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Let A be a 3×3 matrix such that A[123023011]=[001100010] Then A−1 is:
(a) Given A[123023011]=[001100010]
Applying C1↔C3
A[321320110]=[100001010]
Again Applying C2↔C3
A[312302101]=[100010001]
pre-multiplying both sides by A−1
A−1 A[312302101]=A−1[100010001]
I[312302101]=A−1I=A−1
(∵A−1 A=I and I= Identity matrix )
Hence, A−1=[312302101]
If P=[1α3133244] is the adjoint of a 3×3 matrix A and |A|=4, then α is equal to :
(b)
|P|=1(12−12)−α(4−6)+3(4−6)=2α−6
Now, adjA=P⇒|adjA|=|P|⇒|A|2=|P|⇒|P|=16
⇒2a−6=16⇒a=11
Let P and Q be 3×3 matrices P≠Q. If P3=Q3 and P2Q=Q2P then determinant of (P2+Q2) is equal to :
(c) Given P3=Q3…(1)
and P2Q=Q2P…(2)
Subtracting (1) and (2), we get
P3−P2Q=Q3−Q2P⇒P2(P−Q)+Q2(P−Q)=0⇒(P2+Q2)(P−Q)=0
If |P2+Q2|≠0 then P2+Q2 is invertible.
⇒P−Q=0⇒P=Q
Which gives a contradiction (∵P≠Q)
Hence |P2+Q2|=0
Let A=(100210321). If u1 and u2 are column matrices such that Au1=(100) and Au2=(010), then u1+u2 is equal to :
(d) Let Au1=(100) and Au2=(010)
Then, Au1+Au2=(100)+(010)
⇒A(u1+u2)=(110)…(1)
Also, A=(100210321)
⇒|A|=1(1)−0(2)+0(4−3)=1
We know,
A−1=1|A|adjA⇒A−1=adj(A)(∵|A|=1)
Now, from equation (1), we have
u1+u2=A−1(110)=[100−2101−21](110)=[1−1−1]
If AT denotes the transpose of the matrix A=[00a0bcdef], where a,b,c,d,e and f are integers such that abd≠0, then the number of such matrices for which A−1=AT is
(c) A=[00a0bcdef],|A|=−abd≠0
c11=+(bf−ce),c12=−(−ad)=cd,c13=+(−bd)=−bdc21=−(−ea)=ae,c22=+(−ad)=−ad,c23=−(0)=0c31=+(−ab)=−ab,c32=−(0)=0,c33=0
AdjA=[(bf−ce)ae−abcd−ad0−bd00]
A−1=1|A|(adjA)=1abd[bf−ceae−abcd−ad0−bd00]
AT=[00d0beacf] Now A−1=AT⇒1−abd[bf−ceae−abcd−ad0−bd00]=[00d0beacf]
⇒[bf−ceae−abcd−ad0−bd00]=[00−abd20−ab2d−abde−a2bd−abcd−abdf]
∴bf−ce=ae=cd=0…(i)abd2=ab,ab2d=ad,a2bd=bd…(ii)abde=abcd=abdf=0…(iii)
From (ii), (abd2)⋅(ab2d)⋅(a2bd)=ab⋅ad⋅bd⇒(abd)4−(abd)2=0⇒(abd)2[(abd)2−1]=0
∵abd≠0,∴abd=±1…(iv)
From (iii) and (iv),
e=c=f=0…(v)
From (i) and (v), bf=ae=cd=0…(vi)
From (iv), (v), and (vi), it is clear that a,b,d can be any non-zero integer such that abd=±1. But it is only possible if a=b=d=±1
Hence, there are 2 choices for each a,b and d. there fore, there are 2×2×2 choices for a,b and d. Hence number of required matrices =2×2×2=(2)3
Let A and B be real matrices of the form [α00β] and [0γδ0], respectively.
Statement 1: AB−BA is always an invertible matrix.
