GATE Matrix questions cracked !!!
....NEO: The Matrix. What is the Matrix?
TRINITY: Twelve years ago I met a man, a great man, who said that no
one could be told the answer to that question. That they had to see
it, to believe it.
Her body is against his, her lips very close to his ear.
TRINITY : He told me that no one should look for the answer unless
they have to because once you see it, everything changes. Your life
and the world you live in will never be the same. It's as if you wake
up one morning and the sky is falling.
....
GATE Matrix questions cracked !!!
*************************************
Hi friends,
Matrix is my all time favorite movie and Neo is my hero. So this
post is dedicated to Neo, Trinity , Morpheus and to all friends here
at this forum.
Well, lets come back to *real* world. Today, let us try to crack as
many GATE Matrix questions as possible. I was going through 2003-2007
GATE papers. You can expect two to eight marks from Linear Algebra in
GATE 2008.
Thats it.. lets start...
AamzillaA
http://living-in-shadows.blogspot.com/
************************************************************
Unsolved problems:
*Gate 07 CS\Q25 , Gate 06 CS\Q23 (Kindly help if u know how to solve these)
Doubtful solution:
Gate 04 CS\Q76,2 (I think my solution is correct for this question
since i got an answer. But I am not sure whether my reasoning is
correct, please have a look, please see whether there is another way
of solving this problem )
************************************************************
3)
*G05\Q48,2
Consider the following system of equations in three variables x1,x2 and x3
2 x1 - x2 + 3 x3 = 1
3 x1 + 2 x2 + 5 x3 = 2
-1 x1 + 4 x2 + 1 x1=3
The system of equations have
A) no solution
B) a unique solution
C) more than one but a finite number of solutions
D) an infinite number of solutions
Ans: B
Matrix A =
2 -1 3
3 2 5
-1 4 1
|A| = 14
There fore, rank of Matrix A = 3
Augmented Matrix A|B =
2 -1 3 1
3 2 5 2
-1 4 1 3
The rank of Augmented matrix A|B = 3
Since rank of A is equal to rank of Augmented matrix A|B , the system
has a unique solution.
So answer is B
************************************************************
4)
*G05\Q49,2
What are the eigen values of the following 2X2 matrix ?
2 -1
-4 5
A) -1 and 1
B) 1 and 6
C) 2 and 5
D) 4 and -1
ANS: B
This is a two mark question which takes just 2 seconds to
answer , if you know that the product of eigen values of a matrix is
equal to its determinant.
The determinant of the matrix
2 -1
-4 5
is 6
In the given choices , only choice B has got a product of 6. So answer is B.
The traditional way of solving this is
let # represent lamda (I don't find the lamda symbol in notepad)
| A - #I | = 0
so
Determinant of the following matrix will be zero
(2 - #) -1
-4 (5 - #)
ie (2 - # )(5 - #) - (4) = 0
10 - 2# - 5# + #^2 - 4 = 0
ie 6 - 7# + #^2 = 0
ie #^2 - 7# + 6 = 0
solving this quadratic equation we get # = 1,6
ANS: B
************************************************************
5)
*G04\Q26,1
The number of different n x n symmetric matrices with each element either 0 or 1 is :
A) power(2,n)
B) power(2,n^2)
C) power(2,(n^2+n)/2)
D) power(2,(n^2-n)/2)
ANS:C
This is a cute, tricky and simple question...
Let us follow the KISS principle (KISS : keep it simple , stupid ;-) )
Consider the case where n=2
ie a 2 x 2 matrix A
Matrix A =
a b
c d
Total number of elements in this 2x2 matrix is n*n = 2*2 =4
Number of diagonal elements = n = 2
number of non diagonal elements = n*n - n = 4-2=2
Number of upper triangular elements excluding main diagonal elements is
= [ (total number of elements) - (number of diagonal elements) ] / 2
= [(n*n)-(n)]/2
= [2*2 - 2]/2 = 1
These ( n*n - n )/2 upper triangular spaces can hold either 0 or 1.
Once we choose 0 or 1 in any of these positions, the same should be
written in the corresponding lower triangular space, for the matrix to
be symmetric.
In the n principal diagonal element spaces can hold either 0 or 1
So the total number of spaces which can hold either 0 or 1 is
((n*n)-n)/2 +n = (n*n + n )/2
So number of symmetric matrices are,
2 ^ [(n*n + n )/2 ]
So Answer is C
************************************************************
6)
*G04\Q27,1
Let A,B,C,D be n x n matrices, each with non-zero determinant, If ABCD
= I then B^ (-1),B inverse, is
A) D^(-1)C^(-1)A^(-1)
B) CDA
C) ADC
D) Does not necessarily exist.
