# Linear Algebra

Question 1 |

A | |

B | |

C | |

D |

**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 |

This solution is contributed by **Anil Saikrishna Devarasetty** .

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 |

x=4, y=10 | |

x=5, y=8 | |

x=-3, y=9 | |

x= -4, y=10 |

**GATE CS 2010**

**Linear Algebra**

**Discuss it**

Question 5 |

one | |

two | |

three | |

four |

**Linear Algebra**

**GATE CS 2008**

**Discuss it**

Question 6 |

0 | |

1 | |

2 | |

3 |

**Linear Algebra**

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

**Discuss it**

Question 7 |

1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1is ______

4 | |

5 | |

6 | |

7 |

**Linear Algebra**

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

**Discuss it**

Question 8 |

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

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

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)**

**Discuss it**

**Chirag Manwani**.

Question 9 |

a group | |

a monoid but not a group | |

a semigroup but not a monoid | |

neither a group nor a semigroup |

**Linear Algebra**

**GATE-CS-2005**

**Discuss it**

**A**is correct. This explanation is provided by

**Chirag Manwani**.

Question 10 |

no solution | |

a unique solution | |

more than one but a finite number of solutions | |

an infinite number of solutions |

**Linear Algebra**

**GATE-CS-2005**

**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