Statement 2: AB−BA is never an identity matrix
Let A and B be real matrices such that A=[α00β]
and B=[0γδ0]
Now, AB=[0αγβδ0]
and BA=[0γβδα0]
Statement-1:
AB−BA=[0γ(α−β)δ(β−α)0]|AB−BA|=(α−β)2γδ≠0
∴AB−BA is always an invertible matrix.
Hence, statement – 1 is true.
But AB−BA can be identity matrix if γ=−δ or δ=−γ
So, statement – 2 is false.
Consider the following relation R on the set of real square matrices of order 3 .
R={(A,B)∣A=P−1BP for some invertible matrix P}
Statement- 1:R is equivalence relation.
Statement-2 : For any two invertible 3×3 matrices M and N,(MN)−1=N−1M−1
For reflexive
(A,A)∈R
A=P−1AP is true, For P=I, which is an invertible matrix. ∴R is reflexive.
For symmetry
As (A,B)∈R for matrix P
A=P−1BP⇒PAP−1=B⇒B=PAP−1⇒B=(P−1)−1 A(P−1)
∴(B,A)∈R[latex]formatrix[latex]P−1
∴R is symmetric.
For transitivity
A=P−1BP and B=P−1CP
⇒A=P−1(P−1CP)P⇒A=(P−1)2CP2⇒A=(P2)−1C(P2)
∴(A,C)∈R for matrix P2
∴R is transitive.
So R is equivalence
Let A be a 2×2 matrix
Statement -1: adj(adjA)=A
Statement -2 : |adjA|=|A|
(a) We know that |adj(adjA)|=|A|n−2A.
=|A|0A=A
Also |adjA|=|A|n−1=|A|2−1=|A|
∴ Both the statements are true but statement- 2 is not a correct explanation for statement- 1.
Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
(c) ∵ All entries of square matrix A are integers, therefore all cofactors should also be integers.
If detA=±1 then A−1 exists. Also all entries of A−1 are integers.
If A2−A+I=0, then the inverse of A is
(d) Given A2−A+I=0
A−1A2−A−1A+A−1I=A−1.0
(Multiplying A−1 on both sides)
⇒A−1+A−1=0 or A−1=I−A.
Let A=(1−1121−3111). and B=(422−50α1−23). If B is the inverse of matrix A, then α is
(a)
Given that 10B=[422−50α1−23]⇒B=110[422−50α1−23]
Also since, B=A−1⇒AB=I
⇒110[1−1121−3111][422−50α1−23]=[100010001]⇒110[1005−2010−5+α005+α]=[100010001]⇒5−α10=0⇒α=5
Let A=(00−10−10−100). The only correct statement about the matrix A is
(a) A=[00−10−10−100]
clearly A≠0. Also |A|=−1≠0
∴A−1 exists, further (−1)I=[−1000−1000−1]≠A
Also A2=[00−10−10−100][00−10−10−100]
=[100010001]=I
If S is the set of distinct values of ‘ b ‘ for which the following system of linear equations
x+y+z=1x+ay+z=1ax+by+z=0
has no solution, then S is :
(a)
D=|1111a1ab1|=0
⇒1[a−b]−1[1−a]+1[b−a2]=0⇒(a−1)2=0⇒a=1
For a=1, the First two equations are identical ie. x+y+z=1
To have no solution with x+by+z=0 b=1
So b={1}⇒ It is singleton set.
The number of real values of λ for which the system of linear equations
2x+4y−λz=04x+λy+2z=0λx+2y+2z=0
has infinitely many solutions, is :
Since the given system of linear equations has infinitely many solutions.
|24−λ4λ2λ22|=0λ3+4λ−40=0
λ has only 1 real root.
The system of linear equations
x+λy−z=0λx−y−z=0x+y−λz=0
has a non-trivial solution for:
(b) For trivial solution,
|1λ−1λ−1−111−λ|=0⇒−λ(λ+1)(λ−1)=0⇒λ=0,+1,−1
The set of all values of λ for which the system of linear equations :
2x1−2x2+x3=λx12x1−3x2+2x3=λx2−x1+2x2=λx3
has a non-trivial solution,
(a)
2x1−2x2+x3=λx12x1−3x2+2x3=λx2−x1+2x2=λx3}⇒(2−λ)x1−2x2+x3=02x1−(3+λ)x2+2x3=0−x1+2x2−λx3=0
For non-trivial solution,
Δ=0
i.e. |2−λ−212−(3+λ)2−12−λ|=0
⇒(2−λ)[λ(3+λ)−4]+2[−2λ+2]+1[4−(3+λ)]=0⇒λ3+λ2−5λ+3=0⇒λ=1,1,3
Hence λ has 2 values.
