Linear Algebra
Question 1 |
A | |
B | |
C | |
D |
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 |

A | |
B | |
C | |
D |
Discuss it
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 5122Alternative 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 |
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 |
Discuss it
![Rendered by QuickLaTeX.com A = \begin{bmatrix} 2 & 3 \\ x & 4 \\ \end{bmatrix} \newline \newline [A - \lambda i]=0 \newline \newline \begin{bmatrix} 2-\lambda & 3 \\ x & y-\lambda \\ \end{bmatrix} = 0 \newline \newline (2-\lambda)(y-\lambda) - 3x =0 \newline given \hspace{0.2cm} eigen\hspace{0.2cm} values: \lambda=4,8 \newline if \hspace{0.2cm} we \hspace{0.2cm} use \hspace{0.2cm} \lambda=4\hspace{0.2cm} then \hspace{0.2cm} equation, \hspace{0.2cm} -2(y-4)-3x=0 . . . . ...(i) \newline if \hspace{0.2cm} we \hspace{0.2cm} use \hspace{0.2cm} \lambda=8\hspace{0.2cm} then \hspace{0.2cm} equation, \hspace{0.2cm} -6(y-8)-3x=0 . . . . ...(ii) \newline Solving \hspace{0.2cm} equations \hspace{0.2cm} (i) & (ii), we \hspace{0.2cm} get, x=-4,y=10](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-fa0ed4eae40d57b8c049d03a991300f6_l3.png)
Question 5 |
one | |
two | |
three | |
four |
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Question 6 |
0 | |
1 | |
2 | |
3 |
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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 |
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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. |
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Question 9 |

a group | |
a monoid but not a group | |
a semigroup but not a monoid | |
neither a group nor a semigroup |
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Question 10 |
no solution | |
a unique solution | |
more than one but a finite number of solutions | |
an infinite number of solutions |
Discuss it
2 -1 3 3 -2 5 -1 4 1