Free Determinants 01 Practice Test - 12th Grade - Commerce 

Question 1

If f(n)=αn+βn and ∣∣ ∣ ∣∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣∣ ∣ ∣∣=k(1−α)2(1−β)2(α−β)2, then k is equal to

A. 1
B. -1
C. αβ
D. αβγ

SOLUTION

Solution : 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

Question 2

If Δ1=∣∣ ∣∣111abca2b2c2∣∣ ∣∣,Δ2=∣∣ ∣∣1bca1cab1abc∣∣ ∣∣, then 

A. Δ1+Δ2=0
B. Δ1+2Δ2=0
C. Δ1=Δ2
D. Δ1=2Δ2

SOLUTION

Solution : 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
 

Question 3

If α,β are non real numbers satisfying x3−1=0 then the value of ∣∣ ∣∣λ+1αβαλ+β1β1λ+α∣∣ ∣∣ is equal to 

A. 0
B. λ3
C. λ3+1
D. λ3−1

SOLUTION

Solution : 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

 

Question 4

If Dk=∣∣ ∣ ∣∣1nn2kn2+n+1n2+n2k−1n2n2+n+1∣∣ ∣ ∣∣ and ∑nk=1Dk=56, then n equals

A. 4
B. 6
C. 8
D. None of these

SOLUTION

Solution : 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

Question 5

If sin 2x = 1, then ∣∣ ∣∣0cosx−sinxsinx0cosxcosxsinx0∣∣ ∣∣2 equals 

A. 3
B. 0
C. 1
D. None of these

SOLUTION

Solution : 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

Question 6

If A+B+C=π, then ∣∣ ∣ ∣∣sin(A+B+C)sinBcosC−sinB0tanAcos(A+B)−tnaA0∣∣ ∣ ∣∣ is equal to 

A. 1
B. 0
C. -1
D. 2

SOLUTION

Solution : B

Δ=∣∣ ∣∣sinπsinBcosC−sinB0tanAcos(π−C)−tnaA0∣∣ ∣∣=∣∣ ∣∣0sinBcosC−sinB0tanA−cosC−tanA0∣∣ ∣∣
=0 (∵Δ is skew symmetric)

Question 7

Let f(x)=∣∣ ∣∣cosxsinxcosxcox2xsin2x2cos2xcos3xsin3x3cos3x∣∣ ∣∣, then f′(π2) is equal to 

A. 8
B. 6
C. 4
D. 2

SOLUTION

Solution : 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

Question 8

If a,b,c are  non – zero real numbers and if the equations (a-1) x = y + z, (b -1)y = z + x, (c - 1)z = x + y has a non trivial solution, then ab + bc + ca is equal to

A. a +b + c
B. abc
C. 1
D. None of these

SOLUTION

Solution : 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
 

Question 9

∣∣ ∣∣1ab−a1c−b−c1∣∣ ∣∣=

A. 1+a2+b2+c2
B. 1−a2+b2+c2
C. 1+a2+b2−c2
D. 1+a2−b2+c2

SOLUTION

Solution : A

∣∣ ∣∣1ab−a1c−b−c1∣∣ ∣∣=1(1+c2)−a(−a+bc)+b(ac+b)=1+a2+b2+c2.

Question 10

If pλ4+qλ3+rλ+5λ+t=∣∣ ∣∣λ2+3λλ−1λ+3λ+12−λλ−4λ−3λ+43λ∣∣ ∣∣, the value of t is

A. 16
B. 18
C. 17
D. 19

SOLUTION

Solution : B

Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0 in the given equation, we have
t=∣∣ ∣∣0−1312−4−340∣∣ ∣∣=−12+30=18.
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