Question 1 |

If the sum and product of eigenvalues of a 2 \times 2
real matrix \begin{bmatrix}
3 & p\\
p & q
\end{bmatrix} are 4 and -1 respectively, then |p|
is ______ (in integer).

4 | |

2 | |

6 | |

8 |

Question 1 Explanation:

From the property of eigen values,

Sum of eigen values = Trace of matrix

4=3+q

q=1

Product of eigen values = Determinant

\begin{aligned} -1&=\begin{vmatrix} 3 &p \\ p& q \end{vmatrix}\\ 3q-p^2&=-1\\ 3-p^2&=-1\\ p^2&=4\\ p&=\pm 2\\ \Rightarrow |p|&=2 \end{aligned}

Sum of eigen values = Trace of matrix

4=3+q

q=1

Product of eigen values = Determinant

\begin{aligned} -1&=\begin{vmatrix} 3 &p \\ p& q \end{vmatrix}\\ 3q-p^2&=-1\\ 3-p^2&=-1\\ p^2&=4\\ p&=\pm 2\\ \Rightarrow |p|&=2 \end{aligned}

Question 2 |

A is a 3\times 5 real matrix of rank 2. For the set of
homogeneous equations Ax = 0, where 0 is a zero
vector and x is a vector of unknown variables,
which of the following is/are true?

The given set of equations will have a unique solution. | |

The given set of equations will be satisfied by a zero vector of appropriate size. | |

The given set of equations will have infinitely many solutions. | |

The given set of equations will have many but a finite number of solutions. |

Question 2 Explanation:

Zero solution is always a solution of Ax = 0.

Option (b) is correct.

Given A is 3x5 real matrix and r(A) = 2 and Ax = 0 is a system of homogeneous linear equations since r(A) \lt number of unknown the system has infinite solution option (c) is also correct.

Option (b) is correct.

Given A is 3x5 real matrix and r(A) = 2 and Ax = 0 is a system of homogeneous linear equations since r(A) \lt number of unknown the system has infinite solution option (c) is also correct.

Question 3 |

The system of linear equations in real (x, y) given
by

(x\;\;y)\begin{bmatrix} 2 & 5-2\alpha \\ \alpha & 1 \end{bmatrix} =(0\;\;0)

involves a real parameter \alpha and has infinitely many non-trivial solutions for special value(s) of \alpha . Which one or more among the following options is/ are non-trivial solution(s) of (x,y) for such special value(s) of \alpha ?

(x\;\;y)\begin{bmatrix} 2 & 5-2\alpha \\ \alpha & 1 \end{bmatrix} =(0\;\;0)

involves a real parameter \alpha and has infinitely many non-trivial solutions for special value(s) of \alpha . Which one or more among the following options is/ are non-trivial solution(s) of (x,y) for such special value(s) of \alpha ?

x=2,y=-2 | |

x=-1,y=4 | |

x=1,y=1 | |

x=4,y=-2 |

Question 3 Explanation:

\begin{aligned}
\begin{pmatrix}
x &y
\end{pmatrix}\begin{pmatrix}
2 &5-2\alpha \\
\alpha & 1
\end{pmatrix}&=\begin{pmatrix}
0 &0
\end{pmatrix}\\
2x+\alpha y&=0\\
(5-2\alpha )x+y&=0\\
\therefore \begin{pmatrix}
2 & \alpha \\
5-2\alpha & 1
\end{pmatrix}\begin{pmatrix}
x\\y
\end{pmatrix}&=\begin{pmatrix}
0\\0
\end{pmatrix} \;\;\; . . . . (1)
\end{aligned}

To get infinite number of non-trivial solutions \begin{aligned} \begin{vmatrix} 2 &\alpha \\ 5-2\alpha & 1 \end{vmatrix}&=0\\ 2-(5\alpha -2\alpha ^2)&=0\\ (2\alpha -1)(\alpha -2)&=0\\ \therefore \alpha =\frac{1}{2},\alpha &=2\\ \end{aligned}

At \alpha =\frac{1}{2} ; eq. (1) gives (4x+y)=0 . . .(2)

option (B) is satisfied by (2)

At \alpha =2 ; eq. (1) gives (x+y)=0 . . .(3)

option (A) is satisfied by (3)

Both options (A) and (B) are correct

To get infinite number of non-trivial solutions \begin{aligned} \begin{vmatrix} 2 &\alpha \\ 5-2\alpha & 1 \end{vmatrix}&=0\\ 2-(5\alpha -2\alpha ^2)&=0\\ (2\alpha -1)(\alpha -2)&=0\\ \therefore \alpha =\frac{1}{2},\alpha &=2\\ \end{aligned}

At \alpha =\frac{1}{2} ; eq. (1) gives (4x+y)=0 . . .(2)

option (B) is satisfied by (2)

At \alpha =2 ; eq. (1) gives (x+y)=0 . . .(3)

option (A) is satisfied by (3)

Both options (A) and (B) are correct

Question 4 |

If A=\begin{bmatrix}
10 &2k+5 \\
3k-3 & k+5
\end{bmatrix} is a symmetric matrix, the
value of k is ___________.

8 | |

5 | |

-0.4 | |

\frac{1+\sqrt{1561}}{12} |

Question 4 Explanation:

A=\begin{bmatrix}
10 & 2k+5\\
3k-3 & k+5
\end{bmatrix}

(2k + 5) = (3k - 3)

k=8

(2k + 5) = (3k - 3)

k=8

Question 5 |

Consider an n x n matrix A and a non-zero n x 1 vector p. Their product Ap=\alpha ^2p, where \alpha \in \mathbb{R} and \alpha \notin \{-1,0,1\}. Based on the given information, the eigen value of A^2 is:

\alpha | |

\alpha ^2 | |

\sqrt{\alpha } | |

\alpha ^4 |

Question 5 Explanation:

Given, A P=\alpha^{2} P

By comparison with A X=\lambda X \Rightarrow

\Rightarrow \quad \lambda=\alpha^{2}

Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.

By comparison with A X=\lambda X \Rightarrow

\Rightarrow \quad \lambda=\alpha^{2}

Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.

There are 5 questions to complete.

56 no please check

Dear Sangram Kumar

Thank you for your suggestions. We have updated the correction suggested by You.

Great job sir 👍

In question a12 should be -i

In question 24

The answer for question no 4 should be A and it is marked as B

Option B is correct answer.

Sir in question no 24 first row second couloum should be -i

In the Question no 1 check matrix once again

Question 1 is wrong the matrix in the question is wrong