If a,b,c are non-zero real numbers and if the system of 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 equals:
Given system of equations can be written as
(a−1)x−y−z=0−x+(b−1)y−z=0
−x−y+(c−1)z=0
For non-trivial solution, we have
|a−1−1−1−1b−1−1−1−1c−1|=0R2→R2−R3|a−1−1−10b−c−1−1c−1|=0C2→C2−C3|a−10−10b+c−c−1−cc−1|=0 Apply R3→R3−R1|a−10−10b+c−c−a−cc|=0⇒(a−1)[bc+c2−c2]−1[a(b+c)]=0⇒(a−1)[bc]−ab−ac=0⇒abc−bc−ab−ac=0⇒ab+bc+ca=abc
The number of values of k, for which the system of equations :
(k+1)x+8y=4kkx+(k+3)y=3k−1
has no solution, is
(b) From the given system, we have
k+1k=8k+3≠4k3k−1(∵ System has no solution) ⇒k2+4k+3=8k⇒k=1,3
If k=1 then 81+3≠4.12 which is false and if k=3 then 86≠4.39−1 which is true, therefore k=3
Hence for only one value of k. System has no solution.
Consider the system of equations : x+ay=0,y+az=0 and z+ax=0. Then the set of all real values of ‘ a ‘ for which the system has a unique solution is:
(b) Given system of equations is homogeneous which is
x+ay=0y+az=0z+ax=0
It can be written in matrix form as
A=(1a001aa01)
Now, |A|=[1−a(−a2)]=1+a3≠0
So, system has only trivial solution.
Now, |A|=0 only when a=−1
So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution.
Hence set of all real values of ‘ a ‘ is R−{−1}.
Statement-1: The system of linear equations
x+(sinα)y+(cosα)z=0x+(cosα)y+(sinα)z=0x−(sinα)y−(cosα)z=0
has a non-trivial solution for only one value of α lying in the interval (0,π2)
Statement-2: The equation in α
|cosαsinαcosαsinαcosαsinαcosα−sinα−cosα|=0
has only one solution lying in the interval (0,π2).
(c) Δ1=|1sinαcosα1cosαsinα1−sinαcosα|
=|0sinα−cosαcosα−sinα0cosα+sinαsinα−cosα1−sinαcosα|
=(sinα−cosα)2−(cos2α−sin2α)=sin2α+cos2α−2sinα⋅cosα−cos2α+sin2α=2sin2α−2sinα⋅cosα=2sinα(sinα−cosα)
Now, sinα−cosα=0 for only
α=π4 in (0,π2)Δ1=2(sinα)×0=0
since value of sinα is finite for α∈(0,π2)
Hence non-trivivial solution for only one value of α in (0,π2)
|cosαsinαcosαsinαcosαsinαcosα−sinα−cosα|=0⇒|0sinαcosα0cosαsinα2cosα−sinα−cosα|=0⇒2cosα(sin2α−cos2α)=0∴cosα=0 or sin2α−cos2α=0
But cosα=0 not possible for any value of α∈(0,π2) ∴sin2α−cos2α=0⇒sinα=−cosα, which is also not possible for any value of α∈(0,π2) Hence, there is no solution.
If the system of linear equations:
x1+2x2+3x3=6x1+3x2+5x3=92x1+5x2+ax3=b
is consistent and has infinite number of solutions, then :
(d) Given system of equations can be written in matrix form as AX=B where
A=(12313525a) and B=(69b)
Since, system is consistent and has infinitely many solutions
∴(adj.A)B=0
⇒(3a−2515−2a110−aa−6−2−1−11)(69b)=(000)⇒−6−9+b=0⇒b=15 and 6(10−a)+9(a−6)−2(b)=0⇒60−6a+9a−54−30=0⇒3a=24⇒a=8 Hence, a=8,b=15.
