UNIT I
MATRICES
Problem 1. Find the eigen values and eigen vectors of the matrix 2 2 3 A 2 1 6 1 2 0 Solution: The characteristic equation is | A - I | = 0.
2 -
2
2
1 -
3 6
1
2
0 -
i.e.,
0
i.e., (-2 - ) [-(1 - ) -12] - 2[-2 - 6] -3[-4 + 1 - ] = 0 2 i.e., (-2 - ) [ - -12] + 4 + 12 + 9 + 3 = 0 3 2 i.e., + - 21 - 45 = 0 3 2 Now, (-3) + (-3) - 21(-3) - 45 = -27 + 9 + 63 – 45 45 = 0 -3 is a root of equation (1). 3 2 Dividing + - 21 - 45 by + 3
3 1 0
1
3
21 45 6
45
1 2 15 0 Remaining roots are given by 2 - 2 - 15 = 0 i.e., ( + 3) ( - 5) = 0 i.e., = -3, 5. The eigen values are -3, -3, 5
2 λ 2 - 3 x1 0 2 1 - λ - 6 x2 0 The eigen vectors of A are given by - 1 0 - 2 - λ x3 Case 1
Now
=
-3 2 3
2 - 1
1 1 3 - 6 ~ 2 - 1 -2 3 1 ~ 0 0 2
- 3
x1 + 2x2 - 3x3 = 0
2
- 3
4
-6
- 2 3 2 - 3 0 0 0 0
(1)
2
Matrices
Put Then
x2 = k 1, x3 = k 2 x1 = 3k 2 - 2k 1
3k 2 2k 1 k The general eigen vectors corresponding to = -3 is 1 k 2 3 When k 1 = 0, k 2 = 1, we get the eigen vector 0 1 2 When k 1 = 1, k 2 = 0, we get the eigen vector 1 0 3 2 Hence the two eigen vectors corresponding to = -3 are 0 and 1 . 1 0 These two eigen vectors corresponding to = -3 are linearly independent. Case 2 = 5 2 5 2 - 3 7 2 - 3
2 - 1
- 5
2 -4 - 1 - 2 1 2 ~ 0 8 0 0
1- 5 - 6 ~ -2
- 5 5 16 0 -6
-x1 - 2x2 - 5x3 = 0 -8x2 - 16x3 = 0 A solution is x3 = 1, x2 = -2, x1 = -1
1 Eigen vector corresponding to = 5 is 2 . 1
1 1 2 Problem 2. Find the characteristic equation of 2 1 3 and verify Cayley 3 2 3 Hamilton Theorem. Hence find the inverse of the matrix.
3
Matrices
1 1 2 Solution: Let A 2 1 3 Characteristic eqn. of A is 3 2 3 3 2 1 1 3 9 9 1 26 0 i.e 3 2 19 26 0 By Cayley-Hamilton theorem A3 A2 19 A 26I 0 . Verification:
1 1 2 1 1 A2 A.A 2 1 3 2 1 3 2 3 3 2 2 7 1 9 9 10 2 A3 A2 . A 5 10 7 21 3
2
2 9 3 5 9 3 10 7 1 2 16 1 3 43 2 3 67
Substituting in the characteristic equation 2 7 19 16 21 45 9
43 16 67 5 9 10 38 67 45 104 10 7 21 57
19 19 38
7 10 21 21 45 16 67 45 104 38 26
0 57 0 57
0 0 0 26 0 0 0 0 0 26 0 0 0 0
0
Hence verified. Now to find the inverse of the matrix A, premultiply the characteristic equation by A1 A2 A 19 I 26 A1 0
A1
1
19I A A 26 2
19 0 0 1 1 0 19 0 2 26 0 0 19 3
1 1 2
2 7 9 5 5 1 3 5 9 10 3 9 7 26 3 10 7 21 7 5 1 2 9
1 0 3 Problem 3. Given A 2 1 1 , use Cayley-Hamilton Theorem to find the inverse of A 1 1 1 4
and also find A
Solution: The characteristic equation of A is 1 λ 0 3 2
1 λ
1
1
1 0 1 λ
i.e., (1-) [(1 - ) (1 - ) -1] + 3[-2 - (1 - )] = 0
4
Matrices
3
i.e., (1 - ) - (1 - ) – 6 -3 + 3 = 0 2 3 i.e., 1 - 3 + 3 - – 1 + - 9 + 3 = 0 3 2 i.e., - + 3 + - 9 = 0 3 2 i.e., - 3 - + 9 = 0 3 2 By Cayley-Hamilton theorem, A -3A – A + 9I = 0 -1 -1 2 -1 To find A , multiplying by A , A -3A - I + 9A = 0
