MTH 102: Linear Algebra

Department of Mathematics and Statistics Indian Institute of Technology - Kanpur

Problem Set 2

Problems marked (T) are for discussions in Tutorial sessions.

1. (T) A square matrix P is called a permutation matrix if each row and column of P contains
exactly one
 1 and the
 rest of the entries are 0. Determine all the 3×3 permutation matrices. Now,
0 0 6
let A = 1 2 3 . Write down the permutation matrix P such that P A is upper triangular.
0 4 5
Which permutation matrices P1 and P2 make P1 AP2 lower triangular?

Solution: To permute rows of A, left multiply by permutation matrix

   
0 1 0 1 2 3
P = 0 0 1 yielding P A = 0 4 5 .
1 0 0 0 0 6
   
0 0 1 6 0 0
Right multiplication by P2 = 0 1 0 permutes columns of A yielding AP2 = 3 2 1 .
1 0 0 5 4 0
Finally, rearrange rows of AP2 by left multiplying
   
1 0 0 6 0 0
P1 = 0 0 1 resulting is P1 AP2 = 5 4 0 .
0 1 0 3 2 1

2. (T) Decide if they are row-equivalent:

   
1 2 0 1
(a) ,
4 8 1 2

Solution:
No, first one is singular and the second one is nonsingular.
1 0 2 1 0 2
" # " #
(b) 3 −1 1 , 0 2 10
5 −1 5 2 0 4

Solution: Show that their RREFs(Row reduced echelon forms) are same.
       
1 0 2 1 0 2 1 0 2 1 0 2
R2 ↔R3 R ←R3 −R1 R2 ←−2R2
 3 −1 1  − −−−−→  5 −1 5  −−3−−−− −−→  0 −1 −5  −−− −−−−→  0 2 10  .
R2 ←R2 −5R1 R3 ←R3 −R2
5 −1 5 3 −1 1 2 −1 −1 2 0 4
2 1 −1
" #  
1 0 2
(c) 1 1 0 ,
0 2 10
4 3 −1
Solution:
No, two matrices must have the same size, in order to be row equivalent.
 2

3. Supply two examples each and explain their geometrical meaning.

(a) Two linear equations in two variables with exactly one solution.
Solution:o
x+y =2
. They represent two lines in R2 intersecting at a point.
x−y =0
(b) Two linear equations in two variables with infinitely many solutions.
Solution: o
x+y =2
. They represent the same line in R2 .
2x + 2y = 4
(c) Two linear equations in two variables with no solutions.
Solution: o
x+y =2
. They represent two parallel lines in R2 , no intersection.
2x + 2y = 1
(d) Three linear equations in two variables with exactly one solution.
Solution:
x+y =2 o
x − y = 0 . They represent three lines in R2 with a single point in common.
2x − y = 1
(e) Three linear equations in two variables with no solutions.
Solution:
x+y =2 o
x − y = 0 . They represent three lines in R2 with no point in common.
2x + 2y = 1
4. Suppose that x and y are two distinct solutions of the system Ax = b. Prove that there are
infinitely many solutions to this system, by showing that λx + (1 − λ)y is also a solution, for
each λ ∈ R. Do you have a geometric interpretation?
5. Give examples of three matrices of size 3 × 4 that are in Row Reduced Echelon Form (RREF)
but are different from the ones given in Question 6. Give examples of matrices that are not in
RREF, specifying the reasons.
1 ∗ ∗ 0 0 1 ∗ 0 0 0 1 0
" # " # " #
6. Is it possible to have RREF ([A|b]) = 0 0 0 1 , 0 0 0 1 or 0 0 0 1 ? Give reasons
0 0 0 0 0 0 0 0 0 0 0 0
for your answer.
7. Let B be a square invertible matrix. Then, prove that the system Ax = b and BAx = Bb are
row-equivalent.

Solution:
As B is invertible, there exists elementary matrices Ei ’s such that B = E1 E2 · · · Ek . Thus, the
system BAx = Bb is obtained from Ax = b by k elementary row operations.
Conversely, as B −1 = Ek−1 Ek−1
−1
· · · E1−1 and inverse of elementary matrices are also elementary
matrices, we do obtain Ax = b from BAx = Bb by k elementary row operations. Thus, the
above two systems are row equivalent.
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8. [T] Suppose Ax = b and Cx = b have same solutions for every b. Is it true that A = C?

