Question 1 |
Cholesky decomposition is carried out on the following square matrix [\mathrm{A}].
[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]
Let \mathrm{I}_{\mathrm{ij}} and \mathrm{a}_{\mathrm{ij}} be the (i, j)^{\text {th }} elements of matrices [L] and [A], respectively. If the element I_{22} of the decomposed lower triangular matrix [\mathrm{L}] is 1.968 , what is the value (rounded off to the nearest integer) of the element a_{22} ?
[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]
Let \mathrm{I}_{\mathrm{ij}} and \mathrm{a}_{\mathrm{ij}} be the (i, j)^{\text {th }} elements of matrices [L] and [A], respectively. If the element I_{22} of the decomposed lower triangular matrix [\mathrm{L}] is 1.968 , what is the value (rounded off to the nearest integer) of the element a_{22} ?
5 | |
7 | |
9 | |
11 |
Question 1 Explanation:
We know, cholesky decomposition,
A=LL^{T}
Where, L= lower tringular matrix
\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]
\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]
On comparison on both sides,
\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}
and,\;\; L_{11} L_{21}=-5
\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}
and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}
=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}
=6.998 \approx 7
A=LL^{T}
Where, L= lower tringular matrix
\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]
\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]
On comparison on both sides,
\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}
and,\;\; L_{11} L_{21}=-5
\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}
and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}
=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}
=6.998 \approx 7
Question 2 |
For the matrix
[A]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array}\right]
which of the following statements is/are TRUE?
[A]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array}\right]
which of the following statements is/are TRUE?
[A]\{X\}=\{b\} has a unique solution | |
[A]\{X\}=\{b\} does not have a unique solution | |
[A] has three linearly independent eigenvectors | |
[\mathrm{A}] is a positive definite matrix |
Question 2 Explanation:
|A|=\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|
=1(2-1)+1(-1)+0
=1-1=0
So A X=B does not have unique solution because \rho(A) \lt 3
|A-\lambda|=0 \Rightarrow\left|\begin{array}{ccc}1-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda\end{array}\right|=0
(1-\lambda)((2-\lambda)(1-\lambda)-1)+1(-1+\lambda)+0=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1\right)-1+\lambda=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1-1\right)=0
(\lambda-1)\left(\lambda^{2}-3 \lambda\right)=0
\lambda=0,1,3
Matrix A has three distinct Eigen values so have three linearly independent eigen vectors. so option (C) is correct.
Given matrix is symmetric matrix with real value entries. Hence A is not a positive definite matrix. because
|1|=1 \gt 0
\left|\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right|=2-1=1>0
\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|=0 (which is not positive)
Hence option (D) is incorrect.
=1(2-1)+1(-1)+0
=1-1=0
So A X=B does not have unique solution because \rho(A) \lt 3
|A-\lambda|=0 \Rightarrow\left|\begin{array}{ccc}1-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda\end{array}\right|=0
(1-\lambda)((2-\lambda)(1-\lambda)-1)+1(-1+\lambda)+0=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1\right)-1+\lambda=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1-1\right)=0
(\lambda-1)\left(\lambda^{2}-3 \lambda\right)=0
\lambda=0,1,3
Matrix A has three distinct Eigen values so have three linearly independent eigen vectors. so option (C) is correct.
Given matrix is symmetric matrix with real value entries. Hence A is not a positive definite matrix. because
|1|=1 \gt 0
\left|\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right|=2-1=1>0
\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|=0 (which is not positive)
Hence option (D) is incorrect.
Question 3 |
For the matrix
[A]=\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right]
Which of the following statements is/are TRUE?
[A]=\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right]
Which of the following statements is/are TRUE?
