#$&*
course Mth 279
4/29 1
Query 22 Differential Equations*********************************************
Question: Find the values for which the matrix
[ t + 1, t; t, t + 1]
pictured as:
is invertible.
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Your solution:
I believe that the matrix the way it is set up where the main diagonal is t+1 and the other elements are t, that for any value of t. This is because no matter what t is the matrix would always be reducible to the identity matrix using row reductions and other methods.
Ok, I’ve rethought this one and now see I was worng(I left the previous answer so you can see my mistake). To find the answer to this we first calculate the determinant
det[given matrix] = [(t+1)*(t+1)] - [t*t] = [t^2+2t+1] - [t^2] = 2t + 1
Knowing matrix is invertible only when determinant is not equal to zero, we see this matrix is invertiable for all t EXCEPT t = (-1/2). This is because when t=(-1/2)
det[given matrix] = 2*(-1/2) + 1 = -1 + 1 = 0, which makes matrix un-invertible(not sure real word).
Confidence rating:
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Given Solution:
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Self-critique rating:
@& Good.*@
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Question: Find the limit as t -> 0 of the matrix
[ sin(t) / t, t cos(t), 3 / (t + 1); e^(3 t), sec(t), 2 t / (t^2 - 1) ]
pictured as
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Your solution:
lim t0 = 1 0 3
1 1 0
Note that I had to use l’hopital rule for calculating sin(t)/t, where sin(t)/t = cos(t)/1 (took derivative of sin(t) and t), now it is easy to see as t0 cos(t)1 so we have a limit of 1/1 as t0.
Everything else was fairly standard in regards to finding the limit.
Confidence rating:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& Very good.*@
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Question: Find A ' (t) and A ''(t), where the derivatives are with respect to t and the matrix is
A = [ sin(t), 3 t; t^2 + 2, 5 ]
pictured as
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Your solution:
A’ = cos(t) 3
2t 0
A’’ = -sin(t) 0
2 0
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Write the system of equations
y_1 ' = t^2 y_1 + 3 y_2 + sec(t)
y_2 ' = sin(t) y _1 + t y_2 - 5
in the form
y ' = P(t) y + g(t),
where P(t) is a 2 x 2 matrix and y and g(t) are 2 x 1 column vectors.
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Your solution:
t^2 3 * y_1 + sec(t) = y_1’
sin(t) t y_2 -5 y_2’
Confidence rating:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: If
A '' = [1, t; 0, 0]
with
A(0) = [ 1, 1; -2, 1]
A(1) = [-1, 2; -2, 3 ]
then what is the matrix A(t)?
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Your solution:
I’m confused, this is what I’ve got
A’(t) = Int(A '') = t+c_11 t^2/2 + c_12 = t t^2/2 + C
0+c_21 0+c_22 0 0
A(t) = Int(A’(t)) = t^2/2+c_11 t^3/3 + c_12 = t^2/2 t^3/3 + C
0+c_21 0+c_22 0 0
?????Not sure exactly how to find bottom elements(row 2)????????
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& A'(t) = integral(A(t)) = [t+c_11, t^2 / 2 + c_12; c_21, c_22 ] .
A(0) = [c_11, c_12; c_21, c_22] = [ 1, 1; -2, 1] so
c_11 = 1,
c_12 = 1
c_21 = 2
c_22 = 1
and
A ' (t) = [ t + 1, t^2 / 2 + 1; -2, 1 ]
Integrating once more to find A(t) we get
A(t) = [ t^2 / 2 + t + d_11, t^3 / 6 + t + d_12; -2 t + d_21, t + d_22 ]
A(1) = [3/2 + d_11, 7/6 + d_12; -2 + d_21, 1 + d_22 ] = [-1, 2;-2, 3]
so
3/2 + d_11 = -1
7/6 + d_12 = 2
-2 + d_21 = -2
1 + d_22 = 3
We easily evaluate the d_ij coefficients to obtain our matrix A(t).*@
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