Question 1
Which one of the following does NOT equal to
 A A B B C C D D
GATE CS 2013    Linear Algebra
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Question 1 Explanation:

First of all, you should know the basic properties of determinants before approaching For these kind of problems. 1) Applying any row or column transformation does not change the determinant 2) If you interchange any two rows, sign of the determinant will change

A = | 1 x x^2 | | 1 y y^2 | | 1 z z^2 |

To prove option (b)

=> Apply column transformation C2 -> C2+C1

C3 -> C3+C1

=> det(A) = | 1 x+1 x^2+1 | | 1 y+1 y^2+1 | | 1 z+1 z^2+1 |

To prove option (c),

=> Apply row transformations R1 -> R1-R2

R2 -> R2-R3

=> det(A) = | 0 x-y x^2-y^2 | | 0 y-z y^2-z^2 | | 1 z z^2 |

To prove option (d),

=> Apply row transformations R1 -> R1+R2

R2 -> R2+R3

=> det(A) = | 2 x+y x^2+y^2 | | 2 y+z y^2+z^2 | | 1 z z^2 |

This solution is contributed by Anil Saikrishna Devarasetty .

 Question 2
Let A be the 2 × 2 matrix with elements a11 = a12 = a21 = +1 and a22 = −1. Then the eigenvalues of the matrix A19 are
 A A B B C C D D
GATE CS 2012    Linear Algebra
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Question 2 Explanation:
```A =  1    1
1   -1

A2 = 2   0
0   2

A4 = A2 X A2
A4 = 4   0
0   4

A8 = 16   0
0    16

A16 = 256   0
0    256

A18 = A16 X A2
A18 = 512   0
0    512

A19 = 512   512
512  -512

Applying Characteristic polynomial

512-lamda   512
512       -(512+lamda)  =   0

-(512-lamda)(512+lamda) - 512 x 512 = 0

lamda2 = 2 x 5122 ```
 Question 3
Consider the matrix as given below. Which one of the following options provides the CORRECT values of the eigenvalues of the matrix?
 A 1, 4, 3 B 3, 7, 3 C 7, 3, 2 D 1, 2, 3
GATE CS 2011    Linear Algebra
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Question 3 Explanation:

The Eigen values of a triangular matrix are given by its diagonal entries. We can also calculate (or verify given answers) using characteristic equation obtained by |M - λI| = 0.

1-λ    2     3

0     4-λ    7           =       0

0      0     3-λ

Which means

(1-λ)(4-λ)(3-λ) = 0

 Question 4
Consider the following matrix A = If the eigenvalues of A are 4 and 8, then
 A x=4, y=10 B x=5, y=8 C x=-3, y=9 D x=-4, y=10
GATE CS 2010    Linear Algebra
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Question 4 Explanation:
 Question 5
```How many of the following matrices have an eigenvalue 1?
```
 A one B two C three D four
Linear Algebra    GATE CS 2008
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Question 5 Explanation:

 Question 6
If the matrix A is such that then the determinant of A is equal to
 A 0
Linear Algebra    GATE-CS-2014-(Set-2)
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Question 6 Explanation:
This is a numerical answer question of gate paper, in which no options are provided, and the answer is to given by filling a numeral into a text box provided. In the question, matrix A is given as the product of 2 matrices which are of order 3 x 1 and 1 x 3 respectively. So after multiplication of these matrices, matrix A would be a square matrix of order 3 x 3. So, matrix A is : 2 18 10 -4 -36 -20 7 63 35 Now, we can observe by looking at the matrix that row 2 can be made completely zero by using row 1, this is to be done by using the row operation of matrix which here is : R2 <- R2 + 2R1 After applying above row operation in the matrix, the resultant matrix would be: 2 18 10 0 0 0 7 63 35 i.e. Row 2 has become zero now. And if a square matrix has a row or column with all its elements as 0, then its determinant is 0. ( A property of a square matrix ) Hence answer is 0. Note: Determinant is defined only for square matrices, and it is a number which encodes certain properties of a matrix, for ex: a square matrix with determinant 0 does not has its inverse matrix.
 Question 7
The product of the non-zero eigenvalues of the matrix
```1 0 0 0 1
0 1 1 1 0
0 1 1 1 0
0 1 1 1 0
1 0 0 0 1```
is ______
 A 4 B 5 C 6 D 7
Linear Algebra    GATE-CS-2014-(Set-2)
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Question 7 Explanation:
The characteristic equation is : | A - zI | = 0 , where I is an identity matrix of order 5. i.e. determinant of the below shown matrix to be 0. 1-z 0 0 0 1 0 1-z 1 1 0 0 1 1-z 1 0 0 1 1 1-z 0 1 0 0 0 1-z Now solve this equation to find values of z. Steps to solve : 1) Expand the matrix by 1st row. (1-z) [ ( 1-z , 1 , 1, 0 ) (1 , 1-z, 1, 0) (1, 1, 1-z, 0) (0, 0, 0, 1-z ) ] + 1. [ ( 0, 1-z, 1, 1 ) ( 0, 1, 1-z, 1) ( 0, 1, 1, 1-z )( 1, 0, 0, 0) ] Note: ( matrix is represented in brackets, row wise ) 2) Expand both of the above 4x4 matrices along the last row. (1-z)(1-z) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] + 1.(-1) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] 3) Apply row transformations to simplify above matrices ( both the matrices are same). C1 <- C1 + C2 + C3 R2 <- R2- R1 R3 <- R3 - R1 result is : (1-z)(1-z) [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] - 1. [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] 4) Solve the matrix by expanding 1st column. result is : (1-z)(1-z)(z)(z) - (3-z)(z)(z) = 0 solve further to get : z^3 ( 3-z ) ( z-2 ) = 0 hence z = 0 , 0 , 0 , 3 , 2 Therefore product of non zero eigenvales is 6.
 Question 8
Which one of the following statements is TRUE about every
 A If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. B If the trace of the matrix is positive, all its eigenvalues are positive. C If the determinant of the matrix is positive, all its eigenvalues are positive. D If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.