The eigenvalues of [A]^T
are same as the
eigenvalues of [A] | |
The eigenvalues of [A]^{-1}are the reciprocals of
the eigenvalues of [A] | |
The eigenvectors of [A]^T
are same as the
eigenvectors of [A] | |
The eigenvectors of [A]^{-1} are same as the
eigenvectors of [A] |
Question 3 Explanation:
A x=\lambda x \ldots (i)
A^{T} x=\lambda x \ldots (ii)
A and A^T both have same eigen values and eigen vectors.
A x=\lambda x \ldots(i)
\Rightarrow \quad A^{-1} A x=A^{-1}(\lambda x)=\lambda A^{-1} x
\Rightarrow \quad x=\lambda A^{-1} x
A^{-1} x=\frac{1}{\lambda} x
So, eigen value and eigen vector of A^{-1} is \frac{1}{\lambda} x and \mathrm{x}.
A^{T} x=\lambda x \ldots (ii)
A and A^T both have same eigen values and eigen vectors.
A x=\lambda x \ldots(i)
\Rightarrow \quad A^{-1} A x=A^{-1}(\lambda x)=\lambda A^{-1} x
\Rightarrow \quad x=\lambda A^{-1} x
A^{-1} x=\frac{1}{\lambda} x
So, eigen value and eigen vector of A^{-1} is \frac{1}{\lambda} x and \mathrm{x}.
Question 4 |
The state equation of a second order system is
\dot{x}(t)=A x(t), x(0) is the initial condition.
Suppose \lambda_{1} and \lambda_{2} are two distinct eigenvalues of A and v_{1} and v_{2} are the corresponding eigenvectors. For constants \alpha_{1} and \alpha_{2}, the solution, x(t), of the state equation is
\dot{x}(t)=A x(t), x(0) is the initial condition.
Suppose \lambda_{1} and \lambda_{2} are two distinct eigenvalues of A and v_{1} and v_{2} are the corresponding eigenvectors. For constants \alpha_{1} and \alpha_{2}, the solution, x(t), of the state equation is
\sum_{i=1}^{2} \alpha_{i} e^{\lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{2 \lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{3 \lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{4 \lambda_{i} t} v_{i} |
Question 5 |
Let x be an n \times 1 real column vector with length l=\sqrt{x^{T} x}. The trace of the matrix P=x x^{T} is
l^{2} | |
\frac{l^{2}}{4} | |
l | |
\frac{l^{2}}{2} |
Question 5 Explanation:
Given,
l=\sqrt{x^{T} x}, P=\left(x x^{T}\right)_{n \times n}
Let
\begin{aligned} (x)_{n \times 1} & =\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{array}\right] \\ l & =\sqrt{x^{T} x}=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots x_{n}^{2}} \\ P & =x x^{T} \\ &=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \cdot \\ x_{n} \end{array}\right]\left[x_{1} x_{2} x_{3} \ldots x_{n}\right]\\ P&=\left[ \begin{array}{lllll} x_{1}^{2} & & & & \\ & x_{1}^{2} & & & \\ & & - & & \\ & & & - & \\ & & & & x_{n}^{2} \end{array} \right] \end{aligned}
Trace of P=x_{1}^{2}+x_{2}^{2}+ . . .+ x_{n}^{2}=l^2
l=\sqrt{x^{T} x}, P=\left(x x^{T}\right)_{n \times n}
Let
\begin{aligned} (x)_{n \times 1} & =\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{array}\right] \\ l & =\sqrt{x^{T} x}=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots x_{n}^{2}} \\ P & =x x^{T} \\ &=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \cdot \\ x_{n} \end{array}\right]\left[x_{1} x_{2} x_{3} \ldots x_{n}\right]\\ P&=\left[ \begin{array}{lllll} x_{1}^{2} & & & & \\ & x_{1}^{2} & & & \\ & & - & & \\ & & & - & \\ & & & & x_{n}^{2} \end{array} \right] \end{aligned}
Trace of P=x_{1}^{2}+x_{2}^{2}+ . . .+ x_{n}^{2}=l^2
There are 5 questions to complete.
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Update question no 74 …it will be option d