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Test:	DELA - 2019 Spring - Midterm II-1
Number of Questions:	6
Time:

$\hspace{-6mm} \mbox{Let} \ H_1 \ \mbox{be the set of all vectors of the form}\\[2mm]\left [\begin{array}{c}a\\b\\c\end{array}\right ]\quad\mbox{where} \ a, b, c \ \mbox{are real numbers such that} \ a^3+b^3+c^3=0\\[2mm]\mbox{and} \ H_2 \ \mbox{be the set of all vectors of the form}\\[2mm]\left [\begin{array}{c}a\\b\end{array}\right ]\quad\mbox{where} \ a, b \ \mbox{are real numbers such that} \ a^3+b^3=0\\[2mm]\mbox{Then}$

	$H_1 \ \mbox{is a vector space, but} \ H_2 \ \mbox{is not a vector space.}$
	$H_1 \ \mbox{is not a vector space, but} \ H_2 \ \mbox{is a vector space.}$
	$H_1 \ \mbox{is not a vector space and} \ H_2 \ \mbox{is not a vector space.}$
	$H_1 \ \mbox{is a vector space and} \ H_2 \ \mbox{is a vector space.}$

$\mbox{Any} \ n \ \mbox{linearly dependent vectors in an} \ n \ \mbox{dimensional space} \ V$

	$\mbox{None of the above}$
	$\mbox{must also span} \ V$
	$\mbox{All of the above}$
	$\mbox{are a basis for} \ V$
	$\mbox{are linearly independent}$

$\hspace{-6mm} \mbox{Consider the following transformation}\\[2mm]{\bf x}=\left [\begin{array}{c}x_1\\[3mm]x_2\\[3mm]x_3\end{array}\right ]\quad\longrightarrow\quad T({\bf x})=\left [\begin{array}{c}x_1+x_2+x_3\\[3mm]x_1+x_2x_3\\[3mm]x_2\end{array}\right ]$

	$T \ \mbox{is linear, since} \ T({\bf x}_1+{\bf x}_2) = T({\bf x}_1)+T({\bf x}_2)$
	$T \ \mbox{is not linear, since} \ T(c{\bf x})\not = cT({\bf x})$
	$T \ \mbox{is linear, since} \ T({\bf x})=A{\bf x}\ \mbox{where} \ A=\left [\begin{array}{rrr}1&1&1\\[3mm]1&1&-1\\[3mm]0&1&0\end{array}\right ]$
	$\mbox{None of the above}$
	$T \ \mbox{is linear, since} \ T({\bf x})=A{\bf x}\ \mbox{where} \ A=\left [\begin{array}{rrr}1&1&0\\[3mm]1&1&1\\[3mm]1&-1&0\end{array}\right ]$

$\hspace{-6mm} \mbox{Find the determinant of the matrix}\\[2mm]A=\left [\begin{array}{ccc}\sin x&-\cos x&1\\[2mm]\cos x&\sin x&1\\[2mm]1&0&1\end{array}\right ]$

	$1-\sin x -\cos x$
	$1-\sin x +\cos x$
	$1+\sin x -\cos x$
	$\mbox{None of the above}$
	$1+\sin x +\cos x$

$\hspace{-6mm} \mbox{Let}\\[2mm]A=\left [\begin{array}{rr}11&13\\[2mm]14&12\end{array}\right ]\quad\mbox{and}\quad B=\left [\begin{array}{rr}15&17\\[2mm]18&16\end{array}\right ]\\[2mm]\mbox{Then}$

	$AB-BA=\left [\begin{array}{rr}1&0\\[2mm]0&1\end{array}\right ]$
	$AA^{-1}-BB^{-1}=\left [\begin{array}{rr}0&0\\[2mm]0&0\end{array}\right ]$
	$AB-BA=\left [\begin{array}{rr}0&0\\[2mm]0&0\end{array}\right ]$
	$AA^{-1}-BB^{-1}=\left [\begin{array}{rr}1&0\\[2mm]0&1\end{array}\right ]$
	$\mbox{None of the above}$

$\mbox{Let} \ A=\left [\begin{array}{rr}2&1\\[2mm]0&2\end{array}\right ], \ \mbox{then} \ e^{At} \ \mbox{is}$

	$\mbox{None of the above}$
	$e^{2t}\left [\begin{array}{cc}1&t\\[2mm]0&1\end{array}\right ]$
	$e^{2t}\left [\begin{array}{cc}2&t\\[2mm]0&1\end{array}\right ]$
	$e^{2t}\left [\begin{array}{cc}1&2t\\[2mm]0&1\end{array}\right ]$
	$e^{2t}\left [\begin{array}{cc}2&t\\[2mm]0&2\end{array}\right ]$

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