Solution: First, as Ax = b and Cx = b have same solutions, A and C have same shapes, that
is, same number of rows and columns, as well as same null space (follows from taking b = 0).
Now, if we take b to be the first column of A then x = [ 1 0 . . . 0 ]t solves Ax = b and
therefore also solves Cx = b which in turn implies that the first columns of A and C are same.
Same argument holds for other columns as well. Thus A = C.

9. Find the coefficients a, b, c, d so that the graph of y = ax3 + bx2 + cx + d passes through
(1, 2), (−1, 6), (2, 3) and (0, 1).

Solution:
This is as good as solving a+b+c = 1, a−b+c = 5, 4a+2b+c = 1. Apply Gaussian elimination.

10. Find matrices A and B with the given property or explain why you can not:
 
1  
0
(a) The only solution to Ax = 2 is x =
  .
1
3
Solution: Any A satisfying the given equation has to be a 3 × 2 matrix. The linear system
has a unique solution when rank of A is 2 and [ 1 2 3 ]t ∈ {Ax : x ∈ R2 }. Among many
possibilities, one such A is  
1 1
0 2 .
0 3
 
  1
0
(b) The only solution to Bx = is x = 2.
1
3
Solution: Any B satisfying the given equation has to be a 2 × 3 matrix which implies that
the null space of B, N (B), can not be trivial and hence we either have an infinitely many
solutions (when [0 1]t lies in the column space of B) or no solution. Thus, finding a B that
exhibits a unique solution is not possible.

11. Apply Gauss elimination to solve the system 2x+y +2z = 3 3x−y +4z = 7 and 4x+3y +6z = 5.

Solution:
     
2 1 2 3 2 1 2 3 2 1 2 3
R2 ←R2 −(3/2)R1 R ←R −2R1
−−−−−−−−−−→  0 −5/2 1 5/2  −−3−−−3−−−→
 3 −1 4 7  −  0 −5/2 1 5/2 
4 3 6 5 4 3 6 5 0 1 2 −1
 
2 1 2 3
R3 ←R3 +(2/5)R2
−−−−−−−−−−−→  0 −5/2 1 5/2 
0 0 12/5 0
 4

We can thus obtain the solution to the given linear system by solving the equivalent system

2x + y + 2z = 3
(−5/2)y + z = 5/2
(12/5)z = 0

The solution is x = 2, y = −1 and z = 0.

 
1 2 2
12. (T) Using Gauss Jordan method, find the inverse of  2 1 2  .
2 2 1

Solution:
   
1 2 2 1 0 0 1 2 2 1 0 0
R2 ←R2 −2R1 R3 ←R3 −(2/3)R2
 2 1 2 0 1 0 −−−−−−−−→  0 −3 −2 −2 1 0  −−−−−−−−−−−→
R3 ←R3 −2R1
2 2 1 0 0 1 0 −2 −3 −2 0 1
   
1 2 2 1 0 0 1 2 2 1 0 0
R3 ←(−3/5)R3
 0 −3 −2 −2 1 0  −−−−−−−−−→  0 −3 −2 −2 1 0 
0 0 −5/3 −2/3 −2/3 1 0 0 1 2/5 2/5 −3/5
 
1 2 0 1/5 −4/5 6/5
R2 ←R2 +2R3 R2 ←(−1/3)R2
−−−−−−−−→ 0 −3 0 −6/5 9/5 −6/5  −−−−−−−−−→

R1 ←R1 −2R3
0 0 1 2/5 2/5 −3/5
   
1 2 0 1/5 −4/5 6/5 1 0 0 −3/5 2/5 2/5
R1 ←R1 −2R2
 0 1 0 2/5 −3/5 2/5  − −−−−−−−→  0 1 0 2/5 −3/5 2/5 
0 0 1 2/5 2/5 −3/5 0 0 1 2/5 2/5 −3/5
 
−3/5 2/5 2/5
Thus, the inverse is  2/5 −3/5 2/5 .
2/5 2/5 −3/5

13. (T) Let Bn×n be a real skew-symmetric matrix. Show that I − B is nonsingular.