Linear Algebra    GATE-CS-2014-(Set-3)
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 Question 9
Consider the set H of all 3 × 3 matrices of the type where a, b, c, d, e and f are real numbers and abc ≠ 0. Under the matrix multiplication operation, the set H is
 A a group B a monoid but not a group C a semigroup but not a monoid D neither a group nor a semigroup
Linear Algebra    GATE-CS-2005
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Question 9 Explanation:
Because identity matrix is identity & as they define abc != 0, then it is non-singular so inverse is also defined
 Question 10
Consider the following system of equations in three real variables xl, x2 and x3 2x1 - x2 + 3x3 = 1 3x1- 2x2 + 5x3 = 2 -x1 + 4x2 + x3 = 3 This system of equations has
 A no solution B a unique solution C more than one but a finite number of solutions D an infinite number of solutions
Linear Algebra    GATE-CS-2005
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Question 10 Explanation:
The determinant value of following matrix is non-zero, therefore we have a unique solution.
``` 2   -1   3
3   -2   5
-1    4   1  ```
 Question 11
What are the eigenvalues of the following 2 × 2 matrix?
 A -1 and 1 B 1 and 6 C 2 and 5 D 4 and -1
Linear Algebra    GATE-CS-2005
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Question 11 Explanation:
The eigenvalues of A are precisely the real numbers λ that satisfy the equation det(A - &lambda I) = 0. Let us find determinant value of A - &lambda I;
```2-λ     -1
-4       5-λ
```
We get the equation as λ2 -7 λ + 6 = 0 which gives us eigenvalues as 6 and 1.
 Question 12
Let A, B, C, D be n × n matrices, each with non-­zero determinant. If ABCD = 1, then B-1 is
 A D-1C-1A-1 B CDA C ADC D Does not necessarily exist
Linear Algebra    GATE-CS-2004
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Question 12 Explanation:
This solution is contributed by Anil Saikrishna Devarasetty
 Question 13
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 of these
Linear Algebra    GATE-CS-2004
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Question 13 Explanation:
This solution is contributed by Anil Saikrishna Devarasetty.
 Question 14
In an M'N 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}
Linear Algebra    GATE-CS-2004
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Question 14 Explanation:
Suppose a < b, for example let a = 3, b= 5, then we can put non-zero entries only in 3 rows and 5 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we can't put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed. So we can fill only "a" non-zero entries if a < b, similarly if b < a, we can put only "b" non-zero entries. So answer is ≤min(a,b), because whatever is less between a and b, we can put atmost that many non-zero entries. So option (D) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2004.html
 Question 15
Consider the following system of linear equations Notice that the second and the third columns 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
Linear Algebra    GATE-CS-2003
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Question 15 Explanation:
This solution is contributed by Anil Saikrishna Devarasetty.
 Question 16
The rank of the matrix is :
 A 4 B 2 C 1 D 0
Linear Algebra    GATE-CS-2002
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Question 16 Explanation:

Rank of the matrix is defined as the maximum number of linearly independent vectors (or) the number of non-zero rows in its row-echelon matrix. A = | 1 1| | 0 0| Since, the matrix A is already in echelon form, Just count the number of non-zero rows to get the rank of the matrix = 1.

Please refer http://en.wikipedia.org/wiki/Rank_%28linear_algebra%29 This solution is contributed by Anil Saikrishna Devarasetty .
 Question 17
Consider the following statements:
```S1: The sum of two singular n × n matrices may be non-singular
S2: The sum of two n × n non-singular matrices may be singular. ```
Which of the following statements is correct?
 A S1 and S2 are both true B S1 is true, S2 is false C S1 is false, S2 is true D S1 and S2 are both false
Linear Algebra    GATE-CS-2001
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Question 17 Explanation:
Singular Matrix: A square matrix is singular if and only if its determinant value is 0. S1 is True: The sum of two singular n × n matrices may be non-singular It can be seen be taking following example. The following two matrices are singular, but their sum is non-singular.
```M1 and M2 are singular
M1 =  1  1
1  1

M2 =   1  -1
-1   1

But M1+M2 is non-singular
M1+M2 =  2  0
0  2
```
S2 is True: The sum of two n × n non-singular matrices may be singular
```M1 and M2 are non-singular
M1 =  1  0
0  1

M2 =   -1  0
0  -1

But M1+M2 is singular
M1+M2 =  0  0
0  0
```
 Question 18
The determinant of the matrix is
 A 5 B 0 C 4 D 20
Linear Algebra    GATE-CS-2000
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Question 18 Explanation:
The value of determinant is 2*1*2*1
 Question 19
In the LU decomposition of the matrix
```| 2  2 |
| 4  9 |```
, if the diagonal elements of U are both 1, then the lower diagonal entry l22 of L is
 A 4 B 5 C 6 D 7
Linear Algebra    GATE-CS-2015 (Set 1)
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Question 19 Explanation:
LU decomposition (where 'LU' stands for 'lower upper', and also called LU factorization) factors a matrix as the product of a lower triangular matrix and an upper triangular matrix.
```| 2  2 | =  | l11   0   |  * | 1   u12 |
| 4  9 |    | l21   l22 |    | 0    1  |

