Question 1 |
Consider a matrix A=\begin{bmatrix}
1 & 0 & 0\\
0 & 4 & -2\\
0&1 &1
\end{bmatrix}
The matrix A satisfies the equation 6A^{-1}=A^2+cA+dI where c and d are scalars and I is the identity matrix.
Then (c+d) is equal to
The matrix A satisfies the equation 6A^{-1}=A^2+cA+dI where c and d are scalars and I is the identity matrix.
Then (c+d) is equal to
5 | |
17 | |
-6 | |
11 |
Question 1 Explanation:
Characteristic equation:
\begin{aligned} |A-\lambda I|&=0 \\ \begin{vmatrix} 1-\lambda & 0 &0 \\ 0 & 4-\lambda & 2\\ 0 & -1 & 1-\lambda \end{vmatrix}&=0 \\ (1-\lambda )[(4-\lambda )(1-\lambda )+2] &=0\\ \lambda ^3-6\lambda ^2+11\lambda -6&=0 \end{aligned}
By cayley hamilton theorem
\begin{aligned} A^3-6A^2+11A-6 &=0 \\ A^2-6A+11I&=6A^{-1} \end{aligned}
On comparison : c= -6 and d = 11
Therefore, c + d = -6 + 11 = 5
\begin{aligned} |A-\lambda I|&=0 \\ \begin{vmatrix} 1-\lambda & 0 &0 \\ 0 & 4-\lambda & 2\\ 0 & -1 & 1-\lambda \end{vmatrix}&=0 \\ (1-\lambda )[(4-\lambda )(1-\lambda )+2] &=0\\ \lambda ^3-6\lambda ^2+11\lambda -6&=0 \end{aligned}
By cayley hamilton theorem
\begin{aligned} A^3-6A^2+11A-6 &=0 \\ A^2-6A+11I&=6A^{-1} \end{aligned}
On comparison : c= -6 and d = 11
Therefore, c + d = -6 + 11 = 5
Question 2 |
e^A denotes the exponential of a square matrix A. Suppose \lambda is an eigenvalue and v is the corresponding eigen-vector of matrix A.
Consider the following two statements:
Statement 1: e^\lambda is an eigenvalue of e^A.
Statement 2: v is an eigen-vector of e^A.
Which one of the following options is correct?
Consider the following two statements:
Statement 1: e^\lambda is an eigenvalue of e^A.
Statement 2: v is an eigen-vector of e^A.
Which one of the following options is correct?
Statement 1 is true and statement 2 is false. | |
Statement 1 is false and statement 2 is true | |
Both the statements are correct. | |
Both the statements are false. |
Question 2 Explanation:
Eigen value will change but eigen vector not
change.
Question 3 |
Consider a 3 x 3 matrix A whose (i,j)-th element, a_{i,j}=(i-j)^3. Then the matrix A will be
symmetric. | |
skew-symmetric. | |
unitary | |
null. |
Question 3 Explanation:
for \; i=j\Rightarrow a_{ij}=(i-i)^3=0\forall i
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
Question 4 |
Let A be a 10\times10 matrix such that A^{5} is a null matrix, and let I be the 10\times10 identity matrix. The determinant of \text{A+I} is ___________________.
1 | |
2 | |
4 | |
8 |
Question 4 Explanation:
\begin{aligned} \text{Given}:\qquad A^{5}&=0\\ A x &=\lambda x \\ \Rightarrow \qquad \qquad A^{5} x &=\lambda^{5} x \quad(\because x \neq 0) \\ \Rightarrow \qquad \qquad \lambda^{5} &=0 \\ \Rightarrow \qquad \qquad \lambda &=0 \end{aligned}
Eigen values of A+I given \lambda+1
\because Eigen values of I_{A}=1
Hence |A+I|= Product of eigen values =1 \times 1 \times 1 \times \ldots 10 times
=1
Eigen values of A+I given \lambda+1
\because Eigen values of I_{A}=1
Hence |A+I|= Product of eigen values =1 \times 1 \times 1 \times \ldots 10 times
=1
Question 5 |
Let p and q be real numbers such that p^{2}+q^{2}=1. The eigenvalues of the matrix \begin{bmatrix} p & q\\ q& -p \end{bmatrix}
are
1 and 1 | |
1 and -1 | |
j and -j | |
pq and -pq |
Question 5 Explanation:
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
There are 5 questions to complete.