12th Grade > Mathematics
DETERMINANTS MCQs
Total Questions : 60
| Page 1 of 6 pages
Answer: Option B. -> 0
:
B
dndxn[Δ(x)]=∣∣
∣
∣
∣∣dndxnxndndxnsinxdndxncosxn!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣=∣∣
∣
∣
∣∣n!sin(x+nπ2)cos(x+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣⇒[Δn(x)]x=0=∣∣
∣
∣
∣∣n!sin(0+nπ2)cos(0+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣=0{SinceR1≡R2}.
:
B
dndxn[Δ(x)]=∣∣
∣
∣
∣∣dndxnxndndxnsinxdndxncosxn!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣=∣∣
∣
∣
∣∣n!sin(x+nπ2)cos(x+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣⇒[Δn(x)]x=0=∣∣
∣
∣
∣∣n!sin(0+nπ2)cos(0+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣∣
∣
∣
∣∣=0{SinceR1≡R2}.
Answer: Option D. -> None of these
:
D
Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣∣
∣
∣∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣∣
∣
∣∣=a2b2c2abc∣∣
∣∣bc1ab+acac1bc+abab1ac+bc∣∣
∣∣=abc∣∣
∣
∣∣bc1∑abac1∑abab1∑ab∣∣
∣
∣∣{byC3→C3+C1}=abc.∑ab∣∣
∣∣bc11ca11ab11∣∣
∣∣=0,[SinceC2≡C3].
Trick : Put a=1, b=2, c=3 and check it.
:
D
Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣∣
∣
∣∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣∣
∣
∣∣=a2b2c2abc∣∣
∣∣bc1ab+acac1bc+abab1ac+bc∣∣
∣∣=abc∣∣
∣
∣∣bc1∑abac1∑abab1∑ab∣∣
∣
∣∣{byC3→C3+C1}=abc.∑ab∣∣
∣∣bc11ca11ab11∣∣
∣∣=0,[SinceC2≡C3].
Trick : Put a=1, b=2, c=3 and check it.
Answer: Option C. -> 10
:
C
The cofactor of element 4, in the 2nd row and 3rd column is
=(−1)2+3∣∣
∣∣131801021∣∣
∣∣ = - {1(-2)-3(8-0)+1.16}
=10.
:
C
The cofactor of element 4, in the 2nd row and 3rd column is
=(−1)2+3∣∣
∣∣131801021∣∣
∣∣ = - {1(-2)-3(8-0)+1.16}
=10.
Answer: Option B. -> 18
:
B
Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0in the given equation, we have
t=∣∣
∣∣0−1312−4−340∣∣
∣∣=−12+30=18.
:
B
Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0in the given equation, we have
t=∣∣
∣∣0−1312−4−340∣∣
∣∣=−12+30=18.
Answer: Option A. -> 1
:
A
Δ=∣∣
∣
∣∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣∣
∣
∣∣=∣∣
∣∣1111αβ1α2β2∣∣
∣∣×∣∣
∣∣1111αβ1α2β2∣∣
∣∣
Applying C2→−C2−C3→C3−C1,
∣∣
∣∣1001α−1β−11α2−1β2−1∣∣
∣∣2=(α−1)2(β−1)2(β−α)2=(1−α)2(1−β)2(α−β)2
Hence, k=1
:
A
Δ=∣∣
∣
∣∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣∣
∣
∣∣=∣∣
∣∣1111αβ1α2β2∣∣
∣∣×∣∣
∣∣1111αβ1α2β2∣∣
∣∣
Applying C2→−C2−C3→C3−C1,
∣∣
∣∣1001α−1β−11α2−1β2−1∣∣
∣∣2=(α−1)2(β−1)2(β−α)2=(1−α)2(1−β)2(α−β)2
Hence, k=1
Answer: Option C. -> Are in H. P.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
Answer: Option A. -> a1Δ
:
A
B2=∣∣∣a1c1a3c3∣∣∣=a1c3−c1a3C2=−∣∣∣a1b1a3b3∣∣∣=−(a1b3−a3b1)B3=−∣∣∣a1c1a2c2∣∣∣=−(a1c2−a2c1)C3=∣∣∣a1b1a2b2∣∣∣=(a1b2−a2b1)∣∣∣B2C2B3C3∣∣∣=∣∣∣a1c3−a3c1−(a1b3−a3b1)−(a1c2−a2c1)a1b2−a2b1∣∣∣=∣∣∣a1c3−a1b3−a1c2a1b2∣∣∣+∣∣∣a1c3a3b1−a1c2−a2b1∣∣∣+∣∣∣−a3c1−a1b3a2c1a1b2∣∣∣+∣∣∣−a3c1a3b1a2c1−a2b1∣∣∣=a12(b2c3−b3c2)+a1b1(−c3a2+a3c2)+a1c1(−a3b2+a2b3)+c1b1(a3a2−a2a3)=a1Δ.
:
A
B2=∣∣∣a1c1a3c3∣∣∣=a1c3−c1a3C2=−∣∣∣a1b1a3b3∣∣∣=−(a1b3−a3b1)B3=−∣∣∣a1c1a2c2∣∣∣=−(a1c2−a2c1)C3=∣∣∣a1b1a2b2∣∣∣=(a1b2−a2b1)∣∣∣B2C2B3C3∣∣∣=∣∣∣a1c3−a3c1−(a1b3−a3b1)−(a1c2−a2c1)a1b2−a2b1∣∣∣=∣∣∣a1c3−a1b3−a1c2a1b2∣∣∣+∣∣∣a1c3a3b1−a1c2−a2b1∣∣∣+∣∣∣−a3c1−a1b3a2c1a1b2∣∣∣+∣∣∣−a3c1a3b1a2c1−a2b1∣∣∣=a12(b2c3−b3c2)+a1b1(−c3a2+a3c2)+a1c1(−a3b2+a2b3)+c1b1(a3a2−a2a3)=a1Δ.
Answer: Option B. -> bd
:
B
Δ2Δ1=∣∣∣10cd∣∣∣∣∣∣10ab∣∣∣=∣∣∣10c+adbd∣∣∣=bd.
:
B
Δ2Δ1=∣∣∣10cd∣∣∣∣∣∣10ab∣∣∣=∣∣∣10c+adbd∣∣∣=bd.
Answer: Option C. -> Are in H. P.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
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