Solution:
Let if possible I − B be singular. Then, the system (I − B)x = 0 has a non-trivial solution, say
x0 6= 0. Hence, Bx0 = x0 . Also, xt0 Bx0 ∈ R and hence

xt0 Bx0 = (xt0 Bx0 )t = xt0 B t x0 = −xt0 Bx0 .

Thus, xt0 Bx0 = 0. But, 0 = xt0 Bx0 = xt0 (Bx0 ) = xt0 x0 and hence x0 = 0.

14. (T) For two n × n matrices A and B, show that det(AB) = det(A)det(B).

Solution: First, suppose that A is singular. Then, AB is singular as well. We therefore have,
det(A) = 0 and det(AB) = 0 which leads to det(AB) = 0 = det(A)det(B).
 5

Now we assume that A is non-singular. Recall that, for a non-singular square matrix, the
reduced row echelon form is the identity matrix, I. In other words, there exist elementary
matrices Es , Es−1 , . . . , E2 , E1 such that

A = Es Es−1 . . . E2 E1 I.

We therefore have
AB = Es Es−1 . . . E2 E1 B.
Thus the problem of showing det(AB) = det(A)det(B) reduces to showing
(a) det(Eij B) = det(Eij )det(B).
(b) det(Ei (c)B) = det(Eij (c))det(B).
(c) det(Eij (c)B) = det(Eij (c))det(B).
Now, for the proof, we use the defining properties 1, 2 and 3 discussed in class. We have
(a) det(Eij B) = −det(B) (property 2) = det(Eij )detB (using det(Eij ) = −det(I) = −1, follows
from properties 2 and 1).
(b) det(Ei (c)B) = c det(B) (property 3a) = det(Ei (c))det(B) (using det(Ei (c)) = c det(I) = c,
follows from properties 3a and 1).
(c) det(Eij (c)B) = det(B) (property 6) = det(Eij (c))det(B) (using det(Eij (c)) = 1, follows
from property 6 on the identity matrix).
The result now follows

det(AB) = det(Es Es−1 . . . E2 E1 B) = det(Es )det(Es−1 . . . E2 E1 B) = · · ·

= det(Es )det(Es−1 ) . . . det(E2 )det(E1 B)
= det(Es )det(Es−1 ) . . . det(E2 )det(E1 )det(B)
= det(Es )det(Es−1 ) . . . det(E2 E1 )det(B) = · · ·
= det(Es Es−1 . . . E2 E1 )det(B) = det(A)det(B).

15. For an n × n matrix A = [aij ], prove that det(A) = det(AT ).

Solution:
Recall that, a square matrix can be reduced to an upper form using elementary row operations.
In other words, there exist elementary matrices Es , Es−1 , . . . , E2 , E1 such that

Es Es−1 . . . E2 E1 A = U.

where U is an upper triangular matrix and each Ek , 1 ≤ k ≤ s, is either an elementary matrix

Eij or Eij (c) for some i, j and c. We therefore have

EA = U

with det(E) = det(E T ) = ±1. We also have det(U ) = det(U T ) for the upper triangular matrix
U . Thus
det(U )
det(A) = ,
det(E)
 6

and
AT E T = U T
yields
det(U T ) det(U )
det(AT ) = = = det(A).
det(E T ) det(E)
16. Let A be an n × n matrix. Prove that

(a) If A2 = 0 then A is singular.

Solution: 0 = det(A2 ) = (det A)2 ⇒ det A = 0.
(b) If A2 = A, A 6= I then A is singular.
Solution: If A is nonsingular, then it is invertible. Then A = A−1 A2 = A−1 A = I.

17. Consider the system Ax = b. Let RREF ([A|b]) be one of the matrices in Question 6. Now,
recall the matrices Aj ’s, for 1 ≤ j ≤ 3 (defined to state the Cramer’s rule for solving a linear
system), that are obtained by replacing the j-th column of A by b. Then, we see that the above
system has NO solution even though det(A) = 0 = det(Aj ), for 1 ≤ j ≤ 3.

