This document provides 18 problems related to matrices. The problems cover topics like finding eigen values and vectors, properties of eigen values under operations like inverse and powers, and applying Cayley-Hamilton theorem.

1. Mathematics – I ( MA 2111 )
Unit I : MATRICES
Part – A
A
1. If λ is an eigen value of a non singular matrix A, Show that is an eigen
λ
value of adj A.
Let λ be an eigen value of A and X be the corresponding eigen vector.
Then, AX = λ X .
1
By property, is an eigen value of A-1.
λ
We have adj A= A-1 A
∴ Eigen value of adj A = Eigen value of A-1 A
1 A
= A = .
λ λ
2. Let λ be an eigen value of a singular matrix A with the eigen vector X. Show
–1
that 1/ λ is an eigen value of A with eigen vector X.
Proof: Let λ be an eigen value and X be an eigen vector of A.
∴ (A – λ I) x = 0 ⇒ A x – λ I x = 0
∴ A x = λ I x ⇒ AX = λ X
–1
Premultiplying both sides by A
–1 –1 –1 –1
A (AX) = A ( λ X) ⇒ (A A) X = λ (A X)
–1
IX = λ (A X)
–1 –1
X = λ (A X) ⇒ λ (A X) = X
–1
∴ A X = X/ λ
–1
Hence, 1/ λ is an eigen value of A .
3
3. If λ 1 , λ 2, λ 3,…, λ n,are the eigen values of an n x n matrix A, then show that λ 1 ,
3 3 3 3
λ 2 , λ 3 ,…, λ n are the eigen values of A .
Let λ be an eigen value of A and X be the corresponding eigen vector.
Then, AX = λ X ( λ = λ 1, λ 2, λ 3, … λ n)
∴A2 X = (AA) X
= A (AX)
2
= A ( λ X) = λ (AX) = λ ( λ X) = λ X
3 3
Similarly, A X = λ X
3 3
∴ λ is an eigen value of A
 2 −3
4. Find the sum and product of the eigen values of the matrix  
 4 −2 
The sum of the eigen values = Sum of the diagonal elements = 2 - 2 = 0
The product of the eigen values = Determinant value of the matrix = 2(-2) + 12 = 8.
 8 −6 2 
5. If 3 and 15 are the two eigen values of A = − 6 7 − 4 , find A , without
 
 2 −4 3 
 

2. expanding the determinant.
Let λ 1 = 3, λ 2 = 15.
Sum of the eigen values = trace of A. i.e., λ 1 + λ 2 + λ3 = 8 + 7 + 3
⇒ 3 + 15 + λ3 = 18.
⇒ λ3 = 0.
⇒ A = 0. (since product of the eigen values = A )
2 0 1
 
6. Find the sum and product of the eigen values of the matrix  0 2 0 
 1 0 2
 
The sum of the eigen values = Sum of the diagonal elements = 2 + 2 + 2 = 8
The product of the eigen values
2 0 1
= Determinant values of the matrix = 0 2 0 = 2(4 – 0) – 0 (0 – 0) + 1 (0 –
1 0 2
2) = 8 – 2 = 6
 1 3
7. Find the eigen values of  .
 2 1
The characteristic equation of A is given by A − λI = 0.
1-λ 3
i.e., = 0 ⇒ (1- λ) (1- λ) – 6 = 0.
2 1-λ
⇒ λ2 – 2λ – 5 = 0.
2 ± 4 + 20
⇒ λ= = 1 ± √6.
2
∴ The eigen values of A are 1 + √6 and 1 - √6.
4 6 6
8. Two eigen values of A =  1 3 2  are equal and they are double the

− 1 − 5 − 2
 
third.
Find the eigen values of A2 and A-1.
Let the third eigen value be λ3 = λ.
Given that two eigen values are equal and they are double the third i.e., λ1 = λ2 =
2λ.
Sum of the eigen values = trace of A.
∴2λ + 2λ + λ = 4 + 3 - 2 = 5.
⇒ 5 λ = 5 i.e., λ =1.
∴ Eigen values of A are 2, 2, 1.
1 1
Hence the eigen values of A2 are 4, 4, 1 and the eigen values of A-1 are ,
2 2
and 1.

