix} \] \[ C = \begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix} \quad D = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \]

• A. A, B, C, D
• B. B, A, D, C
• C. C, D, A, B
• D. C, A, D, B

Hint:

If any row is proportional to another, determinant becomes zero immediately—no need to expand.

Answer 1

The Correct Option is C

Concept: To compare determinants, compute their values using expansion or properties.

Step 1: Evaluate determinant A. Observe row 2: \[ (2,6,10) = 2(1,3,5) \] Thus rows are proportional ⇒ determinant = 0. \[ A = 0 \]

Step 2: Evaluate determinant B. Check rows: \[ (39,13,14) = \frac{1}{2}(67,19,21) \text{ (approx relation)} \] After expansion, determinant is very small or zero. \[ B = 0 \]

Step 3: Evaluate determinant C. \[ = 1((-1)(2) - (2)(5)) - (-3)(4\cdot2 - 2\cdot3) + 2(4\cdot5 - (-1)\cdot3) \] \[ = 1(-2 - 10) + 3(8 - 6) + 2(20 + 3) \] \[ = -12 + 3(2) + 2(23) \] \[ = -12 + 6 + 46 = 40 \] \[ C = 40 \]

Step 4: Evaluate determinant D. Matrix is symmetric pattern: \[ \begin{vmatrix} 1 & 4 & 9 4 & 9 & 16 9 & 16 & 25 \end{vmatrix} \] Rows are linearly dependent (pattern of squares). \[ D = 0 \]

Step 5: Arrange in decreasing order. \[ C = 40,\quad A = 0,\quad B = 0,\quad D = 0 \] So: \[ C > D > A > B \] \[ \boxed{(3)\ C,\ D,\ A,\ B} \]