Properties of Determinants
Trending Questions
Q.
Is the determinant of a transpose the same?
Q. ∣∣
∣
∣∣b2−abb−cbc−acab−a2a−bb2−abbc−acc−aab−a2∣∣
∣
∣∣=
- abc(a+b+c)
- 3a2b2c2
- \N
- None of these
Q. Let a, b and c be such that b(a+c) ≠ 0.
If ∣∣ ∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣ ∣∣+∣∣ ∣ ∣∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n−1b(−1)nc∣∣ ∣ ∣∣=0, then the value of n is
If ∣∣ ∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣ ∣∣+∣∣ ∣ ∣∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n−1b(−1)nc∣∣ ∣ ∣∣=0, then the value of n is
- zero
any even integer
- any odd integer
- any integer
Q.
What does determinant of mean?
Q. In a triangle ABC, the value of the determinant ∣∣
∣
∣
∣∣sinA2sinB2sinc2sin(A+B+C)sinB2cosA2cos(A+B+C2)tan(A+B+C)sinC2∣∣
∣
∣
∣∣ is less than or equal to
- 12
- 14
- 18
- None of these
Q. If ∣∣
∣∣y+zxyz+xzxx+yyz∣∣
∣∣=k(x+y+z)(x−z)2, then k =
- 2xyz
- 1
- xyz
- x2y2z2
Q. If λ is a non real cube root of -2, then the value of ∣∣
∣∣12λ1λ13λ222λ1∣∣
∣∣
is equal to
is equal to
- -11
- -12
- -13
- \N
Q. Let d∈R, and
A=⎡⎢⎣−24+d(sinθ)−21(sinθ)+2d5(2sinθ)−d(−sinθ)+2+2d⎤⎥⎦,
θ∈[0, 2π]. If the minimum value of det(A) is 8, then a value of d is:
A=⎡⎢⎣−24+d(sinθ)−21(sinθ)+2d5(2sinθ)−d(−sinθ)+2+2d⎤⎥⎦,
θ∈[0, 2π]. If the minimum value of det(A) is 8, then a value of d is:
- −5
- 2(√2+2)
- −7
- 2(√2+1)
Q.
What do determinants tell us?
Q. If ⎡⎢⎣1sinθ1−sinθ1sinθ−1−sinθ1⎤⎥⎦ ; then for all
θ∈(3π4, 5π4), det(A) lies in the interval :
θ∈(3π4, 5π4), det(A) lies in the interval :
- (0, 32]
- (1, 52]
- (32, 3]
- [52, 4)
Q. Let a > 0, d > 0. Find the value of the determinant
∣∣ ∣ ∣ ∣ ∣∣1a1a(a+d)1(a+d)(a+2d)1(a+d)1(a+d)(a+2d)1(a+2d)(a+3d)1(a+2d)1(a+2d)(a+3d)1(a+3d)(a+4d)∣∣ ∣ ∣ ∣ ∣∣ is
∣∣ ∣ ∣ ∣ ∣∣1a1a(a+d)1(a+d)(a+2d)1(a+d)1(a+d)(a+2d)1(a+2d)(a+3d)1(a+2d)1(a+2d)(a+3d)1(a+3d)(a+4d)∣∣ ∣ ∣ ∣ ∣∣ is
- 4da(a+d)(a+2d)(a+3d)(a+4d)
- 4da(a+d)2(a+2d)3(a+3d)2(a+4d)
- 4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)
- \N
Q.
Subtract from the sum of and
Q.
Subtract from
Q. The value of the determinant ∣∣
∣∣1ab+c1bc+a1ca+b∣∣
∣∣ is
- a+b+c
- (a+b+c)2
- \N
- 1+a+b+c
Q. ∣∣
∣∣111abca3b3c3∣∣
∣∣=
- a3+b3+c3−3abc
- a3+b3+c3+3abc
- (a+b+c)(a−b)(b−c)(c−a)
- None of these
Q.
