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}
Question 6 |
The number of purely real elements in a lower triangular representation of the given 3x3
matrix, obtained through the given decomposition is
\begin{bmatrix} 2 &3 &3 \\ 3& 2 & 1\\ 3& 1 & 7 \end{bmatrix}= \begin{bmatrix} a_{11} &0 &0 \\ a_{12}& a_{22} & 0\\ a_{13}& a_{23} & a_{33} \end{bmatrix}\begin{bmatrix} a_{11} &0 &0 \\ a_{12}& a_{22} & 0\\ a_{13}& a_{23} & a_{33} \end{bmatrix}^T
\begin{bmatrix} 2 &3 &3 \\ 3& 2 & 1\\ 3& 1 & 7 \end{bmatrix}= \begin{bmatrix} a_{11} &0 &0 \\ a_{12}& a_{22} & 0\\ a_{13}& a_{23} & a_{33} \end{bmatrix}\begin{bmatrix} a_{11} &0 &0 \\ a_{12}& a_{22} & 0\\ a_{13}& a_{23} & a_{33} \end{bmatrix}^T
5 | |
6 | |
8 | |
9 |
Question 6 Explanation:
As per GATE official answer key MTA (Marks to ALL)
\begin{aligned} \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} I_{11} & 0 &0 \\ I_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} u_{11} &u_{12} &u_{13} \\ 0 &u_{22} &u_{23} \\ 0 & 0 &u_{33} \end{bmatrix} \\ \text{consider, }u_{11}&=u_{22}=u_{33}=1 \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & 0 &0 \\ l_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} 1 &u_{12} &u_{13} \\ 0 &1 &u_{23} \\ 0 & 0 &1 \end{bmatrix} \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & l_{11}u_{12} &l_{11}u_{13} \\ l_{21}& l_{21}u_{12}+l_{22} & l_{21}u_{13}+ l_{22}u_{23}\\ l_{31} &l_{31}u_{12}+l_{32} & l_{31}u_{13}+ l_{32}u_{23}+l_{33}\end{bmatrix} \\ l_{11}&=2 \; \; l_{11}u_{12}=3 \; \; l_{11}u_{13}=3 \; \; \\ l_{21}&=3 \; \; 2u_{12}=3 \; \; 2u_{13}=3 \\ l_{31}&=3\; \; u_{12}=\frac{3}{2} \; \; u_{13}=\frac{3}{2} \\ l_{21}u_{12}+l_{22}&=2\; \; l_{21}u_{13}+l_{22}u_{23}=1 \\ (3)\left ( \frac{3}{2} \right )+l_{22}&=2 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{5}{2} \right )u_{23}=1 \\ l_{22}&=-\frac{5}{2} \; \; u_{23}=-\frac{7}{5} \\ l_{31}u_{12}+l_{32}&=1 \; \; l_{31}u_{13}+l_{32}u_{23}+l_{33}=7\\ (3)(\frac{3}{2})+l_{32}&=1 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{7}{2} \right )\left ( \frac{7}{5} \right )+l_{33}=7 \\ l_{32}&=-\frac{7}{2} \; \; l_{33}=-\frac{74}{10}\\ L&=\begin{bmatrix} 2 &0 &0 \\ 3 &-5/2 &0 \\ 3 &-7/2 & 74/10 \end{bmatrix} \end{aligned}
The number of purely real elements of lower triangular matrix are 9.
\begin{aligned} \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} I_{11} & 0 &0 \\ I_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} u_{11} &u_{12} &u_{13} \\ 0 &u_{22} &u_{23} \\ 0 & 0 &u_{33} \end{bmatrix} \\ \text{consider, }u_{11}&=u_{22}=u_{33}=1 \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & 0 &0 \\ l_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} 1 &u_{12} &u_{13} \\ 0 &1 &u_{23} \\ 0 & 0 &1 \end{bmatrix} \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & l_{11}u_{12} &l_{11}u_{13} \\ l_{21}& l_{21}u_{12}+l_{22} & l_{21}u_{13}+ l_{22}u_{23}\\ l_{31} &l_{31}u_{12}+l_{32} & l_{31}u_{13}+ l_{32}u_{23}+l_{33}\end{bmatrix} \\ l_{11}&=2 \; \; l_{11}u_{12}=3 \; \; l_{11}u_{13}=3 \; \; \\ l_{21}&=3 \; \; 2u_{12}=3 \; \; 2u_{13}=3 \\ l_{31}&=3\; \; u_{12}=\frac{3}{2} \; \; u_{13}=\frac{3}{2} \\ l_{21}u_{12}+l_{22}&=2\; \; l_{21}u_{13}+l_{22}u_{23}=1 \\ (3)\left ( \frac{3}{2} \right )+l_{22}&=2 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{5}{2} \right )u_{23}=1 \\ l_{22}&=-\frac{5}{2} \; \; u_{23}=-\frac{7}{5} \\ l_{31}u_{12}+l_{32}&=1 \; \; l_{31}u_{13}+l_{32}u_{23}+l_{33}=7\\ (3)(\frac{3}{2})+l_{32}&=1 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{7}{2} \right )\left ( \frac{7}{5} \right )+l_{33}=7 \\ l_{32}&=-\frac{7}{2} \; \; l_{33}=-\frac{74}{10}\\ L&=\begin{bmatrix} 2 &0 &0 \\ 3 &-5/2 &0 \\ 3 &-7/2 & 74/10 \end{bmatrix} \end{aligned}
The number of purely real elements of lower triangular matrix are 9.