Statement 1: If the system of equations x+ky+3z=0, 3x+ky−2z=0,2x+3y−4z=0 has a non-trivial solution, then the value of k is 312.
Statement 2: A system of three homogeneous equations in three variables has a non-trivial solution if the determinant of the coefficient matrix is zero.
Given system of equations is
x+ky+3z=03x+ky−2z=02x+3y−4z=0
Since, system has non-trivial solution
∴|1k33k−223−4|=0
⇒1(−4k+6)−k(−12+4)+3(9−2k)=0⇒4k+33−6k=0⇒k=332
Hence, statement – 1 is false. Statement- 2 is the property. It is a true statement.
If the system of equations
x+y+z=6x+2y+3z=10x+2y+λz=0
has a unique solution, then λ is not equal to
(d) Given system of equations is
x+y+z=6x+2y+3z=10x+2y+λz=0
It has unique solution.
∴|11112312λ|≠0⇒1(2λ−6)−1(λ−3)+1(2−2)≠0⇒2λ−6−λ+3≠0⇒λ−3≠0⇒λ≠3
If the trivial solution is the only solution of the system of equations
x−ky+z=0kx+3y−kz=03x+y−z=0
then the set of all values of k is:
(a)
x−ky+z=0kx+3y−kz=03x+y−z=0
The given system of equations will have non trivial solution, if
|1−k1k3−k31−1|=0
⇒1(−3+k)+k(−k+3k)+1(k−9)=0⇒k−3+2k2+k−9=0⇒k2+k−6=0⇒k=−3,k=2
So the equation will have only trivial solution, when k∈R−{2,−3}
The number of values of k for which the linear equations 4x+ky+2z=0,kx+4y+z=0 and 2x+2y+z=0 possess a non-zero solution is
(a)
Δ=0⇒|4k2k41221|=0⇒4(4−2)−k(k−2)+2(2k−8)=0⇒8−k2+2k+4k−16=0k2−6k+8=0⇒(k−4)(k−2)=0⇒k=4,2
Consider the system of linear equations;
x1+2x2+x3=32x1+3x2+x3=33x1+5x2+2x3=1
The system has
(c) D=|121231352|=0D1=|321331152|≠0
⇒ Given system, does not have any solution.
⇒ No solution
Let a,b,c be any real numbers. Suppose that there are real numbers x,y,z not all zero such that x=cy+bz,y=az+cx, and z=bx+ay. Then a2+b2+c2+2abc is equal to
(d)
The given equations are
−x+cy+bz=0cx−y+az=0bx+ay−z=0
∵x,y,z are not all zero
∴ The above system should not have unique (zero) solution
⇒Δ=0⇒|−1cbc−1aba−1|=0⇒−1(1−a2)−c(−c−ab)+b(ac+b)=0⇒−1+a2+b2+c2+2abc=0⇒a2+b2+c2+2abc=1
The system of equations
αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1
has infinite solutions, if α is
(a)
αx+y+z=α−1x+αy+z=α−1x+y+zα=α−1Δ=|α111α111α|
=α(α2−1)−1(α−1)+1(1−α)=α(α−1)(α+1)−1(α−1)−1(α−1)
For infinite solutions, Δ=0
⇒(α−1)[α2+α−1−1]=0⇒(α−1)[α2+α−2]=0
⇒(α−1)[α2+2α−α−2]=0⇒(α−1)[α(α+2)−1(α+2)]=0
(α−1)=0,α+2=0⇒α=−2,1;
But α≠1.
∴α=−2
If the system of linear equations
x+2ay+az=0;x+3by+bz=0;
x+4cy+cz=0 has a non – zero solution, then a,b,c.
