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