For a fixed positive integer n, let D =
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.
Then f (100) is equal to
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
For what real values of k, the system of equations x + 2y + z = 1; x + 3y + 4z = k; x + 5y + 10z = k2 has solution? Find the solution of each case.
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
If each element of a determinant of third order with values A is multiplied by 3, then the value of newly formed determinant is
then x is equal to
Find the value of the determinant
Evaluate the determinant without expansion as far as possible.