MATH-225 Final Exam — Fall 2021 — Solutions
Q 1. Let A be a 3 × 3-matrix, B a 3 × 2-matrix, and z a nonzero 3 × Assume 1-matrix.
that:
1 0
(i) B 6= B.
0 0
0 0
(ii) B 6= B.
0 1
−2 0
(iii) B = AB.
0 1
(iv) det(A) = 2.
(v) z is in Null(BT ).
(vi) A = A T .
Then:
(a) (4 points) Compute det(A − 2I).
(b) (4 points) Compute rank(B).
(c) (4 points) ComputeT B Az.
(d) (4 points) Compute Az in
terms of z.
1 1
(e) (4 points) If B = 1 1 then compute A.
2 −1
Solution. Let B = x y . Then x 6= 0 by (ii),y 6= 0 by (i),Ax = −2x by (iii),
and Ay = y by (iii). Hence x is an eigenvectorAofassociated with the eigenvalue −2
and y is an eigenvector Aofassociated with the eigenvalueTherefore 1. det(A − λI) =
(−2 − λ)(1 − λ)(λ 0 − λ) for some0.λ By (iv), we have 2 = (−2 − 0)(1 − 0−0)(λ
0) = −2λ0.
Hence 0λ = −1.
(a) det(A − 2I) = (−2 − 2)(1 − 2)(−1 − 2) = −12.
(b) The vectors x,y correspond to distinct eigenvalues Hence of A. x, y are linearly inde-
pendent.Therefore rank(B) = rank( x y ) = 2.
T
T T T T −2 0 −2 0 0
(c) B Az = B A z = (AB) z = B z= BT z = , (d) Now
0 1 0 1 0
rank(BT ) = 2 and BT is a 2 × 3-matrix.Hence dim(Null(B T
)) = 3 − 2 = 1.We have z
T T
is in Null(B ) and z is not zero.Hence Null(B) = Span(z).By part (c) we know, Az is
in Null(BT ). Hence Az = λz for some λ.Moreover,z is not in Row(B). Hence z is an
eigenvector of A associated
withTherefore
−1. Az = −z.
1 1
(e) Assume B = 1 1 . Then
2 −1
−1
1 1 2 1 1 2 1 ),
Null(BT ) = Null( ) = Null( ) = Span(
1 1 −1 0 0 −3
0
1 1 −1 −2 0 0 1 1 2 −1 1 −2
hence A = 1 1 1 0 1 0 16 2 2 −2 = 12 1 −1 −2 .
2 −1 0 0 0 −1 −3 3 0 −2 −2 −2
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, 1
Q 2. (20 points) Solve the following first order ODE for
e
. y>
x dy
1+ + 1 + ln(y) = 0.
y dx
Solution.The given Diff.Eq is not in a linear from inHowever,
y(x). one can interchange
the variables x and y, i.e x = x(y).
Interchanging the variables, the equation reads as
dx x
(1 + ln(y)) = − 1 + .
dy y
Since y >−1e, then dividing to the nonzero factor 1+ln(y), this equation can be rewritte
as the following first order linear equation for x
dx 1 1
+ x=− .
dy y(1 + ln(y)) 1 + ln(y)
Now, one can apply the integrating factor method to solve this Diff.Eq.
The integrating
factor is R 1
ρ = e y(1+ln(y)) dy = 1 + ln(y).
Multiplying the above differential equation by its integrating factor, we obtain
dx 1 d
(1 + ln(y)) + x = −1 =⇒ ((1 + ln(y)) x) = −1.
dy y dy
Integrating this equation, we find the general solution in the implicit form as
x + x ln(y) + y = C.
2
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Q 1. Let A be a 3 × 3-matrix, B a 3 × 2-matrix, and z a nonzero 3 × Assume 1-matrix.
that:
1 0
(i) B 6= B.
0 0
0 0
(ii) B 6= B.
0 1
−2 0
(iii) B = AB.
0 1
(iv) det(A) = 2.
(v) z is in Null(BT ).
(vi) A = A T .
Then:
(a) (4 points) Compute det(A − 2I).
(b) (4 points) Compute rank(B).
(c) (4 points) ComputeT B Az.
(d) (4 points) Compute Az in
terms of z.
1 1
(e) (4 points) If B = 1 1 then compute A.
2 −1
Solution. Let B = x y . Then x 6= 0 by (ii),y 6= 0 by (i),Ax = −2x by (iii),
and Ay = y by (iii). Hence x is an eigenvectorAofassociated with the eigenvalue −2
and y is an eigenvector Aofassociated with the eigenvalueTherefore 1. det(A − λI) =
(−2 − λ)(1 − λ)(λ 0 − λ) for some0.λ By (iv), we have 2 = (−2 − 0)(1 − 0−0)(λ
0) = −2λ0.
Hence 0λ = −1.
(a) det(A − 2I) = (−2 − 2)(1 − 2)(−1 − 2) = −12.
(b) The vectors x,y correspond to distinct eigenvalues Hence of A. x, y are linearly inde-
pendent.Therefore rank(B) = rank( x y ) = 2.
T
T T T T −2 0 −2 0 0
(c) B Az = B A z = (AB) z = B z= BT z = , (d) Now
0 1 0 1 0
rank(BT ) = 2 and BT is a 2 × 3-matrix.Hence dim(Null(B T
)) = 3 − 2 = 1.We have z
T T
is in Null(B ) and z is not zero.Hence Null(B) = Span(z).By part (c) we know, Az is
in Null(BT ). Hence Az = λz for some λ.Moreover,z is not in Row(B). Hence z is an
eigenvector of A associated
withTherefore
−1. Az = −z.
1 1
(e) Assume B = 1 1 . Then
2 −1
−1
1 1 2 1 1 2 1 ),
Null(BT ) = Null( ) = Null( ) = Span(
1 1 −1 0 0 −3
0
1 1 −1 −2 0 0 1 1 2 −1 1 −2
hence A = 1 1 1 0 1 0 16 2 2 −2 = 12 1 −1 −2 .
2 −1 0 0 0 −1 −3 3 0 −2 −2 −2
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:43:59 GMT -05:00
https://www.coursehero.com/file/149648236/MATH225FINALFALL2021SOLPDF/
, 1
Q 2. (20 points) Solve the following first order ODE for
e
. y>
x dy
1+ + 1 + ln(y) = 0.
y dx
Solution.The given Diff.Eq is not in a linear from inHowever,
y(x). one can interchange
the variables x and y, i.e x = x(y).
Interchanging the variables, the equation reads as
dx x
(1 + ln(y)) = − 1 + .
dy y
Since y >−1e, then dividing to the nonzero factor 1+ln(y), this equation can be rewritte
as the following first order linear equation for x
dx 1 1
+ x=− .
dy y(1 + ln(y)) 1 + ln(y)
Now, one can apply the integrating factor method to solve this Diff.Eq.
The integrating
factor is R 1
ρ = e y(1+ln(y)) dy = 1 + ln(y).
Multiplying the above differential equation by its integrating factor, we obtain
dx 1 d
(1 + ln(y)) + x = −1 =⇒ ((1 + ln(y)) x) = −1.
dy y dy
Integrating this equation, we find the general solution in the implicit form as
x + x ln(y) + y = C.
2
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:43:59 GMT -05:00
https://www.coursehero.com/file/149648236/MATH225FINALFALL2021SOLPDF/