Ans : B
Since |B| is not equal to zero, B ^ (-1 ) exists.
We know,
ABCD = I
(ABCD)^ -1 = D^-1 C ^-1 B ^-1 A^-1 = I^-1 =I
D . (D^-1 C ^-1 B ^-1 A^-1) = C ^-1 B ^-1 A^-1= D. I = D
C. (C ^-1 B ^-1 A^-1) = B ^-1 A^-1 = CD
(B ^-1 A^-1).A=B ^-1= CDA
So B ^-1= CDA
So answer is B
************************************************************
7)
*G04\Q71,2
How many Solutions does the following system of linear equations have ?
-x + 5y =-1
x -y =2
x+3y =3
A) Infinitely many
B) two distinct solutions
C) Unique
D) None
Ans : C
Method 1:
Let Matrix A =
-1 5
1 -1
1 3
The rank of matrix A = 2
Let the augmented matrix A|B =
-1 5 -1
1 -1 2
1 3 3
The determinant of Augmented Matrix A|B is Zero
There fore the rank of the Augmented Matrix is less than 3 and we find it as 2
Since the rank of the matrix A is equal to rank of augmented matrix is
equal to the number of variables there will be a unique solution.
Method 2:
We have 3 equations and two variables ,
-x + 5y =-1 ------------- equation 1
x -y =2---------------- equation 2
x+3y =3 --------------equation 3
Substituting 2 in 3 we get x = 9/4 and y= 1/4
lets substitute these values in equation 1
-x + 5y =-1
-9/4 +5/4 = -1
So the system of equation is consistent and has a unique solution. I
liked this second solution since its less slower.
************************************************************
8)
*G04\Q76,2
In an MxN matrix such that all non-zero entries are covered in "a"
rows and "b" columns. Then the maximum number of non-zero entries
,such that no two are on the same row or column is,
A) < = a+b
B) < = max (a,b)
C) < = min (M-a,N-b)
D) < = min (a,b)
ANS: is probably D
################
WARNING !!!
Friends, this could be a possibly wrong explanation. Please check
yourself whether my argument is correct. If you know to solve this
problem in a different way, kindly post a comment.
################
Let M=2 , N= 3
Lets consider a 2 x 3 matrix , A
Matrix A =
x1 x2 x3
y1 y2 y3
The condition that no two non-zero elements are on the same row or
column is, satisfied by the elements present on any diagonal of a
square matrix inside this parent matrix.
ie non zero elements can take the place of either x1,y2 or x2,y1 or
x2,y3 or x3,y2
There fore, the maximum number of non-zero entries ,such that no two
are on the same row or column =
min( a,b ) =min (2,3) = 2
So I think the choice is D
************************************************************
9)
*G03\Q41,2
Consider the following system of linear equations
| 2 1 -4 | | x | | a |
| 4 3 -12 | | y | = | 5 |
| 1 2 -8 | | z | | 7 |
Notice that the second and third column of the coefficient matrix are
linearly dependent.
For how many values of "a" , does this system of equations have
infinitely many solutions?
A) 0
B) 1
C) 2
D) infinitely many.
Ans : C
It is mentioned in the question that second and third column of the
coefficient matrix are linearly dependent.
ie second column * -4 = third column
So without checking we can say that the coefficient matrix is
singular. ie its determinant is zero.
if Coefficient matrix, A=
| 2 1 -4 |
| 4 3 -12 |
| 1 2 -8 |
Determinant of coefficient matrix |A| = 0
We find that the rank of the coefficient matrix , A , is 2
For the system to have infinitely many solutions, rank of coefficient
matrix should be equal to rank of the augmented matrix = 2
Augmented Matrix =
| 2 1 -4 a |
| 4 3 -12 5 |
| 1 2 -8 7 |
For the rank of the Augmented matrix to be zero , we should ensure
that there is no 3x3 matrix in the augmented matrix which is not
singular.
ie for all possible 3x3 matrices in the augmented matrix , their
determinants should be zero.
We can select 3 column from 4 columns in 4C3 ways = 4
There are three possible 3x3 matrices,P1,P2,P3 in the augmented
matrix, apart from the coefficient matrix.
#first matrix,P1=
| 2 1 a |
| 4 3 5 |
| 1 2 7 |
#second matrix,P2=
| 2 a -4 |
| 4 5 -12 |
| 1 7 -8 |
#third matrix,P3=
| a 1 -4 |
| 5 3 -12 |
| 7 2 -8 |
Now we should find the values of "a" for which |P1|=0 , |P2|=0, |P3|=0.
