Q: If f: N Z is defined by f(n) = vmatrix n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 vmatrix , k N , and _ n=1 ^ k f(n) = 98, then k is equal to :

Let S = _ n=1 ^ k f(n), where f(n) = vmatrix n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 vmatrix Since only the first column depends on n, the summation can be taken inside that column: S = vmatrix _ n=1 ^ k n & -1 & -5 \\ _ n=1 ^ k (-2n^2) & 3(2k+1) & 2k+1 \\ _ n=1 ^ k (-3n^3) & 3k(2k+1) & 3k(k+2)+1 vmatrix Using standard summation formulas: _ n=1 ^ k n = k(k+1) 2 _ n=1 ^ k (-2n^2) = - k(k+1)(2k+1) 3 _ n=1 ^ k (-3n^3) = - 3k^2(k+1)^2 4 Substituting: S = vmatrix k(k+1)

Q: Among the statements : I: If | array ccc 1 & & \\ & 1 & \\ & & 1 array |= | array ccc 0 & & \\ & 0 & \\ & & 0 array |, then ^ 2 + ^ 2 + ^ 2 = 3 2 , and II : If | array ccc x^ 2 +x & x+1 & x-2 \\ 2 x^

Statement I: Let D_1 = vmatrix 1 & & \\ & 1 & \\ & & 1 vmatrix = 1 - ^2 - ^2 - ^2 + 2 . D_2 = vmatrix 0 & & \\ & 0 & \\ & & 0 vmatrix = 2 . D_1 = D_2 1 - ^2 - ^2 - ^2 = 0 ^2 + ^2 + ^2 = 1 3 2 . Statement I is false. Statement II: Evaluate by substituting x = 0: determinant = -12, so q = -12. At x = 1: determinant = 12, so p + q = 12 p = 24. p^2 = 576 and 196q^2 = 196 144 = 28224. Since p^2 196q^2, Statement II is false. Both statements are false.

Question 1

If the system of linear equations: x+y+z=6, x+2y+5z=10, 2x+3y+ z= has infinitely many solutions, then the value of + equals:

Accepted Answer

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and _x = _y = _z = 0. = vmatrix 1 & 1 & 1 \ 1 & 2 & 5 \ 2 & 3 & vmatrix = 0 Expanding the determinant: 1(2 - 15) - 1( - 10) + 1(3 - 4) = 0 - 6 = 0 = 6 Now, setting _z = 0: _z = vmatrix 1 & 1 & 6 \ 1 & 2 & 10 \ 2 & 3 & vmatrix = 0 Expanding the determinant: 1(2 - 30) - 1( - 20) + 6(3 - 4) = 0 - 16 = 0 = 16 Therefore, + = 6 + 16 = 22. Answer: 22

Question 2

The sum of all possible values of [0, 2 ], for which the system of equations : x 3 - 8y - 12z = 0 x 2 + 3y + 3z = 0 x + y + 3z = 0 has a non-trivial solution, is equal to :

Accepted Answer

For the given system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. vmatrix 3 & -8 & -12 \ 2 & 3 & 3 \ 1 & 1 & 3 vmatrix = 0 Expanding the determinant along the first row: 3 (9 - 3) - (-8)(3 2 - 3) + (-12)( 2 - 3) = 0 6 3 + 24 2 - 24 - 12 2 + 36 = 0 6 3 + 12 2 + 12 = 0 Dividing by 6, we get: 3 + 2 2 + 2 = 0 Using the multiple angle formulas 3 = 4 ^3 - 3 and 2 = 2 ^2 - 1, we substitute these into the equation: (4 ^3 - 3 )

Question 3

If f: N Z is defined by f(n) = vmatrix n & -1 & -5 \ -2n^2 & 3(2k+1) & 2k+1 \ -3n^3 & 3k(2k+1) & 3k(k+2)+1 vmatrix , k N , and _ n=1 ^ k f(n) = 98, then k is equal to :

Accepted Answer

Let S = _ n=1 ^ k f(n), where f(n) = vmatrix n & -1 & -5 \ -2n^2 & 3(2k+1) & 2k+1 \ -3n^3 & 3k(2k+1) & 3k(k+2)+1 vmatrix Since only the first column depends on n, the summation can be taken inside that column: S = vmatrix _ n=1 ^ k n & -1 & -5 \ _ n=1 ^ k (-2n^2) & 3(2k+1) & 2k+1 \ _ n=1 ^ k (-3n^3) & 3k(2k+1) & 3k(k+2)+1 vmatrix Using standard summation formulas: _ n=1 ^ k n = k(k+1) 2 _ n=1 ^ k (-2n^2) = - k(k+1)(2k+1) 3 _ n=1 ^ k (-3n^3) = - 3k^2(k+1)^2 4 Substituting: S = vmatrix k(k+1)

Question 4

Consider the system of linear equations in x, y, z: x + 2y + tz = 0, 6x + y + 5tz = 0, 3x + t^2 y + f(t) z = 0, where f: R R is a differentiable function. If this system has infinitely many solutions

Accepted Answer

For a homogeneous system of linear equations to have infinitely many solutions, the determinant of its coefficient matrix must be zero. The coefficient matrix is: = vmatrix 1 & 2 & t \ 6 & 1 & 5t \ 3 & t^2 & f(t) vmatrix Expanding the determinant along the first row, we get: = 1(1 f(t) - 5t t^2) - 2(6 f(t) - 3 5t) + t(6 t^2 - 3 1) = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t = -11f(t) + t^3 + 27t Since the system has infinitely many solutions for all t R , we must have = 0 for all t R . -11f(t) +