18. (T) Let A be an invertible square matrix with integer entries. Show that A−1 has integer entries
if and only if det(A) = ±1.

Solution: If det(A) = ±1, then entries of A−1 are integers as A−1 = det(A)
1
C T , where C is
the matrix of co-factors. If entries of A−1 are integers then 1/det(A) = det(A−1 ) ∈ Z. As
det(A) ∈ Z, the result follows.

19. Let A be an n × n matrix. Prove that the following statements are equivalent:

(a) det(A) 6= 0.
(b) A is invertible.
(c) The homogeneous system Ax = 0 has only the trivial solution.
(d) The row-reduced echelon form of A is In .
(e) A is a product of elementary matrices.
(f) The system Ax = b has a unique solution for every .
¯
(g) The system Ax = b is consistent for every b.

19a =⇒ 19b
Ct
By, definition, whenever det(A) 6= 0, A−1 = , where C is the co-factor matrix.
det(A)
19b =⇒ 19a
As A is invertible, AA−1 = In and hence det(A) det(A−1 ) = det(In ) = 1. Hence, det(A) 6= 0.
19b =⇒ 19c
As A is invertible, A−1 A = In . Let x0 be a solution of the homogeneous system Ax = 0. Then,

x0 = In x0 = (A−1 A)x0 = A−1 (Ax0 ) = A−1 0 = 0.

 7

Thus, 0 is the only solution of the homogeneous system Ax = 0.

19c =⇒ 19d
Let xt = [x1 , x2 , . . . , xn ]. As 0 is the only solution of the linear system Ax = 0, the final equations
are x1 = 0, x2 = 0, . . . , xn = 0. These equations can be rewritten as

1 · x1 + 0 · x2 + 0 · x3 + · · · + 0 · xn = 0
0 · x1 + 1 · x2 + 0 · x3 + · · · + 0 · xn = 0
0 · x1 + 0 · x2 + 1 · x3 + · · · + 0 · xn = 0
.. .
. = ..
0 · x1 + 0 · x2 + 0 · x3 + · · · + 1 · xn = 0.

That is, the final system of homogeneous system is given by In · x = 0. Or equivalently, the
row-reduced echelon form of the augmented matrix [A 0] is [In 0]. That is, the row-reduced
echelon form of A is In .
19d =⇒ 19e
Suppose that the row-reduced echelon form of A is In . Then there exist elementary matrices
E1 , E2 , . . . , Ek such that
E1 E2 · · · Ek A = In . (1)
Now, the matrix Ej−1 is an elementary matrix and is the inverse of Ej for 1 ≤ j ≤ k. Therefore,
successively multiplying Equation (1) on the left by E1−1 , E2−1 , . . . , Ek−1 , we get
−1
A = Ek−1 Ek−1 · · · E2−1 E1−1

and thus A is a product of elementary matrices.

19e =⇒ 19b
Suppose A = E1 E2 · · · Ek ; where the Ei ’s are elementary matrices. As the elementary matrices
are invertible and the product of invertible matrices is also invertible, we get the required result.
19b =⇒ 19f
Observe that x0 = A−1 b is the unique solution of the system Ax = b.
19f =⇒ 19g
The system Ax = b has a solution and hence by definition, the system is consistent.
19g =⇒ 19b
For 1 ≤ i ≤ n, define ei = (0, . . . , 0, 1
|{z} , 0, . . . , 0)t , and consider the linear system
ith position
Ax = ei . By assumption, this system has a solution, say xi , for each i, 1 ≤ i ≤ n. Define a
matrix B = [x1 , x2 , . . . , xn ]. That is, the ith column of B is the solution of the system Ax = ei .
Then
AB = A[x1 , x2 , . . . , xn ] = [Ax1 , Ax2 , . . . , Axn ] = [e1 , e2 , . . . , en ] = In .
Therefore, the matrix A is invertible.
 8

20. Let A be an n × n matrix. Then prove that det(A) = 0 if and only if the system Ax = 0 has a
non-trivial solution.

21. Let A be an n × n matrix. Then, the two statements given below cannot hold together.

(a) The system Ax = b has a unique solution for every b.

(b) The system Ax = 0 has a non-trivial solution.