Solving |P1|=0 , we get a = 1/5
Solving |P2|=0,we get a = 1/5
Solving |P3|=0, we get a = 0
So for a= 1/5 and a = 0 we can say that the system has infinite solutions.
So Answer is C
************************************************************
2)
*G06\Q23,2
F is an n x n real matrix. b is an n x 1real vector. Suppose there are
two n x 1 vectors , u and v such that u ! = v , and Fu =b and Fv = b .
Which one of the following statements is false ?
A) Determinant of F is Zero
B) There is an infinite number of solutions to Fx =b
C) There is an x ! = 0 such that Fx = 0
D) F must have two identical rows.
************************************
Friends, Any ideas to crack this one ???
**********************************************
1)
*G07\Q25,2 (eigen value of a matrix)
Let A be a 4x4 matrix with eigenvalues -5,-2,1,4. Which of the
following is an eigen value of
| A I |
| I A |
where I is a 4x4 identity matrix.
A) -5
B) -7
C) 2
D) 1
Ans: C
Man.... never expected I would get out on the first ball... still no
idea how to solve this one.. lets come back to this after seeing other
problems...Any one who knows how to solve this problem please help...
************************************************************
Thats all for now guys,
Stay tuned for more posts...And thanks to all friends, especially to
Karthik and Chandler for giving valuable insight in to Math Logic
questions.
Bye,
AamzillaA
http://living-in-shadows.blogspot.com/
************************************************************
These are the solutions provided by friends....
Ashish solved this one
G 07 Q.25
Let Matrix A =
-5 0 0 0 with eigen values -5,-2,1,4(given).
0 -2 0 0
0 0 1 0
0 0 0 4
We have to find the eigen values of matrix
A I
I A
Therefore,characteristic equation=
A-L I
I A-L
so, (A-L) pow(2)-I pow(2)=0
=> (A-L+I)(A-L-I)=0
=> A+I=L or A-I=L
A+I values are -5+1,-2+1,1+1,4+1 =-4,-1,2,5
A-I values are -5-1 ,-2-1, 1-1, 4-1 =-6,-3,0,3.
so 8 eigen values are -4,-1,2,5,-6,-3,0,3.
2 is there so answer is C.
I don't know whether i m right or even the option is (C) or not.
Today I have tried it again due to this post.
hey AamzillA,where to get the reliable correct options of G 07 and other papers too.
TRINITY: Twelve years ago I met a man, a great man, who said that no
one could be told the answer to that question. That they had to see
it, to believe it.
Her body is against his, her lips very close to his ear.
TRINITY : He told me that no one should look for the answer unless
they have to because once you see it, everything changes. Your life
and the world you live in will never be the same. It's as if you wake
up one morning and the sky is falling.
....
GATE Matrix questions cracked !!!
*************************************
Hi friends,
Matrix is my all time favorite movie and Neo is my hero. So this
post is dedicated to Neo, Trinity , Morpheus and to all friends here
at this forum.
Well, lets come back to *real* world. Today, let us try to crack as
many GATE Matrix questions as possible. I was going through 2003-2007
GATE papers. You can expect two to eight marks from Linear Algebra in
GATE 2008.
Thats it.. lets start...
AamzillaA
http://living-in-shadows.blogspot.com/
************************************************************
Unsolved problems:
*Gate 07 CS\Q25 , Gate 06 CS\Q23 (Kindly help if u know how to solve these)
Doubtful solution:
Gate 04 CS\Q76,2 (I think my solution is correct for this question
since i got an answer. But I am not sure whether my reasoning is
correct, please have a look, please see whether there is another way
of solving this problem )
************************************************************
3)
*G05\Q48,2
Consider the following system of equations in three variables x1,x2 and x3
2 x1 - x2 + 3 x3 = 1
3 x1 + 2 x2 + 5 x3 = 2
-1 x1 + 4 x2 + 1 x1=3
The system of equations have
A) no solution
B) a unique solution
C) more than one but a finite number of solutions
D) an infinite number of solutions
Ans: B
Matrix A =
2 -1 3
3 2 5
-1 4 1
|A| = 14
There fore, rank of Matrix A = 3
Augmented Matrix A|B =
2 -1 3 1
3 2 5 2
-1 4 1 3
The rank of Augmented matrix A|B = 3
Since rank of A is equal to rank of Augmented matrix A|B , the system
has a unique solution.
So answer is B
************************************************************
4)
*G05\Q49,2
What are the eigen values of the following 2X2 matrix ?
2 -1
-4 5
A) -1 and 1
B) 1 and 6
C) 2 and 5
D) 4 and -1
ANS: B
This is a two mark question which takes just 2 seconds to
answer , if you know that the product of eigen values of a matrix is
equal to its determinant.
