MAT 371 LNEAR ALGEBRA [EXAM 1]
◦ This document provides answers to selected questions.
◦ It is suggested that you practice writing answers with justifications.
◦ Each solution is followed by some notes that is for your own edification.
Problem 2.
Solve the system of linear equations :
2x1 + x2 − x3 = 5
3x1 − x2 + x3 = 0.
Solution The augmented matrix is given by
2 1 −1 5
.
3 −1 1 0
This matrix in its reduced row echelon form looks like :
r1 → 15 (r1 +r2 )
1 0 0 1
1
2 1 −1 5
r2 →−r2 +3r1
−−−−−−−→
−−−−−−−→
3 −1 1 0
3 −1 1 0
0
0
1
0 1
−1 3
The boxed entires are the pivot entries. Since the third column is a non-pivot column, the
variable x3 is the free variable while x1 and x2 are basic variables. Moreover, we get
x1 = 1, x2 = x3 + 3, x3 ∈ R.
NOTES
The coefficient matrix
2 1 −1
3 −1 1
has rank 2. Therefore, there will be one free variable and you do expect infinitely many
solutions.
Problem 3.
Compute the double sum
3 X
5
X
sin
i=1 j=3
ijπ
2
.
Solution This is a double sum that has 9 = 3 + 3 + 3 terms :
3 X
3πi
4πi
5πi
sin
+ sin
+ sin
.
2
2
2
i=1
We get
+ sin 4π
+ sin 5π
+ sin 6π
+ sin 8π
+ sin 10π
+ sin 9π
+ sin 12π
+ sin 15π
sin 3π
2
2
2
2
2
2
2
2
2
= −1 + 0 + 1 + 0 + 0 + 0 + 1 + 0 + (−1) = 0.
NOTES This requires you to realize that there are nine numbers which need to be evaluated before adding them.
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Problem 4.
Find out how many solutions the following system of linear equations has :
x1 + 3x2 + 3x3 = 5
3x1 − x2 + x3 = 0
2x1 + x2 + 2x3 = 3.
Solution We write down the augmented

1
 3
2
matrix

3 3 5
−1 1 0  .
1 2 3
The reductions are as follows :





1
1 3 3 5
1 3
3
5
r2
r2 →− 10
r2 →r2 −3r1
 0
 3 −1 1 0  −


0 −10 −8 −15
−−−−−→
−−−−−−→ 
r3 →r3 −2r1
r3 →r3 +2r2
2 1 2 3
0 −5 −4 −7
0
3
1
0
3
5
4
5
3
2
0
1
− 10



.
Since the augmented column is a pivot column, this is an inconsistent system and has no
solutions.
NOTES
In the solution above, when you arrive at the second matrix


1 3
3
5
 0 −10 −8 −15 
0 −5 −4 −7
it should be clear that this can have no solutions as the second equation and third equation
clearly contradict each other.
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Problem 5.
Compute A3 , where
A=
0 3
2 4
.
Solution The matrix A2 is given by
0 3
0 3
0×0+3×2 0×3+3×4
6 12
2
A =
·
=
=
.
2 4
2 4
2×0+4×2 2×3+4×4
8 22
Similarly, A3 is computed as follows
0 3
6 12
24 66
3
2
·
=
.
A =A·A =
2 4
8 22
44 112
NOTES It doesn’t matter whether we compute A(A · A) or (A · A)A to compute A3 - this
is due to the associativity of the matrix multiplication.
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Problem 6.
Compute the expression 2AB − 7C, where
0 3
7 −2
−5 2
A=
, B=
, and C =
.
2 4
3 −1
4 0
Solution The matrix 2AB is given by
0 3
7 −2
9 −3
18 −6
2AB = 2
·
=2
=
.
2 4
3 1
26 −8
52 −16
Therefore, the matrix 2AB − 7C is given by
18 −6
−5 2
18 −6
−35 14
53 −20
2AB−7C =
−7
=
−
=
.
52 −16
4 0
52 −16
28 0
24 −16
NOTES There are two basic rules one is (implicitly) using here (i) order of matrices matter in multiplication;
(ii) scaling a matrix means multiplyin every entry by that scalar.
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Problem 7.
We consider the matrix


0 3 3
A =  2 −4 0  .
4 −5 3
(a) Compute its rank and find the matrix B in reduced row echelon form that is
row equivalent to A.
(b) We consider the function A from R3 to R3 associated to the matrix A. Is the
function A one-to-one? Is it onto?
Solution
(b) Since the rank is 3, we conclude that the associated linear map A : R3 → R3 is both
onto and one-to-one.
NOTES The rank of A is the key ingredient here. Since both the domain and codomain
are R3 , the answer to (b) follows immediately depending on what the rank is. In particular,
it is going to be yes for both if rank is 3 or no for both if rank is 0, 1 or 2.
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Problem 8.
Find all possible values of t such that the following system of linear equations is
inconsistent :
3x2 + 2x3 = t + 1
3x1 − 3x2 + x3 = t2
x1 + x3 = t.
Solution The augmented matrix is given by