l21 *  u12 + l22 * 1  = 9  ------ (1)

We need to find l21 and u12
l21 *  1 + l22 * 0  = 4
l21 = 4

l11 * U12 + 0 * 1 = 2
l11 = 2
U12 = 1

Putting value of l21 and u12 in (1), we get
4 * 1 + l22 * 1 = 9
l22 = 5
```
 Question 20
Consider the following 2 × 2 matrix A where two elements are unknown and are marked by a and b. The eigenvalues of this matrix are –1 and 7. What are the values of a and b?
```A = | 1  4 |
| b  a |```
 A a = 6, b = 4 B a = 4, b = 6 C a = 3, b = 5 D a = 5, b = 3
Linear Algebra    GATE-CS-2015 (Set 1)
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Question 20 Explanation:
```The character equation for given matrix is

| 1-λ    4 | = 0
|  b    a-λ|

(1-λ)*(a-λ) - 4b = 0

Putting λ = -1,
=> (1 - (-1)) * (a - (-1)) - 4b = 0
=> 2a + 2 - 4b = 0
=> 2b - a = 1

Putting λ = 7,
=> (1 - 7) * (a - 7) - 4b = 0
=> -6a + 42 - 4b = 0
=> 2b + 3a = 21

Solving the above two equations, we get
a = 5, b = 3