The determinant of the matrix
2 -1
-4 5
is 6
In the given choices , only choice B has got a product of 6. So answer is B.
The traditional way of solving this is
let # represent lamda (I don't find the lamda symbol in notepad)
| A - #I | = 0
so
Determinant of the following matrix will be zero
(2 - #) -1
-4 (5 - #)
ie (2 - # )(5 - #) - (4) = 0
10 - 2# - 5# + #^2 - 4 = 0
ie 6 - 7# + #^2 = 0
ie #^2 - 7# + 6 = 0
solving this quadratic equation we get # = 1,6
ANS: B
************************************************************
5)
*G04\Q26,1
The number of different n x n symmetric matrices with each element either 0 or 1 is :
A) power(2,n)
B) power(2,n^2)
C) power(2,(n^2+n)/2)
D) power(2,(n^2-n)/2)
ANS:C
This is a cute, tricky and simple question...
Let us follow the KISS principle (KISS : keep it simple , stupid ;-) )
Consider the case where n=2
ie a 2 x 2 matrix A
Matrix A =
a b
c d
Total number of elements in this 2x2 matrix is n*n = 2*2 =4
Number of diagonal elements = n = 2
number of non diagonal elements = n*n - n = 4-2=2
Number of upper triangular elements excluding main diagonal elements is
= [ (total number of elements) - (number of diagonal elements) ] / 2
= [(n*n)-(n)]/2
= [2*2 - 2]/2 = 1
These ( n*n - n )/2 upper triangular spaces can hold either 0 or 1.
Once we choose 0 or 1 in any of these positions, the same should be
written in the corresponding lower triangular space, for the matrix to
be symmetric.
In the n principal diagonal element spaces can hold either 0 or 1
So the total number of spaces which can hold either 0 or 1 is
((n*n)-n)/2 +n = (n*n + n )/2
So number of symmetric matrices are,
2 ^ [(n*n + n )/2 ]
So Answer is C
************************************************************
6)
*G04\Q27,1
Let A,B,C,D be n x n matrices, each with non-zero determinant, If ABCD
= I then B^ (-1),B inverse, is
A) D^(-1)C^(-1)A^(-1)
B) CDA
C) ADC
D) Does not necessarily exist.
Ans : B
Since |B| is not equal to zero, B ^ (-1 ) exists.
We know,
ABCD = I
(ABCD)^ -1 = D^-1 C ^-1 B ^-1 A^-1 = I^-1 =I
D . (D^-1 C ^-1 B ^-1 A^-1) = C ^-1 B ^-1 A^-1= D. I = D
C. (C ^-1 B ^-1 A^-1) = B ^-1 A^-1 = CD
(B ^-1 A^-1).A=B ^-1= CDA
So B ^-1= CDA
So answer is B
************************************************************
7)
*G04\Q71,2
How many Solutions does the following system of linear equations have ?
-x + 5y =-1
x -y =2
x+3y =3
A) Infinitely many
B) two distinct solutions
C) Unique
D) None
Ans : C
Method 1:
Let Matrix A =
-1 5
1 -1
1 3
The rank of matrix A = 2
Let the augmented matrix A|B =
-1 5 -1
1 -1 2
1 3 3
The determinant of Augmented Matrix A|B is Zero
There fore the rank of the Augmented Matrix is less than 3 and we find it as 2
Since the rank of the matrix A is equal to rank of augmented matrix is
equal to the number of variables there will be a unique solution.
Method 2:
We have 3 equations and two variables ,
-x + 5y =-1 ------------- equation 1
x -y =2---------------- equation 2
x+3y =3 --------------equation 3
Substituting 2 in 3 we get x = 9/4 and y= 1/4
lets substitute these values in equation 1
-x + 5y =-1
-9/4 +5/4 = -1
So the system of equation is consistent and has a unique solution. I
liked this second solution since its less slower.
************************************************************
8)
*G04\Q76,2
In an MxN matrix such that all non-zero entries are covered in "a"
rows and "b" columns. Then the maximum number of non-zero entries
,such that no two are on the same row or column is,
A) < = a+b
B) < = max (a,b)
C) < = min (M-a,N-b)
D) < = min (a,b)
ANS: is probably D
################
WARNING !!!
Friends, this could be a possibly wrong explanation. Please check
yourself whether my argument is correct. If you know to solve this
problem in a different way, kindly post a comment.