Question 5

If the system of equations: x+y+z=5 x+2y+3z=9 x+3y+ z= has infinitely many solutions, then the value of + is:

Accepted Answer

For the system to have infinitely many solutions, = 0 and _x = _y = _z = 0. = vmatrix 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & vmatrix = 1(2 - 9) - 1( - 3) + 1(3 - 2) = 2 - 9 - + 3 + 1 = - 5 Setting = 0 gives = 5. Now, calculating _z: _z = vmatrix 1 & 1 & 5 \ 1 & 2 & 9 \ 1 & 3 & vmatrix = 1(2 - 27) - 1( - 9) + 5(3 - 2) _z = 2 - 27 - + 9 + 5 = - 13 Setting _z = 0 gives = 13. For = 5 and = 13, we can verify that _x = 0 and _y = 0 as well. Therefore, + = 5 + 13 = 18. Answer: 18

Question 6

If the system of equations x + 5y + 6z = 4, 2x + 3y + 4z = 7, x + 6y + az = b has infinitely many solutions, then the point (a, b) lies on the line

Accepted Answer

For the given system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, = 0. = vmatrix 1 & 5 & 6 \ 2 & 3 & 4 \ 1 & 6 & a vmatrix = 0 Expanding along the first row: 1(3a - 24) - 5(2a - 4) + 6(12 - 3) = 0 3a - 24 - 10a + 20 + 54 = 0 -7a + 50 = 0 a = 50 7 For infinitely many solutions, we must also have _z = 0. _z = vmatrix 1 & 5 & 4 \ 2 & 3 & 7 \ 1 & 6 & b vmatrix = 0 Expanding along the first row: 1(3b - 42) - 5(2b - 7) + 4(12 - 3) = 0 3b -

Question 7

Let , R be such that the system of linear equations x + 2y + z = 5 2x + y + z = 5 8x + 4y + z = 18 has no solution. Then is equal to :

Accepted Answer

The given system of linear equations is: x + 2y + z = 5 2x + y + z = 5 8x + 4y + z = 18 For the system to have no solution, the determinant of the coefficient matrix D must be zero. D = vmatrix 1 & 2 & 1 \ 2 & 1 & \ 8 & 4 & vmatrix = 0 Expanding along the first row: 1( - 4 ) - 2(2 - 8 ) + 1(8 - 8) = 0 - 4 - 4 + 16 = 0 12 - 3 = 0 = 4 To verify the no solution condition, substitute = 4 into the third equation: 8x + 4y + 4 z = 18 4(2x + y + z) = 18 2x + y + z = 9 2 However, the second equation is

Question 8

The system of linear equations x+y+z=6 2 x+5 y+a z=36 x+2 y+3 z=b has

Accepted Answer

Coefficient matrix determinant: = 1(15-2a) - 1(6-a) + 1(4-5) = 8 - a. For a = 8, = 0, so unique solution is not possible. With a = 8: R_2 - 2R_1 gives 3y + 6z = 24 y + 2z = 8. R_3 - R_1 gives y + 2z = b - 6. For consistency: b - 6 = 8 b = 14. When a = 8, b = 14: the system has infinitely many solutions.

Question 9

Accepted Answer

Statement I: Let D_1 = vmatrix 1 & & \ & 1 & \ & & 1 vmatrix = 1 - ^2 - ^2 - ^2 + 2 . D_2 = vmatrix 0 & & \ & 0 & \ & & 0 vmatrix = 2 . D_1 = D_2 1 - ^2 - ^2 - ^2 = 0 ^2 + ^2 + ^2 = 1 3 2 . Statement I is false. Statement II: Evaluate by substituting x = 0: determinant = -12, so q = -12. At x = 1: determinant = 12, so p + q = 12 p = 24. p^2 = 576 and 196q^2 = 196 144 = 28224. Since p^2 196q^2, Statement II is false. Both statements are false.

Question 10

Let n be the number obtained on rolling a fair die. If the probability that the system x- n y+z=6 x+( n -2) y+( n +1) z=8 ( n -1) y+z=1 has a unique solution is k 6 , then the sum of k and all possibl

Accepted Answer

The coefficient matrix has determinant (A) = -(n-1)(n-2). The system has a unique solution when (A) 0, which requires n 1 and n 2. For a fair die, n \ 1,2,3,4,5,6\ , so the unique solution occurs for n \ 3,4,5,6\ . The probability is 4 6 = 2 3 , giving k = 4. The sum of k and all possible values of n where unique solution exists is 4 + (3+4+5+6) = 22.

Question 11

If the system of equations 3 x+y+4 z=3 2 x+ y-z=-3 x+2 y+z=4 has no solution, then the value of is equal to :

Accepted Answer

For no solution = 0 vmatrix 3 & 1 & 4 \ 2 & & -1 \ 1 & 2 & 1 vmatrix = 0 3( + 2) + 1(-1 - 2) + 4(4 - ) = 0 19 - = 0 = 19 & for = 19 _x = vmatrix 3 & 1 & 4 \ -3 & 19 & -1 \ 4 & 2 & 1 vmatrix = 3(21) + 1(-1) + 4(-82) 0 no solution for = 19

Determinants — Previous Year Questions & Solutions