```
 Question 21
The larger of the two eigenvalues of the matrix
```⎡ 4   5 ⎤
⎣ 2   1 ⎦```
is ____
 A 5 B 6 C 7 D 8
Linear Algebra    GATE-CS-2015 (Set 2)
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Question 21 Explanation:
```The character equation for given matrix is

| 4-λ    5 | = 0
| 2     1-λ|

(4-λ)*(1-λ) - 10 = 0
λ2 - 5λ - 6 = 0
(λ+1)*(λ-6) = 0
λ = -1, 6

Greater of two Eigenvalues is 6. ```
 Question 22
Perform the following operations on the matrix
```⎡ 3  4   45⎤
⎢ 7  8  105⎥
⎣13  2  195⎦
1. Add the third row to the second row.
2. Subtract the third column from the first column. ```
The determinant of the resultant matrix is _____________.
 A 0 B 1 C 50 D 100
Linear Algebra    GATE-CS-2015 (Set 2)
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Question 22 Explanation:
The given matrix is singular matrix as its last column is a multiple of first column. If we multiply 15 with first column, we get the third column.
 Question 23
In the given matrix, one of the eigenvalues is 1. the eigenvectors corresponding to the eigenvalue 1 are
```⎡ 1 -1  2 ⎤
⎢ 0  1  0 ⎥
⎣ 1  2  1 ⎦```
 A A B B C C D D
Linear Algebra    GATE-CS-2015 (Set 3)
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Question 23 Explanation:
Let z represents the eigenvalues.
```And let the given matrix be A (square matrix of order 3 x3)

The characteristic equation for this is :

AX = zX ( X is the required eigenvector )
AX - zX = 0
[ A - z I ] [X] = 0 ( I is an identity matrix of order 3 )

put z = 1 ( because one of the eigenvalue is 1 )

[ A - 1 I ] [X] = 0

The resultant matrix is :

[ 0 -1 2 ] [x1] [0]
| 0 0 0 ] |x2] =|0|
[ 1 2 0 ] |x3] [0]

Multiplying thr above matrices and getting the equations as:

-x2 + 2x3 = 0 ----------------(1)
x1 + 2x2 = 0-----------------(2)

now let x1 = k, then x2 and x3 will be -k/2 and -k/4
respectively.

hence eigenvector X = { (k , -k/2, -k/4) } where k != 0

put k = -4c ( c is also a constant, not equal to zero ),
we get X = { ( -4c, 2c, 1c ) }, i.e. { c ( -4, 2, 1 ) }

Hence option B.```
 Question 24
If the following system has non-trivial solution,
```  px + qy + rz = 0
qx + ry + pz = 0
rx + py + qz = 0 ```
then which one of the following options is True?
 A p – q + r = 0 or p = q = –r B p + q – r = 0 or p = –q = r C p + q + r = 0 or p = q = r D p – q + r = 0 or p = –q = –r
Linear Algebra    GATE-CS-2015 (Set 3)
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Question 24 Explanation:
For non-trivial solution, |A| should be equal to 0 Hence, Now solve it using matrix rules: (p+q+r) [(q-r)(p-q) - (r-p) (r-p) ] = 0 Either (p+q+r) = 0 or [(q-r)(p-q) - (r-p) (r-p) = 0 From (p+q+r) =0, you can clearly say that option C is the correct one. and for more precise answer, let’s solve second equation: [(q-r)(p-q) - (r-p) (r-p) = 0 (q-r)(p-q) = (r-p) (r-p) and only p = q = r satisfies this equation. So option C is correct one. This explanation has been contributed by Nitika Bansal.
 Question 25
Let a(x, y), b(x, y,) and c(x, y) be three statements with variables x and y chosen from some universe. Consider the following statement:
(∃x)(∀y)[(a(x, y) ∧ b(x, y)) ∧ ¬c(x, y)] Which one of the following is its equivalent?
 A (∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)] B (∃x)(∀y)[(a(x, y) ∨ b(x, y)) ∧¬ c(x, y)] C ¬ (∀x)(∃y)[(a(x, y) ∧ b(x, y)) → c(x, y)] D ¬ (∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)]
Linear Algebra    GATE-IT-2004
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 Question 26
What values of x, y and z satisfy the following system of linear equations?

