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Linear Algebra

Linear Algebra

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Linear Algebra - Linear Algebra

71:

The unit step input u(t) has the state transition matrix

A.	$\begin{bmatrix} 1&\frac{1}{3}(1-e^{-3t})\\0&e^{-3t} \end{bmatrix}$
B.	$\begin{bmatrix} 1&\frac{1}{3}(e^{-t}-e^{-3t})\\0&e^{-t} \end{bmatrix}$
C.	$\begin{bmatrix} 1&\frac{1}{3}(e^{-t}-e^{-3t})\\0&e^{-3t} \end{bmatrix}$
D.	$\begin{bmatrix} 1&(1-e^{-t})\\0&e^{-t} \end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : $(sI-A)^{-1}=\begin{bmatrix} s&1\\0&s+3 \end{bmatrix}^{-1}=\frac{Adj(sI-A)}{\left \| sI-A \right \|}\\=\frac{\begin{bmatrix} s+3&0\\-1&s \end{bmatrix}}{\begin{bmatrix} s&1\\0&s+3 \end{bmatrix}}= \frac{\begin{bmatrix} s+3&0\\-1&s \end{bmatrix}}{s(s+3)}\\=\begin{bmatrix} \frac{1}{s}&\frac{-1}{s(s+3)}\\0&\frac{1}{s+3} \end{bmatrix}\\\therefore \phi (t)=L^{-1}(sI-A)^{-1}\\=\begin{bmatrix} 1&\frac{1}{3}(1-e^{-3t})\\0&e^{-3t} \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

72:

Nullity of the matrix

$A=\begin{bmatrix} -1&4&2\\1&3&2\\-2&1&6\\2&6&4 \end{bmatrix}$

is

A.	1
B.	2
C.	3
D.	4
	Answer Report Discuss
	Option: A Explanation : $A=\begin{bmatrix} -1&4&2\\1&3&2\\-2&1&6\\2&6&4 \end{bmatrix}=\begin{bmatrix} -1&4&2\\0&7&4\\0&7&4\\0&7&4 \end{bmatrix}$ By R₂+R₁; R₃+2R₂ and R₄ + R₃ $\begin{bmatrix} -1&4\\0&7 \end{bmatrix}\\= -7\neq 0$ Hence rank of matrix, A= 2 Nullity of matrix, A= Number of columns of A - rank of A = 3 - 2 = 1 Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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