Solution to Midterm (Winter 2020)
Debopriya Basu (debopriya.basu@concordia.ca)
Midterm for Winter 2020 MATH 204 was scheduled on March 8, 2020
MATH 204: Vectors and Matrices
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Note
• Most of the questions have several methods of getting the correct solution.
• Unless mentioned in the question, you are always free to use any method you have learnt
during the classes.
• I chose to include solutions from the 2020 students’ midterm booklets (with permission)
as these were very neatly done. It also saved me a lot of effort and time this way. :P
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Question Paper
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Solution
1. For A−1 , contruct the Augmented Matrix [A|I]






1 0 −1 1 −2 0
1 2 3 1 0 0
1 2 3 1 0 0
=r3 −r1
1 −2r2
 0 1 2
 0 1 2 0 1 0  R3−→
 0 1 2 0 1 0  R1 =r
0
1 0 
−→
R3 =r3 −r2
1 3 6 0 0 1
0 1 3 −1 0 1
0 0 1 −1 −1 1


1 0 0 0 −3 1
R1 =r1 +r3
3 −2 
−→  0 1 0 2
R2 =r2 −2r3
0 0 1 −1 −1 1


0 −3 1
3 −2 
A−1 =  2
−1 −1 1
2. The solution is due to: the student wanted to remainanonymous
1 2
3. Given (5A − 3I) =
3 4
−1
1 2
4 −2
−2
1
1
⇒ (5A − 3I) =
= (4)(1)−(2)(3)
=
3 4
−3 1
3/2 −1/2
−2
1
3 0
1
1
⇒ 5A =
+
=
3/2 −1/2
0 3
3/2 5/2
1
1
1/5 1/5
1
⇒A= 5
=
3/2 5/2
3/10 1/2
−1
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4. The solution is due to : J. Lemercier
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5. The solution is due to : X. Wang
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6. Given a equation system x+2y = 3 and ax+by = −9, first construct the augmented matrix:
1 2 3
1
2
3
=
, R2 = r2 − ar1
a b −9
0 b − 2a −9 − 3a
(a) For there to be no solution to the system, we require the second entry in the last
row to be 0, but the third entry must be non-zero.
Thus, b − 2a = 0 but −9 − 3a ̸= 0, which implies a ̸= −3.
(b) For there to be exactly one solution, the second entry in last row must be non-zero,
regardless of the third entry.
So, b − 2a ̸= 0.
(c) For there to be infinitely many solutions, the entries in the last row must all be zero.
As such, b − 2a = 0 and −9 − 3a = 0, which implies a = −3 and b = 2a = −6.
Alternatively, for part (c), you could also take a look at the possible cases left. In this
problem, we already have the case of b − 2a ̸= 0 for unique solution (part (b)). We also
have the case where b − 2a = 0 but −9 − 3a ̸= 0 for no solution (part (a)). The only case
left is that where b − 2a = 0 and −9 − 3a = 0.
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