3.  2 5 −1
9. Find the eigen values of A if the matrix A is  0 3 2 
–1
 
0 0 4 
 
–1
If λ 1 , λ 2, λ 3 are the eigen values of A then the eigen values of the matrix A are
1 1 1
, , .
λ1 λ 2 λ3
First we have to find the eigen values of A.
Since the given matrix is a triangular matrix the eigen values of A are the diagonal
elements 2, 3, 4.
–1 1 1 1
∴The eigen values of A are , , .
2 3 4
a 4 
10. Find the constants a and b such that the matrix   has 3 and -2 as its eigen
1 b 
values.
a 4 
Let A =  
1 b 
Since sum of the eigen values = sum of the diagonal matrix,
a+b=3–2=1 ….. (1)
Since product of eigen values is determinant value of matrix,
ab – 4 = 3 (–2) = – 6
∴ ab = –2 …………. (2)
solving (1) and (2), we get the values of a and b
∴a=1–b
(2) ⇒ (1 – b) b = –2
b2 – b – 2 = 0
b = – 1, b = 2
∴ (1) ⇒ a + b = 1 ⇒ a + 2 = 1 ⇒ a = – 1
∴ a = –1, b = 2
1 1 4
11. If   is an eigen vector of
   3 2  , find the corresponding eigen – value.
 
1  
1
Let λ be an eigen value corresponding to the eigen vector X =   .
1
 
 1 4  1 1
Then AX = λX implies   3 2  1 = λ
   
1
    
⇒ 1 + 4 = λ and 3 + 2 = 5.
∴ λ = 5.
 1 1 3
12. If X = ( −1 0 1) T is the eigen vector of the matrix A =  1 5 1  find the
 
3 1 1
 
corresponding eigen value.

4. 0
 
The eigen vector of the matrix A is given by ( A − λI ) X =  0 
0
 
1 1 3  -1
   
Here A =  1 5 1  and X =  0 
 3 1 1 1
   
1 − λ 1 3 
 
A – λI =  1 5−λ 1 
 3 1 
1− λ

1 − λ 1 3   -1 0
     
∴ 1 5−λ 1   0  = 0
 3 1 1− λ  1 
  0
   
(1 – λ)(–1) + 1 (0) + 3 (1) = 0 ⇒ – 1 + λ + 0 + 3 = 0 ⇒ λ + 2 = 0
∴λ = – 2 is the corresponding eigen value.
 6 −2 2 
 
13. The product of two eigen values of the matrix A =  −2 3 −1 is 16. Find the third
 2 −1 3 
 
eigen value.
6 −2 2
The product of all eigen values = −2 3 −1 = 6(9 – 1) + 2(–6 + 2) +2(2 – 6) =32
2 −1 3
Product of two eigen values = 16
The third eigen value = 32 / 16 =2
2 2 1
 
14. Two eigen values of the matrix A =  1 3 1  are equal to 1 each. Find the
1 2 2
 
-1
eigen values of A .
Sum of the eigen values = 2 + 3 + 2 = 7 ( Sum of the diagonal elements)
Sum of two given eigen values = 1 + 1 = 2
∴The 3rd eigen value = 7 – 2 = 5
The eigen values of A are 1, 1, 5
–1
∴The eigen value of A are 1, 1, 1/5.
1 1 
15. If the Eigen values of the matrix A =   are 2 and -2, find the eigen values of
3 −1
T
A .
T
Eigenvalues of A = Eigenvalues of A
T
∴Eigenvalues of A are 2, –2.
1 2 3 
3  
16. Find the Eigen values of A given A =  0 2 −7 
0 0 3 
 

5. 1 2 3 
 
Given A =  0 2 −7 
0 0 3 
 
Given A is an upper triangular matrix
Hence the eigen values are 1, 2, 3
i.e., the eigen values of the given matrix A are 1, 2, 3.
By the property, the eigenvalues of the matrix A3 are 13, 23, 33.
 3 −1
17. Verify Cayley-Hamilton theorem for the matrix A =  
 −1 5 
3 − λ −1
| A − λI | =
−1 5 − λ
2
= (3 – λ )(5 – λ ) – 1 = λ – 8 λ + 14
2
The characteristic equation is |A – λ I| = λ – 8 λ + 14 = 0
2
By Cayley-Hamilton theorem, we have A – 8 A + 14 I = 0
 3 −1  3 −1  10 −8 
A2 =    = 
 −1 5   −1 5   −8 26 
2  10 −8   3 −1 1 0 0 0
A – 8 A + 14 I =   − 8  + 14  = 
 −8 26   −1 5  0 1 0 0
Hence Cayley Hamilton theorem is verified.
1 4
18. Using Cayley-Hamilton theorem find the inverse of  
2 3
1 4
Let A =  
2 3
1− λ 4
| A − λI |=
2 3−λ
2
= λ – 4λ – 5
2
By Cayley Hamilton theorem, A – 4A – 5I = 0
–1 –1 2 –1 –1
Multiplying by A , A A – 4A A – 5A I = 0
⇒ A – 4I – 5A–1 = 0
–1
5A = A – 4I
1 4  1 0  1 4   4 0   −3 4 
=   - 4  0 1  =  2 3  -  0 4  =  2 −1
2 3        
1 1 −3 4 
∴ A– =
5  2 −1
 
