Matrices & DeterminantshardFree

Matrices & Determinants: Vmatrix Vmatrix Equals

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

Question

U_n = \begin{vmatrix} 1 & k & k \\ 2n & k^2+k+1 & k^2+k \\ 2n-1 & k^2 & k^2+k+1 \end{vmatrix}

and

\displaystyle\sum_{n=1}^k U_n = 110

, then

k

equals:

10

correct

9

6

DNone of these

Solution

Step 1: Sum using linearity in column 1

\sum_{n=1}^k U_n = \begin{vmatrix} k & k & k \\ k(k+1) & k^2+k+1 & k^2+k \\ k^2 & k^2 & k^2+k+1 \end{vmatrix}

using

\sum_{n=1}^k 1 = k

\sum_{n=1}^k 2n = k(k+1)

\sum_{n=1}^k (2n-1) = k^2

. Step 2: Apply

C_2 \to C_2 - C_1

= \begin{vmatrix} k & 0 & k \\ k(k+1) & 1 & k^2+k \\ k^2 & 0 & k^2+k+1 \end{vmatrix}

Step 3: Expand along column 2

= 1 \cdot (-1)^{2+2} \begin{vmatrix} k & k \\ k^2 & k^2+k+1 \end{vmatrix} = k(k^2+k+1) - k \cdot k^2 = k^2+k = k(k+1)

Step 4: Solve

k(k+1) = 110 = 10 \times 11 \implies k = 10

Answer: (1)

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