Linear Algebra

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Linear Algebra - Linear Algebra

61:

Eigen vectors of the matrix

is (are)

A.	(1,0)
B.	(0,1)
C.	(1,1)
D.	(1,-1)
	Answer Report Discuss
	Option: D Explanation : Eigen value are roots of the equation $\begin{bmatrix} 1-a&-1\\-1&1-a \end{bmatrix}=0$ i.e. (1 - a)² - 1 = 0 $\Rightarrow$ a² - 2a = 0 $\Rightarrow$ a = 0 or a = 2 when a = 0 $\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix}=0\\\Rightarrow \begin{bmatrix} x\\y \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$ this gives x = y when a = 2, $\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix}=2\\\Rightarrow \begin{bmatrix} x\\y \end{bmatrix}=\begin{bmatrix} 2x\\2y \end{bmatrix}$ This gives x = -y Hence, eigen vectors corresponding to 0 and 2 are $\begin{bmatrix} x\\x \end{bmatrix}and\begin{bmatrix} x\\-x \end{bmatrix}or\begin{bmatrix} 1\\1 \end{bmatrix}and\begin{bmatrix} 1\\-1 \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

62:

Invers of the matrix

A.	$\begin{bmatrix} \frac{5}{13}&\frac{-1}{13}\\\frac{2}{13}&\frac{3}{13} \end{bmatrix}$
B.	$\begin{bmatrix} \frac{2}{13}&\frac{5}{13}\\\frac{-1}{13}&\frac{3}{13} \end{bmatrix}$
C.	$\begin{bmatrix} \frac{-1}{13}&\frac{5}{13}\\\frac{2}{13}&\frac{3}{13} \end{bmatrix}$
D.	$\begin{bmatrix} \frac{1}{13}&\frac{-5}{13}\\\frac{2}{13}&\frac{3}{13} \end{bmatrix}$
	Answer Report Discuss
	Option: C Explanation : For the matrix $A=\begin{bmatrix} -3&5\\2&1 \end{bmatrix}$ cofactors are a₁₁= 1, a₂₁ = -5, a₁₂= -2, a₂₂= -3 Therefore matrix by cofactors $=\begin{bmatrix} 1&-2\\-5&-3 \end{bmatrix}$ Adjoint A $=\begin{bmatrix} 1&-5\\-2&-3 \end{bmatrix}$ I A I = -3(1) - 5(2) = -3 - 10 = -13 $\therefore A^{-1} =\frac{Adjoint A}{\left \| A \right \|} \begin{bmatrix} \frac{-1}{13}&\frac{5}{13}\\\frac{2}{13}&\frac{3}{13}\end{bmatrix}\\$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

63:

If A and B are square matrices of size n × n ,

then which of the following statement is not true?

A.	det(AB) = det(A) Det(B)
B.	det(kA) = kⁿ det(A)
C.	det(A+B) = det (A) + det (B)
D.	det(A^T) = 1/det(A^-1)
	Answer Report Discuss
	Option: C Explanation : Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

64:

If A be an invertible matrix and inverse of 7A is

then matrix A is

A.	$\begin{bmatrix} 1&\frac{2}{7}\\\frac{4}{7}&\frac{1}{7} \end{bmatrix}$
B.	$\begin{bmatrix} 7&2\\4&1 \end{bmatrix}$
C.	$\begin{bmatrix} 1&-\frac{4}{7}\\-\frac{2}{7}&\frac{1}{7}\end{bmatrix}$
D.	$\begin{bmatrix} 7&4\\2&1\end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : Inverse of 7A= $\begin{bmatrix} -1&2\\4&7 \end{bmatrix}$ $det\left \| \Delta \right \|=\begin{bmatrix} -1&2\\4&7 \end{bmatrix}$ = 7 - 8 = -1 Therefore matrices of cofactor $=det\left \| \Delta \right \|=\begin{bmatrix} -7&-2\\-4&-1 \end{bmatrix}$ and its transpose, $A^{T}=det\left \| \Delta \right \|=\begin{bmatrix} -7&-2\\-4&-1 \end{bmatrix}\\\therefore Inverse = \frac{A^{T}}{\left \| \Delta \right \|}\begin{bmatrix} 7&2\\4&1 \end{bmatrix}\\\therefore A=\frac{1}{7}\left [ \frac{A^{T}}{\left \| \Delta \right \|} \right ]= \begin{bmatrix} 1&\frac{2}{7}\\\frac{4}{7}&\frac{1}{7} \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

65:

If A =

then state transition matrix eAt is

A.	$\begin{bmatrix} 0&e^{-t}\\e^{-t}&0 \end{bmatrix}$
B.	$\begin{bmatrix} e^{t}&0\\0&e^{t} \end{bmatrix}$
C.	$\begin{bmatrix} e^{-t}&0\\0&e^{-t} \end{bmatrix}$
D.	$\begin{bmatrix} 0&e^{t}\\e^{t}&0 \end{bmatrix}$
	Answer Report Discuss
	Option: B Explanation : Here, $\begin{bmatrix} sI-1 \end{bmatrix}^{-1}=\begin{bmatrix} s-1&0\\0&s-1 \end{bmatrix}^{-1}\\Then Z^{-1}\begin{bmatrix} sI-1 \end{bmatrix}^{-1}= Z^{-1}\left \{ \begin{vmatrix} \frac{1}{s-1}&s^{0}\\0&\frac{1}{s-1} \end{vmatrix} \right \}\\=\begin{bmatrix} e^{t}&0\\0&e^{t} \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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