12th Grade > Mathematics
DETERMINANTS MCQs
Total Questions : 60
| Page 3 of 6 pages
Answer: Option A. -> Δ1+Δ2=0
:
A
Δ1=∣∣
∣∣111abca2b2c2∣∣
∣∣, & Δ2=∣∣
∣∣1bca1cab1abc∣∣
∣∣
In Δ2,Transposing the determinant
Δ2=∣∣
∣∣111bccaabaac∣∣
∣∣
C1→aC1,C2→bC2,C3→CC3
Δ2=1abc∣∣
∣∣abcabcabcabca2b2c2∣∣
∣∣
Taking abc common from R2
Δ2=∣∣
∣∣abc111a2b2c2∣∣
∣∣
R1↔R2
Δ2=−∣∣
∣∣111abca2b2c2∣∣
∣∣
⇒Δ2=−Δ1⇒Δ2+Δ1=0
:
A
Δ1=∣∣
∣∣111abca2b2c2∣∣
∣∣, & Δ2=∣∣
∣∣1bca1cab1abc∣∣
∣∣
In Δ2,Transposing the determinant
Δ2=∣∣
∣∣111bccaabaac∣∣
∣∣
C1→aC1,C2→bC2,C3→CC3
Δ2=1abc∣∣
∣∣abcabcabcabca2b2c2∣∣
∣∣
Taking abc common from R2
Δ2=∣∣
∣∣abc111a2b2c2∣∣
∣∣
R1↔R2
Δ2=−∣∣
∣∣111abca2b2c2∣∣
∣∣
⇒Δ2=−Δ1⇒Δ2+Δ1=0
Answer: Option B. -> 0
:
B
∴sin2x=1thenX=π4
Then, ∣∣
∣∣0cosx−sinxsinx0cosxcosxsinx0∣∣
∣∣2=∣∣
∣
∣
∣
∣∣01√2−1√21√201√21√21√20∣∣
∣
∣
∣
∣∣2
=(1√2×1√2×1√2)∣∣
∣∣01−1101110∣∣
∣∣2
=18{0−1(0−1)−1(1)}2
=0
:
B
∴sin2x=1thenX=π4
Then, ∣∣
∣∣0cosx−sinxsinx0cosxcosxsinx0∣∣
∣∣2=∣∣
∣
∣
∣
∣∣01√2−1√21√201√21√21√20∣∣
∣
∣
∣
∣∣2
=(1√2×1√2×1√2)∣∣
∣∣01−1101110∣∣
∣∣2
=18{0−1(0−1)−1(1)}2
=0
Answer: Option B. -> λ3
:
B
x3−1=0∴x=1,ω,ω2
Here, α=ω,β=ω2∣∣
∣
∣∣λ+1ωω2ωλ+ω21ω21λ+ω∣∣
∣
∣∣
Applying C1→C1+C2+C3, then
∣∣
∣
∣∣λωω2λλ+ω21λ1λ+ω∣∣
∣
∣∣
Applying R−2→R2−R1 and R3→R3−R1, then we get
∣∣
∣
∣∣λωω20λ+ω2−ω1−ω201−ωλ+ω−ω2∣∣
∣
∣∣=λ((λ+ω2−ω)(λ+ω−ω2)−(1−ω)(1−ω2))=λ(λ2)=λ3
:
B
x3−1=0∴x=1,ω,ω2
Here, α=ω,β=ω2∣∣
∣
∣∣λ+1ωω2ωλ+ω21ω21λ+ω∣∣
∣
∣∣
Applying C1→C1+C2+C3, then
∣∣
∣
∣∣λωω2λλ+ω21λ1λ+ω∣∣
∣
∣∣
Applying R−2→R2−R1 and R3→R3−R1, then we get
∣∣
∣
∣∣λωω20λ+ω2−ω1−ω201−ωλ+ω−ω2∣∣
∣
∣∣=λ((λ+ω2−ω)(λ+ω−ω2)−(1−ω)(1−ω2))=λ(λ2)=λ3
Answer: Option C. -> 4
:
C
f′(x)=∣∣
∣∣−sinxsinxcosx−2sin2xsin2x2cos2x−3sin3xsin3x3cos3x∣∣
∣∣+∣∣
∣∣cosxcosxcosxcos2x2cos2x2cos2xcos3x3cos3x3cos3x∣∣
∣∣+∣∣
∣∣cosxsinx−sinxcos2xsin2x−4sin2xcos3xsin3x−9sin3x∣∣
∣∣
f′(π2)=∣∣
∣∣−11000−23−10∣∣
∣∣+0+∣∣
∣∣01−1−1000−19∣∣
∣∣=2(1−3)+0+1(9−1)=−4+8=4
:
C
f′(x)=∣∣
∣∣−sinxsinxcosx−2sin2xsin2x2cos2x−3sin3xsin3x3cos3x∣∣
∣∣+∣∣
∣∣cosxcosxcosxcos2x2cos2x2cos2xcos3x3cos3x3cos3x∣∣
∣∣+∣∣
∣∣cosxsinx−sinxcos2xsin2x−4sin2xcos3xsin3x−9sin3x∣∣
∣∣
f′(π2)=∣∣
∣∣−11000−23−10∣∣
∣∣+0+∣∣
∣∣01−1−1000−19∣∣
∣∣=2(1−3)+0+1(9−1)=−4+8=4
Answer: Option A. -> 1+a2+b2+c2
:
A
∣∣
∣∣1ab−a1c−b−c1∣∣
∣∣=1(1+c2)−a(−a+bc)+b(ac+b)=1+a2+b2+c2.
