Linear Algebra

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Linear Algebra - Linear Algebra

66:

For the matrix

one of the eigen values is equal to -2.

which of the following is an eigen vector?

A.	$\begin{bmatrix} 3\\-2\\1 \end{bmatrix}$
B.	$\begin{bmatrix} -3\\2\\-1 \end{bmatrix}$
C.	$\begin{bmatrix} 1\\-2\\3 \end{bmatrix}$
D.	$\begin{bmatrix} 2\\5\\0 \end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : $\left \| P-\lambda I \right \|=\begin{vmatrix} 3-\lambda &-2&2\\0&2-\lambda &1\\0&0&1-\lambda \end{vmatrix}=0\\\therefore \lambda = -2, 3,1$ Putting λ= -2, we get 5x--2y+2z = 0 Putting z = 0, we get 5x = 2y Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

67:

then (a + b) equals

A.	$\frac{7}{20}$
B.	$\frac{3}{20}$
C.	$\frac{19}{60}$
D.	$\frac{11}{20}$
	Answer Report Discuss
	Option: A Explanation : We know, AA^-1 = I $\therefore \begin{bmatrix} 2&-0.1\\0&3 \end{bmatrix}\begin{bmatrix} \frac{1}{2}&a\\0&b \end{bmatrix}=\begin{bmatrix} 1&2a-0.1b\\0&3b \end{bmatrix}\\\Rightarrow 3b=1\\\Rightarrow b=\frac{1}{3}\\0r2a-0.1b=0\\\Rightarrow 2a=\frac{b}{10} \\\Rightarrow a=\frac{1}{60} \\\therefore a + b= \frac{1}{3}+\frac{1}{60}\\=\frac{21}{60}=\frac{7}{20}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

68:

The eigen values and the corresponding eigen vectors of a 2 x 2 matrix are given by

Eigenvalue

Eigenvector

The matrix is

A.	$\begin{bmatrix} 6&2\\2&6 \end{bmatrix}$
B.	$\begin{bmatrix} 4&6\\6&4 \end{bmatrix}$
C.	$\begin{bmatrix} 2&4\\4&2 \end{bmatrix}$
D.	$\begin{bmatrix} 4&8\\8&4 \end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : Given matrix $A=\begin{bmatrix} 6&2\\2&6 \end{bmatrix}$ We know, $\left \| \lambda I-A \right \|=0$ $\Rightarrow \begin{vmatrix} \lambda -6&-2\\-2&\lambda -6 \end{vmatrix}\\\Rightarrow \lambda ^{2}-I2\lambda +32=0$ $\Rightarrow$ $\lambda$ = 4, 8 (eigen values) For $\lambda$ ₁ = 4 $(\lambda_{1}I-A)=\begin{bmatrix} -2&-2\\-2&-2 \end{bmatrix}, \nu _{1}=\begin{bmatrix} 1\\1 \end{bmatrix}$ For $\lambda$ ₂= 8 $(\lambda_{2}I-A)=\begin{bmatrix} 2&-6\\-6&2 \end{bmatrix}, \nu _{2}=\begin{bmatrix} 1\\-1 \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

69:

For the

A.	2
B.	4
C.	6
D.	8
	Answer Report Discuss
	Option: C Explanation : Given matrix $M= \begin{bmatrix} 4&2\\2&4 \end{bmatrix}\\\therefore M-\lambda I=\begin{bmatrix} 4-\lambda &2\\2&4-\lambda \end{bmatrix}$ For the eigen vector $\begin{bmatrix} 101\\101 \end{bmatrix}$ $(4 - \lambda) (101) + 2 (101) = 0$ $= 4 - \lambda + 2 = 0$ = λ = 6 (eigenvalue) Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

70:

A second order system starts with an initial condition of $\begin{bmatrix} 2\\3 \end{bmatrix}$ without any external input. The state transition matrix for the system is given by $\begin{bmatrix} e^{-2t}&0\\0&e^{-t} \end{bmatrix}$ .

The state of the end of 1 second is given by

A.	$\begin{bmatrix} 0.271\\1.100 \end{bmatrix}$
B.	$\begin{bmatrix} 0.135\\0.368 \end{bmatrix}$
C.	$\begin{bmatrix} 0.271\\0.736 \end{bmatrix}$
D.	$\begin{bmatrix} 0.135\\1.100 \end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : The state of the system at time t is, $X(t)=\left [ sI-A \right ]^{-1}\\X(0)=\phi (t)X(0)\\=\begin{bmatrix} e^{-2t}&0\\0&e^{-t} \end{bmatrix}\begin{bmatrix} 2\\3 \end{bmatrix}=\begin{bmatrix} 2e^{-2t}\\3e^{-t} \end{bmatrix}$ At t = 1 $X(1)=\begin{bmatrix} 2e^{-2t}\\3e^{-t} \end{bmatrix}=\begin{bmatrix} 0.271\\1.100 \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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