 A x=6,y=3,z=2 B x=12,y=3,z=—4 C x=6,y=6,z=—4 D x=12,y=—3,z=O
Linear Algebra    GATE-IT-2004
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 Question 27
Let H1,H2,H3,... be harmonic numbers Then, for   can be expressed as
 A nHn+1 - (n + 1) B (n + 1)Hn - n C nHn - n D (n+1)Hn+1—(n+1)
Linear Algebra    GATE-IT-2004
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Question 27 Explanation:
This solution is contributed by Mohit Gupta.
 Question 28
Two eigenvalues of a 3 x 3 real matrix P are (2 + √ -1) and 3. The determinant of P is _____   Note : This question was asked as Numerical Answer Type.
 A 0 B 1 C 15 D -1
Linear Algebra    GATE-CS-2016 (Set 1)
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Question 28 Explanation:
The determinant of a real matrix can never be imaginary. So, if one eigen value is complex, the other eigen value has to be its conjugate.   So, the eigen values of the matrix will be 2+i, 2-i and 3.   Also, determinant is the product of all eigen values. So, the required answer is (2+i)*(2-i)*(3) = (4-i2)*(3) = (5)*(3) = 15.   Thus, C is the required answer.
 Question 29
Let f (x) be a polynomial and g(x) = f (x) be its derivative. If the degree of (f(x) + f(−x)) is 10, then the degree of (g(x) − g(−x)) is _______________ [This Question was originally a Fill-in-the-blanks Question]
 A 8 B 7 C 9 D 10
Linear Algebra    GATE-CS-2016 (Set 2)
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Question 29 Explanation:
f(x) can be either an even function or an odd function. If f(x) was an odd function, then f(x) + f(-x) = 0, but here it has been given that it has degree 10. So, it must be an even function. Therefore, f(x) = f(-x) => f'(x) = -f'(-x) Also, it has been mentioned that g(x) is the derivative of f(x). So, g(x) = f'(x) and g(-x) = -f'(-x) => g(x) - g(-x) = f'(x) - (-f'(-x)) => g(x) - g(-x) = f'(x) + f'(x) => g(x) - g(-x) = 2 * f'(x) But, f'(x) will have degree 9.   Thus, C is the correct option.
 Question 30
Consider the systems, each consisting of m linear equations in n variables.
```I. If m < n, then all such systems have a solution
II. If m > n, then none of these systems has a solution
III. If m = n, then there exists a system which has a solution```
Which one of the following is CORRECT?
 A I, II and III are true B Only II and III are true C Only III is true D None of them is true
Linear Algebra    GATE-CS-2016 (Set 2)
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Question 30 Explanation:
For 'n' variables, we need atleast 'n' linear equations in terms of the given  variable to find value of each variable. If no. of equations > no. of variables then also we can find the value of each variable. Here, m : No. of linear equations n : No. of variables For solution, m > or = n. This solution is contributed by Mohit Gupta.
 Question 31
Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of (A−1)T is _________ [This Question was originally a Fill-in-the-blanks Question]
 A 1/8 B 1 C 1/4 D 2
Linear Algebra    GATE-CS-2016 (Set 2)
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Question 31 Explanation:
Answer: 1/8 Determinant of A = 1*4*2 = 8 which is product of Eigen values. Determinant of A-1 = 1/8 as determinant of inverse of matrix is inverse of determinant. Determinant of Transpose (A-1) = 1/8 since determinant of transpose of matrix remains same as original one.
 Question 32
How many antisymmetric relations are there on a set with n elements ?
 A 2n.3n(n-1)/2 B 2n C n2 D n
Linear Algebra    GATE 2017 Mock
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Question 32 Explanation:
Any subset of diagonal pairs is an antisymmetric relation. In a antisymmetric relation each diagonal pair can appear in 2 ways. In antisymmetric relation each non diagonal combination can appear in 3 ways.
By Product rule,
Number of antisymmetric relations possible on A = 2^n. 3^(n(n-1)/2).
There are 32 questions to complete.

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