PART A
−1 1
1
1.Find the sum and product of the Eigen values of the matrix A= [ 1 −1
1] .
1
1 −1
(Anna University March 1996)
Solution:
We know that,
Sum of the Eigen values = sum of the principal diagonal elements
= -1 -1 – 1
= -3
Product of the Eigen values = | A |
= - 1 (1 – 1) – 1 (-1 -1) + 1 (1 + 1)
=0+2+2
=4
Hence, Sum= -3 and Product =4.
6 −2
2
2. The product of two Eigen values of the matrix A = [−2
3 −1] is 16. Find the third eigen
2 −1
3
value.
(Anna University Jan-2012,June2003,Apr/May 2009)
Solution:
Let Eigen values be 1, 2, 3
Given 1 . 2 = 16
We know that,
Product of the Eigen values = | A |
i.e., 1 .3 = | A |
1 2 3 = 6 (9 – 1) + 2 (-6 + 2) +2 (2 – 6)
1 2 3 = 48 – 8 – 8
163 = 32
32
3 = 16
3 = 2
Hence,the third Eigen value is 2.
8 −6 2
3. Two Eigen values of the matrix A = [−6 7 −4] are 3 and 0. What is the third eigen value?
2 −4
3
What is the product of the Eigen values of A?
Solution:
Given 1 = 3, 2 = 0
To find 3:
We Know that,
Sum of Eigen values = sum of the main diagonal elements
1 +  2 + 3 = 8 + 7 + 3
3 + 0 + 3 = 18
3 = 18 – 3
3 = 15
Product of Eigen values = 1. 2. 3
= 3  0  15
Product of Eigen values = 0
Hence the third eigen value is 15 and Product is 0.
7 4
−4
4. One of the Eigen values of [4 −8 −1] is -9. Find the other two eigen values.
4 −1 −8
Solution:
Let the Eigen values 1, 2, 3
Given 1 = -9
We Know that,
Sum of the Eigen values = sum of the principal diagonal elements
1 +  2 + 3 = 7 – 8 – 8
-9 + 2 + 3 = -9
2 + 3 = 0
3 = -2 ..........................(1)
We Know that,
Product of the Eigen values = | A |
1 23 = 7 (64 - 1) – 4 (-32 + 4) - 4(-4 + 32)
-9 2 3 = 7 (63) – 4 (-28) - 4(28)
7 ×63
2 3 = −9
2 3 = -49 ......................(2)
From (1) and (2)
2 (-2) = -49
−22 = -49
2 = 7
Therefore the other two eigen values are 7, -7.
3 10
5. If 2, 2, 3 are the Eigen values of A = [−2 −3
3
5
Solution:
We Know that,
Eigen values of A = Eigen values of AT
Therefore Eigen values of AT = 2, 2, 3.
5
−4 ] find the eigen values of AT.
7
6. Prove that, a square matrix A and its transpose AT have the same Eigen values. (or)
Prove that, a square matrix A and its transpose AT have the same characteristic values.
Solution:
Let A be a square matrix of order ‘n’
The characteristic equation of A and AT are
|A - I| = 0 ................(1)
And |AT - I| = 0 .................(2)
Since the determinant value is unaltered by the interchange of rows and columns and |A| = |AT|
Therefore, (1) and (2) are identical
Hence, Eigen values of A and AT are the same.
2
7. Find the sum and product of the Eigen values of [0
0
Solution:
1 0
2 1].
0 2
Given matrix is upper triangular matrix.
We Know that,
Eigen values = Main diagonal element
Therefore, Eigen values are 2, 2, 2
Sum of Eigen values
=2+2+2=6
Product of eigen values = 2  2  2 = 8.
8. Prove that if  is an Eigen value of matrix A, then 1/ is the Eigen value of A -1
(or)
If  is an Eigen value of matrix A, what can you say about Eigen value of matrix A-1.
Prove your statement.
(Anna University-May /June 2012)
Solution:
If X be the Eigen vector corresponding to 
 AX = X ....................(1)
Premultiplying both sides by A -1
A -1 AX = A -1 X
IX =  A -1X
X =  A -1X
𝑋

1

It gives
1

= A -1X
𝑋 =A-1X
is an Eigen values of inverse matrix A -1.
3 −1 1
9. Two of the Eigen values of A=[−1 5 −1] are 3 and 6. Find the Eigen values of A-1.
1 −1 3
(Anna University Jan/Feb 2010,Jan 2011)
Solution:
Let Eigen values are 1, 2, 3. Given 1 = 3, 2 = 6, 3 =?
We Know that,
Sum of Eigen values = sum of the main diagonal elements
1 + 2 + 3 = 3 + 5 + 3
3 + 6 + 3 = 11
3 = 11 – 9
3 = 2
1
1
1
1
2
3
Eigen values of A-1 =  ,  , 
1 1 1
= 3,6,2 .
2
10. Find the Eigen values of A-1, if the matrix A is [0
0
5
3
0
−1
2].
4
Solution:
Given matrix is upper triangular matrix.
We Know that,
Eigen values = main diagonal element
= 2, 3, 4
Eigen values of A-1 =
=
1
1
1
, ,
1 2 3
1 1 1
, , .
2 3 4
3
11. If X = [ ] is the eigen vector of matrix A, find the eigen vector of A-1.
2
Solution:
We Know that,
Eigen Vector of the matrix A and A-1 are identical.
3
A-1 is also having the same Eigen vector [ ].
2
12. If  is an Eigen value of an orthogonal matrix then, prove that
1