(d) For homogeneous system of equations to have non zero solution, Δ=0
|12aa13bb14cc|=0C2→C2−2C3
|10a1bb12cc|=0R3→R3−R2,R2→R2−R1
On simplification, 2b=1a+1c
∴a,b,c are in Harmonic Progression.
For any y∈R, let cot−1(y)∈(0,π) and tan−1(y)∈(−π2,π2). Then the sum of all the solutions of the equation tan−1(6y9−y2)+cot−1(9−y26y)=2π3 for 0<|y|<3, is equal to
(c)
Case-I : y∈(−3,0)
tan−1(6y9−y2)+π+tan−1(6y9−y2)=2π3
2tan−1(6y9−y2)=−π3
y2−6√3y−9=0⇒y=3√3−6(∵y∈(−3,0))
Case-II : y∈(0,3)
2tan−1(6y9−y2)=2π3⇒√3y2+6y−9√3=0y=√3 or y=−3√3 (rejected) sum =√3+3√3−6=4√3−6
Let M=(aij),i,j∈{1,2,3}, be the 3×3 matrix such that aij=1 if j+1 is divisible by i, otherwise aij=0. Then which of the following statements is (are) true ?
M=[a11a12a13a21a22a23a31a32a33]=[111101010]
|M|=−1+1=0⇒M is singular so non-invertible
(B) M[a1a2a3]=[−a1−a2−a3]⇒[111101010][a1a2a3]=[−a1−a2−a3]
a1+a2+a3=−a1a1+a3=−a2a2=−a3}⇒a1=0 and a2+a3=0 infinite solutions exists [B] is correct.
Option (D)
M−2I=[111101010]−2[100010001]=[−1111−2101−2]
|M−2I|=0⇒[D] is wrong
Option (C) :
MX=0⇒[111101010][xyz]=[000]
x+y+z=0x+z=0y=0
∴ Infinite solution
[C] is correct
Let R={(a3 bc2 d050):a,b,c,d∈{0,3,5,7,11,13,17,19}}. Then the number of invertible matrices in R is
Let us calculate when |R|=0
Case-I: ad=bc=0
Now ad =0
⇒ Total – (When none of a \& d is 0 )
=82−1=15 ways
Similarly bc=0⇒15 ways
∴15×15=225 ways of ad=bc=0
Case-II: ad=bc≠0
either a=d=b=c OR a≠d,b≠d but ad=bc
7C1=7 ways 7C2×2×2=84 ways
Total 91 ways
∴|R|=0 in 225+91=316 ways |R|≠0 in 84−316=3780
Let for A=[123α31112],|A|=2. If |2adj(2adj(2A))|=32n, then 3n+α is equal to
|A|=2adj(kA)=km−1adjA{m= order of matrix }
adj(2A)=22adjA=4adj(A)adj(2adj(2A))=adj(8adjA)=82adjadj(A)
|2adj2adj(2 A)|=|27adjadj(A)|
=(27)3| A|22=221| A|4=221⋅24⇒225=(32)n⇒225=25n
⇒n=5| A|=2(6−1)−2(2α−1)+3(α−3)=2⇒5−4α+2+3α−9⇒α=−43n+α=11
If the system of equations
2x+y−z=52x−5y+λz=μx+2y−5z=7
has infinitely many solutions, then (λ+μ)2+(λ−μ)2 is equal to
(b)
Δ=0⇒|21−12−5λ12−5|=0
⇒2(25−2λ)−1(−10−λ)−1(4+5)=0⇒51−3x=0⇒λ=17Δx=0
|51−1μ−51772−5|=0⇒5(25−34)−1(−5μ−119)−1(2μ+35)=0⇒−45+5μ+119−2μ−35=0⇒39+3μ=0⇒μ=−13(λ+μ)2+(λ−μ)2=42+(30)2=916
The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9}, is
A=[abcbdecef],a,b,c,d,e,f∈{0,1,2,….9}, Number of matrices =106
For the system of linear equations
2x+4y+2az=bx+2y+3z=42x−5y+2z=8
which of the following is NOT correct?