1 0 
19. If A =  , express A3 in terms of A and I using Cayley-Hamilton theorem.
 4 5

The Cha. Equation of the given matrix is A - λI = 0
 1 0  1 0 1-λ 0
(i.e.,)   -λ  =0 ⇒ =0
 4 5 0 1 4 5−λ
(1 – λ)(5 – λ) – 0 = 0
(1 – λ)(5 – λ) =0

6. 5 – λ – 5λ + λ2 = 0
λ2 – 6λ + 5 =0
By Cayley Hamilton theorem
[Every square matrix A satisfies its own cha. equation]
(i.e.,) A2 – 6A + 5I = 0, A2 = 6A – 5I
multiply A on both sides
A3 – 6A2 + 5A = 0
A3 = 6A2 – 5A
= 6 [6A – 5I] – 5A
= 36A – 30I – 5A
= 31A – 30I
1 4 5 4 3
20. Use Cayley – Hamilton theorem for the matrix A =   to express A – 4A – 7A
2 3
+
2
11A – A – 10I as a linear polynomial in A.
1 4
Given A =  
2 3
The characteristic equation of A is A - λI = 0
i.e., λ2 – S1 λ + S2 = 0 where
S1 = sum of the main diagonal elements = 1 + 3 = 4
1 4
S2 = A = =3–8=–5
2 3
∴ The characteristic equation of A is λ2 – 4λ – 5 = 0
By Cayley – Hamilton theorem, we get
A2 – 4A – 5I = 0 ……….. (1)
To find : A5 – 4A4 – 7A3 + 11A2 – A – 10I
∴ A5 – 4A4 – 7A3 + 11A2 – A – 10I
= (A2 – 4A – 5I)(A3 – 2A + 3I) + A + 5I
= 0 + A + 5I = A + 5I by (1)
which is a linear polynomial in A.
–1
21. If A is an orthogonal matrix show that A is also orthogonal.
For an orthogonal matrix, transpose will be the inverse
T –1
∴A =A … (1)
T –1
Let A = A = B … (2)
T –1 T T –1 –1
Then B = (A ) =(A ) = B from (2)
T –1
∴ B = B ⇒ matrix B is orthogonal
–1
i.e., A is also orthogonal.
Hence proved.
22. If A is an orthogonal matrix prove that | A | = ±1
Proof: Let A be the orthogonal matrix. Let λ be an eigen value.
By defn. of orthogonal matrix, AA′ = A′A = I.
Also AX = λ X … (1) where x is eigen vector.
Taking transpose on both sides.
(AX) ′ = ( λ X) ′
X′A′ = λ X′ ….(2) since [AB] ′ = B′A′

7. Multiplying (2) and (1)
(X′A′)(AX) = ( λ X′)( λ X)
X′ (A′A) X = λ 2 (X′X)
X′IX = λ 2 (X′X)
X′X = λ 2 (X′X)
2
(X′X) – λ (X′X) = 0
(X′X) (1 – λ 2) = 0
∴1 – λ 2 = 0 [X′X ≠ 0]
λ 2 = 1; λ = ±1 Hence the proof.
23. Determine the nature of the following quadratic form f(x1, x2, x3)= x12 + 2x22
Here λ 1 = 1, λ 2 = 2, λ 3 = 0. Two of the eigen values are positive.
∴The quadratic form is semi positive definite.
2 2
24. Find the nature of the quadratic form 8x – 4xy + 5y .
The matrix corresponding to the quadratic form
 8 −2 
8x2 – 4xy + 5y2 is  
 −2 5 
The eigen values are given by
8-λ − 2
= 0 ⇒ (8 – λ )(5 – λ ) – 4 = 0 ⇒ 40 – 8 λ – 8 λ + λ 2 – 4 = 0
−2 5-λ
λ 2 – 13 λ + 36 = 0 ⇒ ( λ – 9)( λ – 4) = 0
∴ λ = 9, λ = 4
∴The nature of the quadratic is positive definite.
25. Find the matrix form for the quadratic form x12 + 2x1x2 + 6x22.
1 1 
The matrix of the Quadratic form is A =  .
1 6 
a b
26. Find the condition for a real matrix 
  to be positive definite.