-1
A =
1 9
1 0 A2 2 1 1 1 4 1 1 A 3 9 0 0 1 3 9 3
3
2
[-A + 3A + I]
4 3 6 1 2 1 1 3 2 4 1 1 1 1 0 2 5 3 6 3 0 9 1 0 0 2 4 6 3 3 0 1 0 2 5 3 3 3 0 0 1 3 1
0
3
3
7 1 1 2
4
To find A : 3 2 We have A - 3A – A + 9I = 0 3 2 i.e., A = 3A + A - 9I Multiplying (1) by A, we get, 4 3 2 A = 3A + A -9A 2 2 = 3(3A + A - 9I) + A - 9A 2 = 10A - 6A - 27I 4 3 6 1 0
(1)
using (1) 3
1 0 0 10 3 2 4 62 1 1 27 0 1 0 0 2 5 1 1 1 0 0 1 7 30 42 18 13 46 6 14 17
0 0 2 Problem 4. . If A 2 1 0 express A6 25 A2 122 A as a single matrix 1 1 3 6 Solution: To avoid higher powers of A like A we use Cayley Hamilton Theorem. Characteristic equation is 3 4 2 5 2 0 By Cayley Hamilton Theorem A3 4 A2 5 A 2I 0 To find A6 25 A2 122 A we will express this in terms of smaller powers of A using the characteristics equation. We know that (Divisor) X (Quotient) + Remainder = Dividend Assuming A3 4 A2 5 A 2I as the divisor we get,
5
Matrices
A3 4 A2 11A 22I A 4 A 5 A 2 I 3
2
A6 0 A5 0 A4 25 A2 122 A 0 I A6 4 A5 5 A4 2 A3
4 A5 5 A4 2 A3 25 A2 122 A 4 A5 16 A4 20 A3 8 A2 11 A4 22 A3 33 A2 122 A 11 A4 44 A3 55 A2 22 A 22 A3 88 A2 100 A 22 A3 88 A2 110 A 44I 10 A 44 I
A6 25 A2 122 A A3 4 A2 5 A 2I A3 4 A2 11A 22I 10 A 44I But A3 4 A2 5 A 2I 0 A6 25 A2 122 A 0 10 A 44I
10 A 44I 0 0 20 44 0 0 20 10 0 0 44 0 10 10 20 0 0 44 0 20 44 20 54 0 10 10 74 44 0 20 20 54 0 10 10 74
Problem 5. If i are the eigen values of the matrix A, then prove that i k i are the eigen values of kA where ‘k’ is a nonzero scalar. m
i
ii. iii.
1 i
are the eigen value of Am and are the eigen values of A1 .
Solution: Let i be the eigen values of matrix A and Xi be the corresponding eigen vectors. Then by defn: AXi iXi......(I ) ( i.e by defn. of eigen vectors) i. Premultiply ( I ) with the scalar k. Then
k AXi k iXi i.e. kA X i k i Xi
k i are the eigen values of kA (comparing with ( I ) i.e by defn.)
6
Matrices
ii. Premultiply ( I ) with A, then A AXi A iXi i.e. A2 X i i AXi
i i Xi from (I) 2
i Xi 3
m
1 y we can prove that A3 Xi i Xi and so on Am Xi i Xi i m are the eigen values of the Am (comparing with ( I ) i.e. by defn.)
iii. Premultiply ( I ) with A1 , then 1
A
AXi A1 iXi
i.e. A1 A Xi i A1 Xi i.e. IXi i A1 Xi i.e. A1 Xi
1 i
1 i
Xi
are the eigen values of A1 (comparing with ( I ) ).
2 0 1 Problem 6. Find the characteristic vectors of 0 2 0 and verify that they are 1 0 2 mutually orthogonal.
2 0 1 Solution: A = 0 2 0 Characteristic equation is 1 0 2 Solving: 1,2,3 Consider the matrix equation A I X 0 Case (i) when
3
6 2 11 6 0
1;
1x1 0 x2 1x3 0 1 1 0 1 x1 0 0 1 0 x 0 i.e. 0 x 1x 0 x 0 2 1 2 3 2 1 0 1 x 0 1x1 0 x2 1x3 0 3 3
equation (1) & (3) are identical.