If a2 +b2 + c2 =-2 and
then f(x) is a polynomial of degree
2
3
0
1
Q. If ∣∣
∣∣a−b−c2a2a2bb−c−a2b2c2cc−a−b∣∣
∣∣
=(a+b+c)(x+a+b+c)2, x≠0 and a+b+c≠0, then x is equal to :
=(a+b+c)(x+a+b+c)2, x≠0 and a+b+c≠0, then x is equal to :
- abc
- 2(a+b+c)
- −(a+b+c)
- −2(a+b+c)
Q. Let A=[aij] and B=[bij] be two 3×3 real matrices such that bij=(3)(i+j−2)aji, where i, j=1, 2, 3. If the determinant of B is 81, then the determinant of A is :
- 19
- 181
- 13
- 3
Q. The total number of distinct x∈R for which
∣∣ ∣ ∣∣xx21+x32x4x21+8x33x9x21+27x3∣∣ ∣ ∣∣=10 is
∣∣ ∣ ∣∣xx21+x32x4x21+8x33x9x21+27x3∣∣ ∣ ∣∣=10 is
Q. Let ω be the complex number cos2π3+isin2π3. Then the number of distinct complex numbers z satisfying ∣∣
∣
∣∣z+1ωω2ωz+ω21ω21z+ω∣∣
∣
∣∣=0 is equal to
- 2
- 1
- \N
- 3
Q. If a+b+c=0, then the solution of the equation ∣∣
∣∣a−xcbcb−xabac−x∣∣
∣∣=0 is
- 0
- ±32(a2+b2+c2)
- 0, ±√32(a2+b2+c2)
- 0, ±√(a2+b2+c2)
Q. If a, b, c are unequal what is the condition that the value of the following determinant is zero
Δ=∣∣ ∣ ∣∣aa2a3+1bb2b3+1cc2c3+1∣∣ ∣ ∣∣
Δ=∣∣ ∣ ∣∣aa2a3+1bb2b3+1cc2c3+1∣∣ ∣ ∣∣
- 1+abc = 0
- a+b+c+1=0
- (a-b) (b-c) (c-a) = 0
- None of these
Q. If a≠b≠c, the value of x which satisfies the equation ∣∣
∣∣0x−ax−bx+a0x−cx+bx+c0∣∣
∣∣=0, is
- x=0
- x=a
- x=b
- x=c
Q. Find the value of the determinant ∣∣
∣∣bccaabpqr111∣∣
∣∣, where a, b and c are respectively the pth, qth and rth terms of a harmonic progression.
- 1
- pqr
- \N
- 1abc
Q. The roots of the equation ∣∣
∣∣14201−2512x5x2∣∣
∣∣ = 0 are
- -1, -2
- -1, 2
- 1, -2
- 1, 2
Q. If n is positive integer, then ∣∣
∣
∣∣n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4∣∣
∣
∣∣ is equal to
- 3
- -1
- -5
- -9
Q. Let D1=∣∣
∣∣xab−10xx21∣∣
∣∣ and D2=∣∣
∣∣cx22a−bx21−10x∣∣
∣∣. If all the roots of the equation (x2−4x−7)(x2−2x−3)=0 satisfy the equation D1+D2=0, then the value of a+4b+c is
Q. For a fixed positive integer n, if
D=∣∣ ∣ ∣∣n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!∣∣ ∣ ∣∣, then [D(n!)3−4] is___
D=∣∣ ∣ ∣∣n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!∣∣ ∣ ∣∣, then [D(n!)3−4] is
- divisible by n.
- divisible by n+1
- divisible by n+2
- divisible by n+3
Q. The roots of the equation ∣∣
∣∣14201−2512x5x2∣∣
∣∣ = 0 are
- -1, -2
- -1, 2
- 1, -2
- 1, 2
Q. If a, b, c are respectively the pth, qth, rth terms of an A.P., the ∣∣
∣∣ap1bq1cr1∣∣
∣∣ =
- 1
- -1
- \N
- pqr