Δ=|242a1232−52|=18(3−a)Δx=|b42a4238−52|=(64+19 b−72a)
For unique solution Δ=0
⇒a≠3 and b∈R
For infinitely many solution :
Δ=Δx=Δy=Δz=0⇒a=3∵Δ=0 and b=8∵Δx=0
Let B=[13α123αα4],α>2 be the adjoint of a matrix A and |A|=2. then
[α−2αα]B[α−2αα] is equal to
Given, B=[13α123αα4]
|B|=41(8−3α)−3(4−3α)+α(α−2α)=4−α2+6α−8=0α=2,4
Given α>2
So, α=2 is rejected
[4−84][134123444][4−84]=−16
Let the determinant of a square matrix A of order m be m−n, where m and n satisfy 4m+n=22 and 17m+ 4n=93. If det(nadj(adj(mA)))=3a5b6c, then a+b+c is equal to
|A|=m−n4 m+n=2217 m+4n=93 m=5,n=2| A|=3
∣2adj(adj5A))|=25|5A|16
=25⋅580| A|16=25⋅580⋅316=311⋅580⋅65
a+b+c=96
Let Pb a square matrix such that P2=I−P. For α,β,γ,δ∈N, if Pα+Pβ=γI−29P and Pα−Pβ=δI−13P, then α+β+γ−δ is equal to:
P2=I−PPα+Pβ=γI−29PPα−Pβ=δI−13PP4=(I−P)2=I+P2−2PP4=I+I−P−2P=2I−3PP8=(P4)2=(2I−3P)2=4I+9P2−12P=4I+9(I−P)−12PP8=13I−21P…(1)P6=P4⋅P2=(2I−3P)(I−P)=2I−5P+3P2=2I−5P+3(I−P)=5I−8P…(2)
(1) +(2)
P8+P6=18I−29P
(1)−(2)P8−P6=8I−13P
Comparing with given equation
α=8,β=6γ=18δ=8α+β+γ−δ=32−8=24
If the system of equations
x+y+az=b2x+5y+2z=6x+2y+3z=3
has infinitely many solutions, then 2a+3b is equal to :
x+y+az=b2x+5y+2z=6x+2y+3z=3
For ∞ solution
Δ=0,Δx=0,Δy=0,Δz=0
Δ=|11a252123|=0⇒11−4−a=0⇒a=7
Δz=|11b256123|=0⇒3−0−b=0⇒b=3
Hence 2a+3b=23
If A=[15λ10],A−1=αA+βI and α+β=−2, then 4α2+β2+λ2 is equal to :
|A−x|=0⇒|1−x5λ10−x|=0⇒x2−11x+10−5λ=0⇒(10−5λ)A−1=−A+11I∴α=−110−5λ and β=+1110−5λ
α+β=−2⇒1010−5λ=−2⇒10−5λ=−5⇒λ=3∴α=15&β=−115∴4a2+β2+λ2=425+12125+32=14 Ans.
Let S be the set of all values of θ∈[−π,π] for which the system of linear equations
x+y+√3z=0−x+(tanθ)y+√7z=0x+y+(tanθ)z=0
has non-trivial solution. Then 120π∑θ=Sθ is equal to
For non trivial solutions
D=0|11√3−1tanθ√711tanθ|=0tan2θ−(√3−1)−√3=0tanθ=√3,−1θ={π3,−2π3,−π4,3π4}120π(Σθ)=120π×π6=20
Let P=[√3212−12√32],A=[1101] and Q=PAPT. If PTQ2007P=[abcd],
then 2a+b−3c−4d equal to
Q=PAPTPT⋅Q2007⋅P=PT⋅Q⋅Q…Q⋅P
=PT(PAPT)(P⋅APT)…(PAPT)P
⇒(PTP)A(PTP)A…A(PTP)
PT⋅P=[√3/2−1/21/2√3/2][−√3/21/2−1/2√3/2]=[1001]=I
∴PT⋅Q2007⋅P=A2007
A2=[1101][1101]=[1201]
∴A2007=[1200701]=[abcd]
a=1,b=2007,c=0,d=12a+b−3c−4d=2+2007−4=2005
Let A=[21012−10−12]
If |adj(adj(adj2 A))|=(16)n, then n is equal to
|A|=2[3]−1[2]=4∴|adj(adj(adj2A))|=|2A|(n−1)3⇒|2A|8=16n⇒(23|A|)8=16n⇒(23×22)8=16n
=240=16n=1610=16n⇒n=10
If A=15!6!7![5!6!7!6!7!8!7!8!9!], then |adj(adj(2A))| is equal to
|A|=15!6!7!5!6!7!|164217561872|R3→R3→R2R2→R2→R1|A|=|184201140116|=2|adjadj(2A)|=|2A|(n−1)2=|2A|4=(23|A|)4=212|A|4⇒216
Let S be the set of values of λ, for which the system of equations
6λx−3y+3z=4λ22x+6λy+4z=1
3x+2y+3λz=λ has no solution.