Solving (1) and (2) using the rule of cross multiplication 1 x1 x2 x3 x1 x2 x3 i.e. X 1 0 0 1 0 1 0 1 1 0 1 1 Case (ii) when
2;
7
Matrices
0 x1 0 x2 1x3 0 1 0 1 x1 0 0 1 0 x 0 i.e. 0 x 0 x 0 x 0 1 2 3 2 1 0 1 x 0 1x1 0 x2 0 x3 0 3
x3 0 i.e. x2 is arbitrary say k
0 0 X 2 k i.e 1 . 0 0 Case (ii) when 3; x1 0 x2 1x3 0 1 0 1 x1 0 0 1 0 x 0 i.e. 0 x 1x 0 x 0 1 2 3 2 1 0 1 x 0 1x1 0 x2 1x3 0 3
x1 0
Solving (1) and (2)
1 X 3 0 1 0 1 1
x1
x2
x3
Thus the eigen values are 1,2,3 and the correspondent eigen vectors are 1 1 0
0 , 1 and 1 0 X 2T X 3 0
0 . 1
To check orthogonallity, X1T X 2 0
X1T X 3 0
X 1 , X 2 , X 3 are mutually orthogonal.
6 6 5 Problem 7. Find the latent vectors of 14 13 10 7 6 4 Solution: Characteristic equation is
3 1 0 1, 1, 1
When 1 (repeated 3 times) we have to find 3 corresponding latent vectors. 7 x1 6 x2 5 x3 0 7 6 5 x1 0
14 12 10 x 0 i.e. 14 x 12 x 10 x 0 1 2 3 2 7 6 5 x 0 7 x1 6 x2 5 x3 0 3
All three equation are identical
.i.e. we get only one equation, but we have to find three vectors that are linearly independent. 0 x2 x3 Assume x1 0 6 x2 5x3 0 i.e. 6 x2 5x3 i.e. X 1 5 5 6 6
8
Matrices
5 Assume x2 0 7 x2 5 x3 0 i.e. 7 x1 5x3i.e.. X 2 0 5 7 7 6 x1 x2 And assume x2 0 7 x2 6 x3 0 i.e. 7 x1 6 x2 0i.e.. X 3 7 6 7 0 x1
x3
X1, X2 and X3 are linearly independent.
1 1 1 Problem 8. Find the eigen vectors of the matrix A 0 2 1 4 4 3 Solution: 1 1 - 1 The characteristic equation of A is 0 2 - 1 0 4 4 3 -
i.e., (1 - ) [(2 - ) (3 - ) - 4] -1[0 + 4] +1[0 + 4(2 - )] = 0 2 i.e., (1 - )( - 5 + 6 - 4) – 4 + 8 - 4 = 0 2 i.e., (1 - )( - 5 + 2) + 4 - 4 = 0 2 i.e., (1 - )( - 5 + 2 + 4) = 0 2 i.e., ( -1)( - 5 + 6) = 0 i.e., ( -1)( - 2)( - 3) = 0 The eigen values of A are = 1, 2, 3. 1 x 1 0 1 - λ 1 0 2 - λ 1 x 0 The eigen vectors are given by 2 4 4 3 - λ x 3 0 Case 1 =
1
0 1 1 4 4 2 0 1 1 ~ 0 1 1 4 4 2 0 0 0 -4x1 + 4x2 + 2x3 = 0 x2 + x3 = 0 A solution is, x3 = 2, x2 = -2, x1= -1 1 Eigen vector X1 = 2 2
9
Matrices
Case 2 =
2
1 1 1 1 1 1 0 0 1 ~ 0 0 1 4 4 1 0 0 0 -x1 + x2 + x3 = 0 x3 = 0 A solution is, x3 = 0, x2 = 1, x1 = 1 1 Eigen vector X2 = 1 0 Case 3 =
3 1
2 1 2 1 1 0 1 1 ~ 0 1 1 4 4 0 0 0 0 -2x1 + x2 + x3 = 0 -x2 + x3 = 0 A solution is, x3 = 1, x2 = 1, x1 = 1
1 Eigen vector X3 = 1 1
2 2 0 Problem 9. Diagonalise the matrix 2 5 0 using orthogonal transformation. 0 0 3 3 2 Solution: Characteristic equation is 10 27 18 0 Solving we get the eigen value as 1,3,6 2 0 1 When 1, X 1 1 ; When 3, X 2 0 ; When 6, X 3 2 0 1 0 2 1 5 0 5 , 0 and 2 Normalizing each vector, we get 1 5 5 1 0 0
10
Matrices
2 0 5 0 Normalized Modal Matrix, N 1 5 1 0
2 5 5 T 2 . N N 0 5 1 0 5 1
1
0
0 1, 2 0 5 5
Then by the orthogonal transformation,
2 5 N AN 0 1 5
2 0 5 2 2 0 5 0 1 2 5 0 0 0 2 0 0 0 3 1 1 5 5 1
0
5 2 . On simplifying, we get 5 0 1
N AN D 1 , 2 , 3
1 0 0 which is diagonal matrix with eigen values along the D 1,3, 6 0 3 0 0 0 6 diagonal (in order).
