Question

For a fixed positive integer n, let D =

             

     

Solution

Correct option is

– 64

Take (n – 1)!, (n + 1)!, (n + 3)! Common from R1, R2, R3 respectively 

                 

Making two zeros in column 1 by applying R3 – R2 and R2 – R1, we have    

        

     

            = 8 [4n + 6 – 4n – 14] = 8 [– 8] = – 64.

SIMILAR QUESTIONS

Q1

 

Then f (100) is equal to

 

Q2
Q3

The number of values of k for which the system of equations

                (k + 1) x + 8y = 4k

                kx + (k + 3) y = 3k – 1

has infinitely many solution is

Q4

For what real values of k, the system of equations x + 2y + z = 1; x + 3y + 4z = k; x + 5y + 10z = khas solution? Find the solution of each case.

Q5

Let aij denoted the element of the ith row and jth column in 3 × 3 determinant (1 ≤ i ≤ 3, 1 ≤ j ≤ 3) and let aij – aij for every I and j. then the determinant has all the principle diagonal elements as

Q6

If each element of a determinant of third order with values A is multiplied by 3, then the value of newly formed determinant is

Q7
Q8

 

then x is equal to

 

Q9

Find the value of the determinant

                    

Q10

Evaluate the determinant without expansion as far as possible.