:
A
∣∣
∣∣1ab−a1c−b−c1∣∣
∣∣=1(1+c2)−a(−a+bc)+b(ac+b)=1+a2+b2+c2.
Answer: Option B. -> abc
:
B
For non trivial solution
∣∣
∣∣a−1−1−111−b1111−c∣∣
∣∣=0
Applying C1→C1−C3 and C2→C2−C3, then
∣∣
∣∣a010−b1cc1−c∣∣
∣∣=0⇒a(−b+bc−c)−0+c(0−b)=0⇒ab+bc+ca=abc
:
B
For non trivial solution
∣∣
∣∣a−1−1−111−b1111−c∣∣
∣∣=0
Applying C1→C1−C3 and C2→C2−C3, then
∣∣
∣∣a010−b1cc1−c∣∣
∣∣=0⇒a(−b+bc−c)−0+c(0−b)=0⇒ab+bc+ca=abc
Answer: Option D. -> None of these
:
D
∴∑nk=1Dk=∣∣
∣
∣∣∑nk=11nn2∑nk=1kn2+n+1n2+n2∑nk=1k−∑nk=11n2n2+n+1∣∣
∣
∣∣
=∣∣
∣
∣∣nnnn2+nn2+n+1n2+nn2n2n2+n+1∣∣
∣
∣∣=56
=Applying C2→C2→C1→C3→C3−C1, then
∣∣
∣∣n00n2+n10n0n+1∣∣
∣∣=56⇒n(n+1)=56=7×8⇒n=7
:
D
∴∑nk=1Dk=∣∣
∣
∣∣∑nk=11nn2∑nk=1kn2+n+1n2+n2∑nk=1k−∑nk=11n2n2+n+1∣∣
∣
∣∣
=∣∣
∣
∣∣nnnn2+nn2+n+1n2+nn2n2n2+n+1∣∣
∣
∣∣=56
=Applying C2→C2→C1→C3→C3−C1, then
∣∣
∣∣n00n2+n10n0n+1∣∣
∣∣=56⇒n(n+1)=56=7×8⇒n=7
Answer: Option B. -> 0
:
B
Δ=∣∣
∣∣sinπsinBcosC−sinB0tanAcos(π−C)−tnaA0∣∣
∣∣=∣∣
∣∣0sinBcosC−sinB0tanA−cosC−tanA0∣∣
∣∣
=0 (∵Δ is skew symmetric)
:
B
Δ=∣∣
∣∣sinπsinBcosC−sinB0tanAcos(π−C)−tnaA0∣∣
∣∣=∣∣
∣∣0sinBcosC−sinB0tanA−cosC−tanA0∣∣
∣∣
=0 (∵Δ is skew symmetric)
Answer: Option D. -> ω
:
D
Since Δ=ω2−2ω2=−ω2. Therefore Δ2=ω4=ω.
:
D
Since Δ=ω2−2ω2=−ω2. Therefore Δ2=ω4=ω.
Answer: Option A. -> A unique solution
:
A
Given system of equation can be written as ⎡⎢⎣1113−12311⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣26−18⎤⎥⎦
On solving the above system we get the unique solution x = -10, y = -4, z = 16.
:
A
Given system of equation can be written as ⎡⎢⎣1113−12311⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣26−18⎤⎥⎦
On solving the above system we get the unique solution x = -10, y = -4, z = 16.
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