6 2 2 Problem 10. Reduce 2 3 1 to a diagonal matrix by orthogonal reduction. 2 1 3 3 2 Solution: Characteristic equation is 12 36 32 0 8, 2, 2 When 8 2 2 2 x1 0 2 5 1 x 0 2 2 1 5 x 0 3 i.e
2 x1 2 x2 2 x3 0 2 x1 5x2 1x3 0 2 x1 1x2 5 x3 0
2 Solving any two equations X 1 1 2 1 1 1 When 2 (repeated twice) 4 2 2 x1 0 2 1 1 x 0 i.e 2 x 2 x 2 x 0 . All the equations are identical. 1 2 3 2 2 1 1 x 0 3 x1
x2
x3
11
Matrices
0 To get one of the vectors, assume x1 0 x2 x3 0 i.e. X 2 1 1 1 1 a X 1T X 2 0 . Therefore X 1 and X 2 are orthogonal. Now assume X 3 b to be mutually c x2
x3
orthogonal with X1 and X2.
a X 1T X 3 0 i.e. 2 1 1 b 0 i.e.2a b c 0 c a b c i.e a 2 2 2 and X 2T X 3 0 i.e. 0 1 1 b 0 i.e.0a b c 0 c 1 X 3 1 . 1 After normalizing these 3 mutually orthogonal vectors, we get the normalized Modal
2 6 Matrix N 1 6 1 6
3 1 1 2 3 1 1 2 3 0
1
Diagonalizing we get
2 2 1 1 1 1 6 6 6 6 2 2 6 6 3 T 1 1 1 1 1 D N AN 0 2 3 1 2 2 6 2 3 1 2 1 3 1 1 1 1 1 3 3 3 6 3 3 on simplifying we get D D 1 , 2 , 3 8 0 0 2 0 0 D 8,
0
0 2 2,
2
12
Matrices
3 1 1 Problem 11. Diagonalise the matrix A 1 3 -1 1 -1 3 Solution:
1 3- 1 The characteristic equation of A is 1 3- -1 0 1 -1 3- 2
i.e., (-1)( - 8 + 16) = 0 The eigen values of A are = 1, 4, 4.
1 x1 0 3-λ 1 1 3-λ -1 x 0 The eigen vectors are given by 2 1 -1 3-λ x 3 0
Case 1 =
1
1 Eigen vector X1 = 1 1 Case 2 =
4
0 Eigen vector X2 = 1 1 a Now assume X 3 b to be mutually orthogonal with X1 and X2. c X 1T X 3 0 i.e. a b c 0 a b c i e . 2 1 1 and X 2T X 3 0 i.e. b c 0
2 X 3 1 . 1 1 0 2 Hence the modal matrix M 1 1 1 1 1 1
13
Matrices
1 0 3 1 The Normalized Modal Matrix is N 1 3 2 1 1 3 2
6 1 6 1 6
2
Diagonalizing, we get
1 1 3 3 1 D N T AN 0 2 2 1 6 6
1 0 3 3 1 3 1 1 1 1 3 1 1 2 3 2 1 1 3 1 1 1 6 3 2 1
6 1 6 1 6
2
1 0 0 0 4 0 = D(1, 4, 4) 0 0 4
Problem 12. Reduce the Quadratic From 10 x12 2 x22 5 x32 6 x2 x3 10 x3 x1 4 x1 x2 into canonical form by orthogonal reduction. Hence find the nature, rank, index and the signature of the Q.F. Find also a nonzero set of values of X which will make the Q.F. vanish.