0 3 2 t+1
 3 −3 1 t2  .
1 0 1
t
The reduction steps are as follows :






0 3 2 t+1
1 0 1
t
1 0
1
t
r1 ↔r3
r2 →r2 −3r1
 3 −3 1 t2  −
−−→  3 −3 1 t2  −−
−−−−→  0 −3 −2 t2 − 3t  .
1 0 1
t
0 3 2 t+1
0 3
2
t+1
Replacing the third row by the sum

1

 0
0
of the second and the third rows, we get

0
1
t

-3 −2
t2 − 3t
.
2
0
0
t − 2t + 1
The system is inconsistent precisely when the last column is a pivot column, i.e., t2 −2t+1 6=
0. But
t2 − 2t + 1 = (t − 1)2 ,
and, therefore, the system is inconsistent if and only if t 6= 1.
NOTES
That the rank of the coefficient

0
 3
1
matrix

3 2
−3 1 
0 1
is 2 follows from the solution above. This means that this system is neither onto, nor
one-to-one. What are the solutions when t = 1 when the system is consistent?
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Problem 9.
Let A be the function from R3 to R3 given by the following formula
A(x1 , x2 , x3 ) = (−2x2 + x3 , 3x1 + x2 + x3 , x1 − x2 + x3 ).
Find a particular triple (b1 , b2 , b3 ) ∈ R3 such that there exists no triple (x1 , x2 , x3 )
with the property that A(x1 , x2 , x3 ) = (b1 , b2 , b3 ).
Solution We write down the matrix associated to A. In terms of this matrix, we are trying
to ensure that


 

0 −2 1
x1
b1
 3 1 1   x 2  =  b2 
1 −1 1
x3
b3
has no solution for some particular choices of

0 −2
 3 1
1 −1
b1 , b2 , b3 . Form the augmented matrix

1 b1
1 b2  .
1 b3
The reduction steps are as follows :



0 −2 1
b1
0 −2 1 b1
1
r2 →r1 + 2 (r2 −3r3 )
b2
 3 1 1 b2  −−

−−−−−−−−→
0 0 0 b1 + 2 −
1 −1 1 b3
1 −1 1
b3
Moving the last row to the top and
we get

1
 0
0

3b3
2
.
pushing the other rows down (this requires two swaps)

−1 1
b3
.
−2 1
b1
b2
3b3
0 0 b1 + 2 − 2
The system having no solutions is equivalent to being inconsistent, i.e.,
b1 +
b2 3b3
−
6= 0.
2
2
A particular triple is given by (0, 0, 1).
NOTES
There are other alternative ways to do this. We observe that
1
3
(−2x2 + x3 ) + (3x1 + x2 + x3 ) − (x1 − x2 + x3 ) = 0.
2
2
Therefore, any element (b1 , b2 , b3 ) in the image of this linear map must satisfy
b2 3b3
−
= 0.
2
2
Consequently, anything that doesn’t satisfy the above will do.
b1 +
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Problem 10.
Let B be the function from R3 to R2 defined by the following formula
B(x1 , x2 , x3 ) = (2x2 + x3 , x1 + 3x2 + x3 ).
Find two distinct triples (c1 , c2 , c3 ) and (d1 , d2 , d3 ) we have an equality B(c1 , c2 , c3 ) =
B(d1 , d2 , d3 ).
Solution Suppose we are trying to find triples such that
B(c1 , c2 , c3 ) = B(d1 , d2 , d3 ) = (b1 , b2 ).
We write down the augmented matrix associated to B :
0 2 1 b1
.
1 3 1 b2
The reduction steps are as follows :
0 2 1 b1
1
r1 ↔r2
−−−−−−
→
3
1 3 1 b2
0
r1 →r1 − 2 r2
0
2
− 12 b2 − 32 b1
1
b1
The third variable x3 is free and we have
x2 = b1 − x3 , x1 = b2 −
b1 x 3
+ .
3
2
Fixing b1 and b2 gives us freedom to choose two different values for x3 and get two distinct
triples while ensuring
B(c1 , c2 , c3 ) = B(d1 , d2 , d3 ) = (b1 , b2 ).
In particular, we may choose b1 = 0 = b2 and x3 = 2e (resp. x3 = 2π) to get triples (e, −e, 2e)
and (π, −π, 2π) respectively.
NOTES The map B has rank 2 and is, therefore, onto. So, any (b1 , b2 ) will work for the
argument above. This may not be the case in general. The zero vector is always in the image
and usually it’s a good idea to try to proceed with (b1 , b2 ) = (0, 0) as illustrated below.
There are other alternative ways to do this. Observe that B(0, 0, 0) = (0, 0). Therefore,
it suffices to find (d1 , d2 , d3 ) 6= (0, 0, 0) such that B(d1 , d2 , d3 ) = (0, 0). Writing down the
sugmented matrix and reducing it, we get
1 0 − 21 0
.
0 2 1 0
Therefore, d3 = t, d2 = − 2t and d1 = 2t . Taking t = 2 gives us (d1 , d2 , d3 ) = (1, −1, 2).
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