Skip to content
Skip to main navigation
Skip to first column
Skip to second column

Test!

Test yourself!

Test

Test:	DELA - 2018 Spring - Midterm II-2
Number of Questions:	9
Time:

$\hspace{-30mm} \mbox{Let} \ V \ \mbox{be the vector space of all solutions of the differential equation}\\[2mm] y^{\prime\prime}+y^{\prime}-6y=0\\[2mm] \hspace*{-24mm} \mbox{Then the dimension of} \ V \ \mbox{is}$

	$5$
	$\mbox{None of the above}$
	$4$
	$2$
	$6$
	$1$
	$0$
	$3$

$\hspace{-10mm} \mbox{Let} \ A \ \mbox{be a square matrix.}$

	$\hspace{-6mm}\mbox{If a multiple of one row (column) of} \ A \ \mbox{is added to another row (column)} \\ \mbox{to produce a matrix} \ B, \ \mbox{then} \ \det A=-\det B.$
	$\hspace{-6mm}\mbox{If one row (column) of} \ A \ \mbox{is multiplied by} \ k \ \mbox{to produce} \ B, \\ \mbox{then} \ \det B=k^2\det A.$
	$\mbox{All of the above.}$
	$\hspace{-6mm}\mbox{If two rows (columns) of} \ A \ \mbox{are interchanged to produce} \ B, \\ \mbox{then} \ \det A=\det B.$
	$\mbox{None of the above.}$

$\hspace{-30mm} \mbox{Let} \ V \ \mbox{be the set of all solutions of the system}\\[2mm] \left \{ \begin{aligned} x+y+z&=0\\[2mm] x+y+2z&=0 \end{aligned} \right .\\[2mm] \hspace*{-24mm} \mbox{Then}$

	$V \ \mbox{is a vector space and its dimension is 3}$
	$\mbox{None of the above}$
	$V \ \mbox{is a vector space and its dimension is 2}$
	$V \ \mbox{is a vector space and its dimension is 1}$
	$V \ \mbox{is a vector space and its dimension is 0}$
	$V \ \mbox{is not a vector space}$

$\hspace{-40mm} \mbox{Let} \ W_1 \ \mbox{be the set of} \ 2\times 2 \ \mbox{matrices}\\[2mm] \left [ \begin{array}{ll} a&b\\[2mm] c&1 \end{array} \right ]\\[2mm] \hspace*{-35mm} \mbox{where} \ a, b, c \ \mbox{are real numbers such that} \ a+b+c=0. \\ \hspace*{-34mm} \mbox{Let also} \ W_2 \ \mbox{be the set of} \ 2\times 2 \ \mbox{matrices}\\[2mm] \left [ \begin{array}{ll} a&b\\[2mm] c&0 \end{array} \right ]\\[2mm] \hspace*{-35mm} \mbox{where} \ a, b, c \ \mbox{are real numbers such that} \ a+b+c=1. \ \mbox{Then}$

	$W_1 \ \mbox{is a vector space, but} \ W_2 \ \mbox{is not a vector space.}$
	$W_1 \ \mbox{is a vector space and} \ W_2 \ \mbox{is a vector space.}$
	$W_1 \ \mbox{is not a vector space and} \ W_2 \ \mbox{is not a vector space.}$
	$W_1 \ \mbox{is not a vector space, but} \ W_2 \ \mbox{is a vector space.}$

$\hspace{-15mm} \mbox{Let}\\[2mm] {\bf v}_1= \left [ \begin{array}{r} 1\\[2mm] -3\\[2mm] 4 \end{array} \right ], \hspace{10mm} {\bf v}_2= \left [ \begin{array}{r} -2\\[2mm] 1\\[2mm] 0 \end{array} \right ], \hspace{10mm} {\bf v}_3= \left [ \begin{array}{r} -4\\[2mm] -3\\[2mm] 8 \end{array} \right ]\\[2mm] \hspace*{-10mm} \mbox{Then}$

	${\bf v}_1, {\bf v}_2, {\bf v}_3 \ \mbox{are linearly dependent and span} \ \mathbb{R}^3$
	$\mbox{None of the above}$
	${\bf v}_1, {\bf v}_2, {\bf v}_3 \ \mbox{are linearly dependent and do not span} \ \mathbb{R}^3$
	${\bf v}_1, {\bf v}_2, {\bf v}_3 \ \mbox{are linearly independent and do not span} \ \mathbb{R}^3$
	${\bf v}_1, {\bf v}_2, {\bf v}_3 \ \mbox{are linearly independent and span} \ \mathbb{R}^3$

$\hspace{-30mm} \mbox{Convert the differential equation}\\[2mm] \dfrac{d^4y}{dt^4}+\left (\dfrac{dy}{dt}\right )^3+y=\sin t\\[2mm] \hspace*{-24mm} \mbox{into a system of first-order equations.}$

