# Linear Algebra

Question 1 |

Which one of the following does NOT equal to

C | |

D | |

B | |

A |

**GATE CS 2013**

**Linear Algebra**

**Discuss it**

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 |

Question 2 |

^{19}are

A | |

B | |

C | |

D |

**GATE CS 2012**

**Linear Algebra**

**Discuss it**

A = 1 1 1 -1 A^{2}= 2 0 0 2 A^{4}= A^{2}X A^{2}A^{4}= 4 0 0 4 A^{8}= 16 0 0 16 A^{16}= 256 0 0 256 A^{18}= A^{16}X A^{2}A^{18}= 512 0 0 512 A^{19}= 512 512 512 -512 Applying Characteristic polynomial 512-lamda 512 512 -(512+lamda) = 0 -(512-lamda)(512+lamda) - 512 x 512 = 0 lamda^{2 = 2 x 5122 }

**Alternative solution:**

det(A) = -2. det(A^19) = (det(A))^19 = -2^19 = lambda1*lambda2. The only viable option is D.Thanks to Matan Mandelbrod for suggesting this solution.

Question 3 |

Which one of the following options provides the CORRECT values of the eigenvalues of the matrix?

1, 4, 3 | |

3, 7, 3 | |

7, 3, 2 | |

1, 2, 3 |

**GATE CS 2011**

**Linear Algebra**

**Discuss it**

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

x=-3, y=9 | |

x= -4, y=10 | |

x=5, y=8 | |

x=4, y=10 |

**GATE CS 2010**

**Linear Algebra**

**Discuss it**

Question 5 |

How many of the following matrices have an eigenvalue 1?

three | |

four | |

two | |

one |

**GATE CS 2008**

**Linear Algebra**

**Discuss it**

Question 6 |

0 | |

1 | |

2 | |

3 |

**GATE-CS-2014-(Set-2)**

**Linear Algebra**

**Discuss it**

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 ______

4 | |

5 | |

6 | |

7 |

**GATE-CS-2014-(Set-2)**

**Linear Algebra**

**Discuss it**

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 eigenvalues is 6.

Question 8 |

Which one of the following statements is TRUE about every

If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. | |

If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. | |

If the trace of the matrix is positive, all its eigenvalues are positive. | |

If the determinant of the matrix is positive, all its eigenvalues are positive. |

**GATE-CS-2014-(Set-3)**

**Linear Algebra**

**Discuss it**

The trace of a matrix is the sum of the elements of the principal diagonal of the matrix.

Fact - The sum of Eigen values of a matrix is equal to it’s trace.

Fact - The product of Eigen values of a matrix is equal to its determinant value.

Since it’s given that the trace is positive and the determinant is negative, there must be atleast one negative Eigen value. In general there can be an odd number of negative Eigen values in this case since the determinant value is negative.

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 group | |

a monoid but not a group | |

a semigroup but not a monoid | |

neither a group nor a semigroup |

**GATE-CS-2005**

**Linear Algebra**

**Discuss it**

Because identity matrix is identity & as they define abc != 0, then it is non-singular so inverse is also defined.

The set of matrices is the set of Upper triangular matrices(H) of size 3*3 with non-zero determinant. Along with the multiplication operator the set forms an Algebraic Structure since it follows the Closure Property. This is because the product of Two Upper Triangular Matrices is also a Upper Triangular Matrix.

The Algebraic Structure also follows the Associative Property since, multiplication of matrices in general follows the Associative Property. Therefore it is a Semi Group.

The Algebraic Structure is also a Monoid, since it has an Identity element, which is the Identity Matrix- I3.

The Algebraic Structure is a Group since every matrix in H has an inverse, since every matrix in H is non-singular (given in question).

The Algebraic Structure is not an Abelian Group since it does not follow the Commutative Property.

Therefore Option **A** is correct.

Question 10 |

no solution | |

a unique solution | |

more than one but a finite number of solutions | |

an infinite number of solutions |

**GATE-CS-2005**

**Linear Algebra**

**Discuss it**

Consider the matrix as A

| A | = 2 ( 2 -20 ) +1 ( 3 + 5 ) + 3 ( 12 + 2 )

= 36 + 8+ 42

= 86

The determinant value of following matrix is non-zero, therefore we have a unique solution.

2 -1 3 3 -2 5 -1 4 1