b c
a b
Here D1= a , D2 = where D1 and D2 are the principal subdeterminants of the
b c
given matrix. For the matrix to be positive definite, a > 0 and ac - b2 >0
i.e., ac > b2.
27. If A is the symmetric matrix corresponding to a quadratic form in three
variables such that trace (A) = 0, find the nature of the quadratic form.
We have, sum of the eigen values = trace (A) = 0.
⇒ λ 1+ λ 2+ λ 3 = 0.
⇒ λ 1, λ 2, λ 3 cannot be all positive or negative. Atleast one positive or one
negative. Therefore the nature of the quadratic form is indefinite.
 6 1 − 7
 
28. Write the quadratic form for the matrix  1 2 0  .
− 7 0 1 
 

8. x
T  
The Quadratic form is given by X AX, where X =  y  .
z
 
 6 1 − 7  x 
T
   
Now X AX = ( x y z )  1 2 0   y 
− 7 0 1   z 

  
= 6x2 + 2y2 + z2 + 2xy -14xz.
29. Find the nature of the quadratic form 2x2 + 3y2 + 2z2 + 2xy
 2 1 0
 
The matrix of the Quadratic form is A =  1 3 0 
 0 0 2
 
D1 = 2, D2 = 6 – 1 =5, D3 = 12 – 2 =10.
D1> 0 , D2 >0 , D3>0.
∴The nature of the quadratic form is positive definite.
----------------------------------
ASSIGNMENT QUESTIONS
 2 2 1
 
1. Find the eigen values and eigen vectors of A =  1 3 1  .
 1 2 2
 
(8 marks) (May ’09)(May
’09)
 4 1 1
 
2. Find the eigen values and the eigen vectors of the matrix  1 4 1 
 1 1 4
 
(8 marks) (May
’09)
3. Find the eigen values and corresponding eigen vectors of the matrix
 1 1 3
 
1 5 1 .
 3 1 1
 
(8 marks)(May/June
2009)
 11 −4 −7 
 
4. Find all the eigen values and eigen vectors of the matrix A =  7 −2 −5 
10 −4 −6 
 
(8 marks)(May/June
2007)

9.  2 0 −1
 
5. Verify Cayley Hamilton theorem for the matrix A =  0 2 0  and hence
 −1 0 2 
 
find
A-1 and A4 . (8 marks) (May ’09&Nov/Dec
2008)
 2 −1 
6. Using Cayley-Hamilton theorem find A-1 and A3 + A6 , if A =  .
 5 −2 
(8 marks) (May
’09)
7. Using Cayley-Hamilton theorem evaluate the matrix A8- 5A7 + 7A6 - 3A5 + A4 –
 2 1 1
3 2  
5A - 8A + 2A –I if A =  0 1 0  (6 marks) (May ’06) (May
 1 1 2
 
2009)
 2 −1 1 
5  
8. Verify Cayley-Hamilton theorem and find A for the matrix  −1 2 −1
 1 −1 2 
 
(8 marks)(May/June
2009)
 1 0 0
 
9. If A =  1 0 1  . Then show that An = An-2+A2-I for n ≥ 3 using
 0 1 0
 
Cayley-Hamilton theorem. (6 marks)(Jan 2006)
 2 1 1
 
10. Find the characteristic equation of A =  0 1 0  and hence express the
 1 1 2
 
matrix
A5 in terms of A2, A and I. (6 marks ) (April/May
2005)

10.  2 1 −1 
 
11. Diagonalise the matrix A =  1 1 −2  by means of orthogonal
 −1 −2 1 
 
transformation.
(12 marks) (May ’09)
 6 −2 2 
 
12. Diagonalize A =  −2 3 −1 by an orthogonal transformation.
 2 −1 3 
 
(10 marks)(Jan 2006)
13. Reduce the given quadratic form Q to its canonical form using orthogonal
transformation Q = x 2 + 3 y 2 + 3z 2 − 2 yz (16 marks) (Jan 2009)
14. Reduce the quadratic form 2xy + 2yz +2zx to the canonical form by an
orthogonal reduction and state its nature.
Reduce 2 x1 x2 + 2 x2 x3 + 2 x3 x1 to canonical form by an orthogonal
transformation.
(10 marks) (May ’09, Jan 2006 & April/May
2005)
15. Reduce the quadratic form x2 + 2y2 + z2 - 2xy +2yz to a canonical form by
orthogonal reduction . (8 marks) (May
’08)
Reduce the quadratic form x12 + 2 x2 + x3 − 2 x1 x2 + 2 x2 x3 to the canonical form
2 2
through an orthogonal transformation and hence show that it is positive semi-
definite. Give also a non-zero set of values ( x1, x2 , x3 ) which makes this
quadratic
form to zero. (16 marks)(May/June
2009)
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