Matrices & DeterminantshardFree

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

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

Question

Let

D_k

be the

k\times k

matrix with zeros on the main diagonal, unity as the element of the 1st row and

(f(k))^{\text{th}}

column, and

k

for all other off-diagonal entries. If

f(x) = x - \{x\}

where

\{x\}

denotes the fractional part function, then the value of

\det(D_2) + \det(D_3)

equals:

32

34

correct

36

44

Solution

Step 1: Evaluate

f(k)

for natural numbers Since

\{k\} = 0

for every

k \in \mathbb{N}

, we have

f(k) = k

. Step 2: Construct

D_2

and compute its determinant

D_2

has

f(2)=2

, so unity is at position

(1,2)

. The remaining off-diagonal entry is

(2,1) = 2

D_2 = \begin{pmatrix}0 & 1\\ 2 & 0\end{pmatrix}, \quad \det(D_2) = 0 - 2 = -2

Step 3: Construct

D_3

and compute its determinant

D_3

has

f(3)=3

, so unity is at position

(1,3)

. All other off-diagonal entries equal

3

D_3 = \begin{pmatrix}0 & 3 & 1\\ 3 & 0 & 3\\ 3 & 3 & 0\end{pmatrix}

Expanding along

R_1

\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(D_2) + \det(D_3) = -2 + 36 = 34

Answer: (2)

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