22. (T) A real square matrix A is said to be orthogonal if AT A = AAT = I. Show that if A is
orthogonal then det(A) = ±1.

Solution: Since AAT = I, we have det(AAT ) = det(I) = 1. Now,

1 = det(AAT ) = det(A)det(AT ) = det(A)det(A) = (det(A))2 .

Thus, det(A) = ±1.

23. Let A = [aij ] be an invertible matrix and let B = [pi−j aij ], for some p 6= 0. Find the inverse of
B and also find det(B).

Solution: Note that B = DAD−1 , where D is a diagonal matrix with dii = pi . Hence,
B −1 = DA−1 D−1 . Also,

det(B) = det(DAD−1 ) = det(D) · det(A) · det(D−1 ) = det(A).

24. Suppose the 4 × 4 matrix M has 4 equal rows all containing a, b, c, d. We know that det(M ) = 0.
The problem is to find by any method
 
 1+a b c d 
 
 a 1 + b c d 
det(I + M ) =  .
 a b 1+c d 

 a b c 1+d 

Solution: Subtracting row 1 from rows 2, 3 and 4, we get

 
 1+a b c d 
 
 −1 1 0 0 
det(I + M ) =  .
 −1 0 1 0 

 −1 0 0 1 

Now, adding columns 2, 3 and 4 to column 1, we get

 
 1+a+c+d a b d 
 
 0 1 0 0 
det(I + M ) =  .
 0 0 1 0 

 0 0 0 1 

Thus, det(I + M ) = 1 + a + b + c + d.
 9

25. (T) For a complex matrix A = [aij ], let Ā = [ aij ] and A∗ = ĀT . Show that det(Ā) = det(A∗ ) =
detA . Therefore if A is Hermitian (that is, A∗ = A) then its determinant is real.

Solution: The first equality follows directly from the statement in Problem 4:
 
T
det(Ā) = det (Ā) = det(A∗ ).

The equality det(Ā) = detA follows from the determinant formula:

X X
det(Ā) = (sgn σ) a1σ(1) a2σ(2) · · · anσ(n) = (sgn σ)a1σ(1) a2σ(2) · · · anσ(n) = det(A).
σ∈Sn σ∈Sn

26. The numbers 1375, 1287, 4191 and 5731 are all divisible by 11. Prove that 11 also divides the
determinant of the matrix  
1 1 4 5
 3 2 1 7 
 .
 7 8 9 3 
5 7 1 1

Solution: Adding 1000 × Row1 , 100 × Row2 , 10 × Row3 to Row4 , we have

   
 1 1 4 5   1 1 4 5 
  
 3 2 1 7   3 2 1 7 
 7 8 9 3 = 7 .
  
   8 9 3 
 5 7 1 1   1375 1287 4191 5731 

Since Row4 is divisible by 11, the determinant is divisible by 11.

x1 x21 x31 x41
 
1
 1
 x2 x22 x32 x42 
27. Compute determinant of   1 x3 x23 x33 x43  .
 1 2 3
x4 x4 x4 x4 4

1 x5 x25 x35 x45 .

Solution: We give the solution for the general case. Let

1 x1 x21 · · · x1n−1
 
 1 x2 x22 · · · xn−1 
 2 
An =  . . . . . . . 

 . . . ... . 
1 xn x2n · · · xnn−1 .

If n = 2, det(A2 ) = x2 − x1 . We will prove that

Y
det(An ) = (xj − xi ).
i<j
 10

Assume the result for n − 1 and define

1 x1 x21 · · · xn−1
 
1
 
1 x2 x22 · · · xn−1
 
 
 2 
F (x) =  . . . ... . .


 . . . ... . 

 2
1 x x ··· x n−1 

n−1
Q
Then F is a polynomial of degree n − 1 with roots x1 , x2 , . . . , xn−1 . So, F (x) = c (x − xi )
i=1
where c is coefficient of xn−1 which is clearly det(An−1 ). Therefore,
n−1
Y
F (x) = det(An−1 ) (x − xi ).
i=1

The result follows for n as

n−1
Y
det(An ) = F (xn ) = det(An−1 ) (xn − xi ).
i=1