	$\mbox{None of the above}$
	$\begin{cases} \dfrac{dx_1}{dt}=x_1\\[2mm] \dfrac{dx_2}{dt}=x_2\\[2mm] \dfrac{dx_3}{dt}=x_3\\[2mm] \dfrac{dx_4}{dt} =\sin t-x_2^3-x_1 \end{cases}$
	$\begin{cases}\dfrac{dx_2}{dt}=x_1\\[2mm] \dfrac{dx_3}{dt}=x_2\\[2mm] \dfrac{dx_4}{dt}=x_3\\[2mm] \dfrac{dx_1}{dt} =\sin t-x_2^3-x_1 \end{cases}$
	$\begin{cases}\dfrac{dx_2}{dt}=x_2\\[2mm] \dfrac{dx_3}{dt}=x_3\\[2mm] \dfrac{dx_4}{dt}=x_4\\[2mm] \dfrac{dx_1}{dt} =\sin t-x_2^3-x_1 \end{cases}$
	$\begin{cases}\dfrac{dx_1}{dt}=x_2\\[2mm] \dfrac{dx_2}{dt}=x_3\\[2mm] \dfrac{dx_3}{dt}=x_4\\[2mm] \dfrac{dx_4}{dt} =\sin t-x_2^3-x_1 \end{cases}$

$\hspace{-6mm} \mbox{Let}\\[2mm] {\bf m}_1=\left [ \begin{array}{rr} 1&0\\ 0&1 \end{array} \right ], \quad {\bf m}_2=\left [ \begin{array}{rr} 1&1\\ 1&0 \end{array} \right ], \quad {\bf m}_3=\left [ \begin{array}{rr} 0&1\\ 1&0 \end{array} \right ], \quad {\bf m}_4=\left [ \begin{array}{rr} 0&1\\ 1&1 \end{array} \right ]\\[2mm] \hspace*{-1mm} \mbox{Find a basis of the vector space spanned by} \ {\bf m}_1, \ {\bf m}_2, \ {\bf m}_3, \ {\bf m}_4.$

	$\mbox{None of the above}$
	$\left \{ {\bf m}_1, \ {\bf m}_2, \ {\bf m}_3, \ {\bf m}_4 \right \}$
	$\left \{ {\bf m}_1, \ {\bf m}_2, \ {\bf m}_3 \right \}$
	$\left \{{\bf m}_1, \ {\bf m}_2 \right \}$
	$\left \{ {\bf m}_1 \right \}$

$\hspace{-30mm} \mbox{Let}\\[2mm] A=\left [ \begin{array}{crr} e^x&0&1\\[2mm] 0&1&2\\[2mm] 0&1&-1 \end{array} \right ]\\[2mm] \hspace*{-25mm} \mbox{Find} \ A^{-1}.$

	$\dfrac{1}{3}\left [ \begin{array}{c} 3e^{-x}\\[2mm] 0\\[2mm] 0 \end{array} \begin{array}{c} -e^{-x}\\[2mm] 1\\[2mm] 1 \end{array} \begin{array}{c} e^{-x}\\[2mm] 2\\[2mm] -1 \end{array} \right ]$
	$\dfrac{1}{3}\left [ \begin{array}{c} 3e^{-x}\\[2mm] 0\\[2mm] 0 \end{array} \begin{array}{c} -3e^{-x}\\[2mm] 1\\[2mm] 1 \end{array} \begin{array}{c} e^{-x}\\[2mm] 2\\[2mm] -1 \end{array} \right ]$
	$\mbox{None of the above}$
	$\dfrac{1}{3}\left [ \begin{array}{c} e^{-x}\\[2mm] 0\\[2mm] 0 \end{array} \begin{array}{c} 3e^{-x}\\[2mm] 1\\[2mm] 1 \end{array} \begin{array}{c} 3e^{-x}\\[2mm] 2\\[2mm] -1 \end{array} \right ]$
	$\dfrac{1}{3}\left [ \begin{array}{c} e^{-x}\\[2mm] 0\\[2mm] 0 \end{array} \begin{array}{c} e^{-x}\\[2mm] 1\\[2mm] 1\end{array} \begin{array}{c} -3e^{-x}\\[2mm] 2\\[2mm] -1 \end{array} \right ]$
	$\dfrac{1}{3}\left [ \begin{array}{c} -e^{-x}\\[2mm] 0\\[2mm] 0 \end{array} \begin{array}{c} -3e^{-x}\\[2mm] 1\\[2mm] 1 \end{array} \begin{array}{c} e^{-x}\\[2mm] 2\\[2mm] -1 \end{array} \right ]$

$\hspace{-30mm} \mbox{Let} \ V \ \mbox{be a vector space of all differentiable functions and let}\\[2mm] T(f)=f^{\prime\prime}+f^{\prime}\\[2mm] \hspace*{-24mm} \mbox{where} \ f^{\prime} \ \mbox{is the derivative of} \ f. \ \mbox{Then}$

	$T \ \mbox{is linear, since} \ T(f_1+f_2)=T(f_1)+T(f_2) \ \mbox{and} \ T(c+f)=c+T(f)$
	$T\ \mbox{ is not linear, since} \ T(f_1f_2)\not =T(f_1)T(f_2)$
	$T \ \mbox{is not linear, since} \ T(cf)\not =cT(f)$
	$T \ \mbox{is not linear, since} \ T(f_1+f_2)\not =T(f_1)+T(f_2)$
	$\mbox{None of the above}$
	$T \ \mbox{is linear, since} \ T(f_1f_2)=T(f_1)T(f_2)$

Choose the answer

Powered by Valery Gromov

Main Menu

Home
Teaching
Calculus Website
Calculus Tests
Precalculus Website
Precalculus Tests
Linear Algebra Website
Linear Algebra Tests
Linear Algebra and Matrix Theory Website
Linear Algebra and Matrix Theory Tests
Differential Equations and Linear Algebra Website
Differential Equations and Linear Algebra Problems
Differential Equations and Linear Algebra Tests
All Tests