is also it’s Eigen value.
(Anna University Nov/Dec 2003)
Solution:
We Know that,
A square matrix said to be orthogonal if AAT= ATA=I
i.e AT = A-1
Let A be an orthogonal matrix
Given  is an Eigen value of A
1

is an Eigen value of A-1
Since AT = A-1
1

is an Eigen value of AT
But the matrices A and AT have the same Eigen values. Since the determinants |A - I| and
| AT - I| are the same.
1
Hence  is also an Eigen value of A.
1 0
) be diagonalized? Why?
0 1
13.Can A= (
(Anna University Jan 2010)
Solution:
The Matrix A can be diagonalized, because the eigen values of A are 1,1.
When  =1, the system of equation (A-I)X has infinite number of solution,
i.e., we can find distinct eigen vectors for  =1,1.
14. If 1 and 2 are the Eigen values of a 22 matrix A, What are the Eigen values of A2 and A-1?
(Anna University May/June 2010)
Solution:
Given, Eigen Values of A = 1 and 2
We Know that,
If the eigen values of A are 1, 2 , 3 , then the eigen values of A2 are 12, 22 , 32 and the
eigen values of A-1 are
𝟏
,
𝟏
𝟏
, .
 𝟏  𝟐 𝟑
Therefore, Eigen Values of A2 = 1 and 4
1
Eigen Values of A-1 = 1 and 2.
15. If sum of two Eigen values of 33 matrix A are equal to the trace of the matrix, then find the
value of determinant of A.
Solution:
(Anna University May/June 2009)
We Know that,
Sum of two Eigen values = Sum of diagonal elements
1+2 = sum of diagonal elements
3=0
1 23 = 0
|A|=0
Hence, the value of the determinant is zero.
16. If the Eigen values of A are 2, 3, 4.Find the Eigen values of A-1 and Adj A.
Solution:
Given Eigen values of A are 2, 3, 4
We know that,
Eigen values of adj A =Eigen values of
Here
|𝑨|

|A|=234 =24
Eigen values of A-1 = 2 , 3,
1 1
1
Eigen values of adj A =
|𝑨|
4

1
1
1
= 242, 243, 244
Eigen values of adj A = 12, 8, 6.
17. State Cayley-Hamilton Theorem.
(Anna University May/June 2010,Jan/Feb 2010,June/July2008)
Statement:
Every square matrix satisfies its own characteristic equation.
18. Define Quadratic form.
Solution:
A homogeneous polynomial of the second degree in any number of variables is called a Quadratic
form
Example: 3x2+2xy+7y2 and 𝑥12 + 2𝑥22 + 4𝑥32 +6 x1x2 + 7 x1x3 + 5 x2x3 are Quadratic forms in 2 and 3
respectively.
19. Show that the Quadratic form 3𝑥12 + 3𝑥22 + 3𝑥32 +62x1x2 - 2 x2x3 is positive definite.
Solution:
The matrix of the Q. F is
𝑐𝑜. 𝑒𝑓𝑓 𝑥 2
Q=
1
2
1
𝑐𝑜. 𝑒𝑓𝑓 𝑥𝑦
𝑐𝑜. 𝑒𝑓𝑓 𝑦 2
𝑐𝑜. 𝑒𝑓𝑓 𝑦𝑥
1
[2 𝑐𝑜. 𝑒𝑓𝑓 𝑧𝑥
2
𝑐𝑜. 𝑒𝑓𝑓 𝑧𝑦
1
2
1
2
𝑐𝑜. 𝑒𝑓𝑓 𝑥𝑧
𝑐𝑜. 𝑒𝑓𝑓 𝑦𝑧
𝑐𝑜. 𝑒𝑓𝑓 𝑧 2 ]
3
1
1
Q= [1
3
− 1]
1
−1
3
|𝐷1 | = 3
1
|𝐷2 | =|3
|= 8
1
3
3
|𝐷3 | =|1
1
1
3
−1
1
− 1| = 3(9-1)-1(3+1)+(-1-3)
3
= 24-4-4 = 16
Here 𝐷1, 𝐷2, 𝐷3 are all positive and 𝐷𝑛 > 0 for all n.
 Q.F is positive definite
20. Write down the Application Of Cayley Hamilton Theorem.
Solution:
We can find
1. Higher integral positive power of the given square matrix
2. Inverse of the given square matrix.