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Matrices & Determinants: Let Matrix Zeros Main Diagonal Unity Element 1st

JEE Maths question with a full step-by-step solution.

Question
Let DkD_k be the k×kk\times k matrix with zeros on the main diagonal, unity as the element of the 1st row and (f(k))th(f(k))^{\text{th}} column, and kk for all other off-diagonal entries. If f(x)=x−{x}f(x) = x - \{x\} where {x}\{x\} denotes the fractional part function, then the value of det⁡(D2)+det⁡(D3)\det(D_2) + \det(D_3) equals:
A3232
B3434correct
C3636
D4444
Solution
Step 1: Evaluate f(k)f(k) for natural numbers Since {k}=0\{k\} = 0 for every k∈Nk \in \mathbb{N}, we have f(k)=kf(k) = k. Step 2: Construct D2D_2 and compute its determinant D2D_2 has f(2)=2f(2)=2, so unity is at position (1,2)(1,2). The remaining off-diagonal entry is (2,1)=2(2,1) = 2.
D2=(0120),det⁡(D2)=0−2=−2D_2 = \begin{pmatrix}0 & 1\\ 2 & 0\end{pmatrix}, \quad \det(D_2) = 0 - 2 = -2
 Step 3: Construct D3D_3 and compute its determinant D3D_3 has f(3)=3f(3)=3, so unity is at position (1,3)(1,3). All other off-diagonal entries equal 33.
D3=(031303330)D_3 = \begin{pmatrix}0 & 3 & 1\\ 3 & 0 & 3\\ 3 & 3 & 0\end{pmatrix}
 Expanding along R1R_1:
det⁡(D3)=−3∣3330∣+1∣3033∣=−3(0−9)+1(9−0)=27+9=36\det(D_3) = -3\begin{vmatrix}3 & 3\\ 3 & 0\end{vmatrix} + 1\begin{vmatrix}3 & 0\\ 3 & 3\end{vmatrix} = -3(0-9) + 1(9-0) = 27 + 9 = 36
 Step 4: Sum
det⁡(D2)+det⁡(D3)=−2+36=34\det(D_2) + \det(D_3) = -2 + 36 = 34
 Answer: (2)
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