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Diff Eq 5.1

 

<== Differential Equations      Matrix Multiplication Hint
Number Question Answer
See the work
1. Compute the following product.
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -3 0 3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
2 -2 -5
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-4
-4
=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
The work

Point Cost: 3
2. Compute the following product.
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -3 -3 4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -1 -3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-3 2
3 1
=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
The work

Point Cost: 3
3. Compute the following product.
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 0 3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-1 2
2 3
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -2 -2 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-3 2
=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
The work

Point Cost: 3
4.
If A=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -1 -4 2 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
1 0 -4
0 -3 4
and B=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 4 4 3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
0 -1 -4
0 -3 2

Then AB=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]

and BA=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
AB=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]

BA=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
The work

Point Cost: 3
5. Match the differential equations and their vector valued function solutions:
It will be good practice to multiply at least one solution out fully, to make sure that you know how to do it, but you can get the other answers quickly by process of elimination and just multiply out one row element.

 1.
y'(t)=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -86 218 -160 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
73 -49 80
111 -138 165
y(t)=

 2.
y'(t)=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 14 0 -4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
2 13 -8
-3 0 25
y(t)=

 3.
y'(t)=
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 15 0 0 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
4 20 -15
4 30 -25
y(t)=


A.
y(t)==
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 0 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
1
1
e^{5 t}

B.
y(t)==
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] 4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
5
1
e^{13 t}

C.
y(t)==
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight] -2 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
1
3
e^{45 t}
1. C
2. B
3. A
The work