Then 12∑l∈S|λ| is equal to _____.
Δ=|6λ−3326λ4323λ|=02λ(9λ2−4)+(3λ−6)+(2−9λ)=018λ3−14λ−4=0(λ−1)(3λ+1)(3λ+2)=0⇒λ=1,−1/3,−2/3
For each values of λ,Δ1=|6λ−34λ226λ132λ|≠0
12(1+13+23)=24
If A is a 3×3 matrix and |A|=2, then |3adj(|3A|A2)| is equal to :
Given |A|=2 Now, |3adj(|3 A|A2)||3 A|=33⋅|A|=33.(2)
Adj. (|3 A|A2)=adj{(33⋅2)A2}=(2⋅33)2(adjA)2=22⋅36⋅(adjA)2
|3adj(|3 A|A2)|=|22⋅3⋅36(adjA)2|=(22⋅37)3|adjA|2=26⋅321(| A|2)2=26⋅321(22)2=210⋅321=210⋅310⋅311
|3adj(|3A|A2)|=610⋅311
For the system of linear equations
2x−y+3z=53x+2y−z=74x+5y+αz=β,
which of the following is NOT correct?
Given
2x−y+3z=53x+2y−z=74x+5y+αz=β
Δ=|2−1332−145α|=7α+35
Δ=7(α+5)
For unique solution Δ≠0
α≠−5
For inconsistent & Infinite solution
Δ=0α+5=0⇒α=−5
Δ1=|5−1372−1β5−5|=−5(β−9)Δ2=|25337−14β−5|=11(β−9)Δ3=|2−1532745β|
For Inconsistent system : –
At least one Δ1,Δ2&Δ3 is not zero α=−5,β=8 option (a) True Infinite solution:
Δ1=Δ2=Δ3=0
From here β−9=0⇒β=9
α=−5& option (d) True
β=9
Unique solution
α≠−5,β=8→ option (c) True
Option (b) False
For Infinitely many solution α must be -5 .
If the system of linear equations
7x+11y+αz=135x+4y+7z=β175x+194y+57z=361
has infinitely many solutions, then α+B+2 is equal to :
7x+11y+αz=135x+4y+7z=β175x+194y+57z=361
4sc condition of Infinite Many solution Δ=0 \& Δx,Δy,Δz=0 check.
After solving we get α+13+2=4
Let A be a 2×2 matrix with real entries such that A′=αA+I, where a∈R−{−1,1}. If det(A2−A)=4, then the sum of all possible values of α is equal to :
AT=αA+IA=αAT+IA=α(αA+I)+IA=α2 A+(α+1)IA(1−α2)=(α+1)I
A=I1−α…(1)
|A|=1(1−α)2…(2)
|A2−A|=|A||A−1|… (3)
A−I=II−α−I=α1−αI|A−I|=(α1−α)2…(4)
Now |A2−A|=4|A||A−I|=4⇒1(1−α)2α2(1−α2)=4⇒α(1−α)2=±2⇒2(1−α)2=±α(C1)2(1−α)2=α
2α2−5α+2=0<α1α2
α1+α2=52
(C2)2(1−α)2=−α
2α2−3α+2=0α∉R
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