################
Let M=2 , N= 3
Lets consider a 2 x 3 matrix , A
Matrix A =
x1 x2 x3
y1 y2 y3
The condition that no two non-zero elements are on the same row or
column is, satisfied by the elements present on any diagonal of a
square matrix inside this parent matrix.
ie non zero elements can take the place of either x1,y2 or x2,y1 or
x2,y3 or x3,y2
There fore, the maximum number of non-zero entries ,such that no two
are on the same row or column =
min( a,b ) =min (2,3) = 2
So I think the choice is D
************************************************************
9)
*G03\Q41,2
Consider the following system of linear equations
| 2 1 -4 | | x | | a |
| 4 3 -12 | | y | = | 5 |
| 1 2 -8 | | z | | 7 |
Notice that the second and third column of the coefficient matrix are
linearly dependent.
For how many values of "a" , does this system of equations have
infinitely many solutions?
A) 0
B) 1
C) 2
D) infinitely many.
Ans : C
It is mentioned in the question that second and third column of the
coefficient matrix are linearly dependent.
ie second column * -4 = third column
So without checking we can say that the coefficient matrix is
singular. ie its determinant is zero.
if Coefficient matrix, A=
| 2 1 -4 |
| 4 3 -12 |
| 1 2 -8 |
Determinant of coefficient matrix |A| = 0
We find that the rank of the coefficient matrix , A , is 2
For the system to have infinitely many solutions, rank of coefficient
matrix should be equal to rank of the augmented matrix = 2
Augmented Matrix =
| 2 1 -4 a |
| 4 3 -12 5 |
| 1 2 -8 7 |
For the rank of the Augmented matrix to be zero , we should ensure
that there is no 3x3 matrix in the augmented matrix which is not
singular.
ie for all possible 3x3 matrices in the augmented matrix , their
determinants should be zero.
We can select 3 column from 4 columns in 4C3 ways = 4
There are three possible 3x3 matrices,P1,P2,P3 in the augmented
matrix, apart from the coefficient matrix.
#first matrix,P1=
| 2 1 a |
| 4 3 5 |
| 1 2 7 |
#second matrix,P2=
| 2 a -4 |
| 4 5 -12 |
| 1 7 -8 |
#third matrix,P3=
| a 1 -4 |
| 5 3 -12 |
| 7 2 -8 |
Now we should find the values of "a" for which |P1|=0 , |P2|=0, |P3|=0.
Solving |P1|=0 , we get a = 1/5
Solving |P2|=0,we get a = 1/5
Solving |P3|=0, we get a = 0
So for a= 1/5 and a = 0 we can say that the system has infinite solutions.
So Answer is C
************************************************************
2)
*G06\Q23,2
F is an n x n real matrix. b is an n x 1real vector. Suppose there are
two n x 1 vectors , u and v such that u ! = v , and Fu =b and Fv = b .
Which one of the following statements is false ?
A) Determinant of F is Zero
B) There is an infinite number of solutions to Fx =b
C) There is an x ! = 0 such that Fx = 0
D) F must have two identical rows.
************************************
Friends, Any ideas to crack this one ???
**********************************************
1)
*G07\Q25,2 (eigen value of a matrix)
Let A be a 4x4 matrix with eigenvalues -5,-2,1,4. Which of the
following is an eigen value of
| A I |
| I A |
where I is a 4x4 identity matrix.
A) -5
B) -7
C) 2
D) 1
Ans: C
Man.... never expected I would get out on the first ball... still no
idea how to solve this one.. lets come back to this after seeing other
problems...Any one who knows how to solve this problem please help...
************************************************************
Thats all for now guys,
Stay tuned for more posts...And thanks to all friends, especially to
Karthik and Chandler for giving valuable insight in to Math Logic
questions.
Bye,
AamzillaA
http://living-in-shadows.blogspot.com/
************************************************************
These are the solutions provided by friends....
Ashish solved this one
G 07 Q.25
Let Matrix A =
-5 0 0 0 with eigen values -5,-2,1,4(given).
0 -2 0 0
0 0 1 0
0 0 0 4
We have to find the eigen values of matrix
A I
I A
Therefore,characteristic equation=
A-L I
I A-L
so, (A-L) pow(2)-I pow(2)=0
=> (A-L+I)(A-L-I)=0
=> A+I=L or A-I=L
A+I values are -5+1,-2+1,1+1,4+1 =-4,-1,2,5
A-I values are -5-1 ,-2-1, 1-1, 4-1 =-6,-3,0,3.
so 8 eigen values are -4,-1,2,5,-6,-3,0,3.
2 is there so answer is C.
I don't know whether i m right or even the option is (C) or not.
Today I have tried it again due to this post.
hey AamzillA,where to get the reliable correct options of G 07 and other papers too.
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