Question 7 |
Consider a 2x2 matrix M=[v_1\;\; v_2], where, v_1\;and \; v_2 are the column vectors. Suppose M^{-1}=\begin{bmatrix} u_1^T\\ u_2^T \end{bmatrix}, where u_1^T \; and \; u_2^T are the row vectors. Consider the following statements:
Statement 1: u_1^T v_1=1\; and \; u_2^Tv_2=1
Statement 2: u_1^T v_2=0\; and \; u_2^Tv_1=0
Which of thefollowing options is correct?
Statement 1: u_1^T v_1=1\; and \; u_2^Tv_2=1
Statement 2: u_1^T v_2=0\; and \; u_2^Tv_1=0
Which of thefollowing options is correct?
Statement 1 is true and statement 2 is false | |
Statement 2 is true and statement 1 is false | |
Both the statements are true | |
Both the statements are false |
Question 7 Explanation:
\begin{aligned} M^{-1}M &=I \\ \Rightarrow \; \begin{bmatrix} u_1^T\\ u_2^T \end{bmatrix}\begin{bmatrix} v_1 & v_2 \end{bmatrix} &= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ \begin{bmatrix} u_1^T v_1 & u_1^T v_2\\ u_2^T v_1 & u_2^T v_2 \end{bmatrix}&=\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \\ u_1^T v_1=1; &u_1^T v_2 =0\\ u_2^T v_1=1; &u_2^T v_2 =1\\ \end{aligned}
Statement (1) and (2) are both correct. Option (C) is correct.
Statement (1) and (2) are both correct. Option (C) is correct.
Question 8 |
The rank of the matrix, M=\begin{bmatrix} 0 &1 &1 \\ 1& 0& 1\\ 1& 1& 0 \end{bmatrix}, is ______
1 | |
2 | |
3 | |
4 |
Question 8 Explanation:
\begin{aligned} M &=\begin{bmatrix} 0 & 1 & 1\\ 1& 0 & 1\\ 1& 1 & 0 \end{bmatrix} \\ &R_1\leftrightarrow R_2 \\ &=\begin{bmatrix} 1 & 0 & 1\\ 0& 1 & 1\\ 1& 1 & 0 \end{bmatrix} \\ &R_3\rightarrow R_3 -R_1 \\ &= \begin{bmatrix} 1 & 0 & 1\\ 0& 1 & 1\\ 0& 1 & -1 \end{bmatrix}\\ &R_3\rightarrow R_3-R_2 \\ &= \begin{bmatrix} 1 & 0 & 1\\ 0& 1 & 1\\ 0& 0 & -2 \end{bmatrix} \end{aligned}
Which is in echelon form
\therefore \;\; \rho (A)=3
Which is in echelon form
\therefore \;\; \rho (A)=3
Question 9 |
M is a 2x2 matrix with eigenvalues 4 and 9. The eigenvalues of M^2 are
4 and 9 | |
2 and 3 | |
-2 and -3 | |
16 and 81 |
Question 9 Explanation:
M is 2 x 2 matrix with eigen values 4 and 9. The eigen values of M^2 are 16 and 81.
Question 10 |
Let A=\begin{bmatrix} 1 & 0&-1 \\ -1&2 & 0\\ 0& 0 & -2 \end{bmatrix} and B=A^{3}-A^{2}-4A+5I, where I is the 3x3 identity matrix. The
determinant of B is ________ (up to 1 decimal place).
0.5 | |
1.0 | |
2.8 | |
2.1 |
Question 10 Explanation:
\begin{aligned} A=\begin{bmatrix} 1 & 0 & -1\\ -1 &2 &0 \\ 0 & 0 & -2 \end{bmatrix}\\ |A-\lambda I|=0\\ \begin{bmatrix} 1-\lambda & 0& -1\\ -1& 2-\lambda & 0\\ 0 & 0 &-2-\lambda \end{bmatrix}=0\\ (1-\lambda )((2-\lambda )(-2-\lambda ))-1(0-0)=0\\ \lambda =1,2,-2\\ \text{Eigen value of A are 1,2,-2}\\ \text{Eigen value of } A^3 \text{ are 1,8,-8}\\ \text{Eigen value of } A^2 \text{ are 1,4,4}\\ \text{Eigen value of } 4A \text{ are 4,8,-8}\\ \text{Eigen value of } 5I \text{ are 5,5,5}\\ A^3-A^2-4A+5I \text{ are 1,1,1}\\ \therefore \;\; |B|=(1)(1)(1)=1 \end{aligned}
There are 10 questions to complete.