10 2 5 , which is a real and symmetric Solution: Matrix of the given Q.F. is A 2 2 3 5 3 5 3 2 matrix. The characteristic equation is 17 42 0 Solving, we get 0, 3, 14 1 1 3 When 0, X 1 5 ; When 3, X 2 1 ; When 14, X 3 1 4 1 2 T T T and X1 , X 2 , X 3 are mutually orthogonal since X1 , X 2 0, X 2 X 3 0 andX3 X 1 0 Normalizing these vectors we get the normalized model matrix
1 42 N 5 42 4 42
14 1 14 2 14
3
1 3 1 3 1 3
14
Matrices
Diagonalising we get D N T AN D 12 , 3 in order
D 0, 3, 14 0 0 0 D 0 3 0 (i.e. the eigen values in order along the principal 0 0 14
i.e
diagonal). Now to reduce the Q.F to C.F (.i.e Canonical form)
y1 Consider the orthogonal transformation X = NY where Y y2 y 3 T
Then the Q.F. X T AX becomes NY A NY Y T N T AN Y = Y T DY since N T AN D 0 0 0 y1
y1 y2 y3 0 3 0 y2 0 0 14 y 3 0 y1 3 y2 14 y3 2
2
2
Thus = 0 y1 3 y2 14 y3 is the Canonical form of the given Q.F. And the equations of 2
2
2
this transformation are got from X= NY.
1 42 x1 x NY 5 2 42 x 3 4 42 x1
1 42 5
x2 x3
y1
42 4 42
1 1
y1
y1
3 1 3
3 1 3 1
y2
3
3 3 14
y3
3
y2
y2
14 y1 1 y2 14 y3 2 14
3
1
14 3 14
y3
y3
To get the non-zero set of values of x which make the Q.F zero we assume values for y1 , y2 and y3 such that the C.F. vanishes.
15
Matrices
i.e 0 y1 3 y2 14 y3 will vanish if y2 0, y3 0 and y1 is any arbitrary value (for 2
2
2
simplicity sake, assume y1 as the denominator of the coeff. of y1 in the equations) let
y1 42 1
42
x1 i.e.
42
1 3
3
0
14
(0)
x1 1 0 0 1
III 1 y x2 5 0 0 5 and x3 4 0 0 4 Thus the set of values of x
i.e 1,
5, 4 will reduce the given Q.F. to zero.
To find the rank, index, signature and nature using canonical form: C.F. is 0 y1 3 y2 14 y3 2
2
2
rank is 2 (no. of terms in C.F) Index is 2 (no. of positive terms) Signature of Q.F. = ( no. of positive terms) – (no. of negative terms) = 2 Nature of the Q.F. is positive semi definite.
Problem 13. Reduce the Q.F. 2 xy 2 yz 2 zx into a form of sum of squares. Find the rank, index and signature of it. Find also the nature of the Q.F. 0 1 1
0 1 1 1 0 Characteristic equation is 3 3 2 0 solving 2, 1, 1 Solution: Matrix of the Q.F. is A 1
1 When 2, X 1 1 1 When 1 (repeated twice) we get identical equations as x1 x2 x3 0 x1 0 x2 x3 0 i.e. x2 x3 i.e. Assume
x2
1
x3 1
0 X 2 1 1
which is orthogonal with X 1.
a Now to find X 3 orthogonal with both X1 and X 2 assume X 3 b c
16
Matrices
if X 2T X 3 0, a b c 0
if X 2T X 3 0, i.e.
a
b
0a b c 0 c
1 1 2 2 X 3 1 i.e. 1 1 1 2
which is orthogonal with X1 and X 2 .
1 0 3 3 2 6 1 1 1 Normalising these vectors we get N and D N AN 3 2 6 1 1 2 3 2 6 2 0 0 = D 1 , 2 , 3 0 1 0 .Consider the orthonormal transformation X = NY 0 0 1 such that Q.F.is reduced to C.F. The Q.F. is reduced as T
X T AX NY A NY
Y T N T AN Y Y T DY 2 0 0 y1 y1 , y2 , y3 , 0 1 0 y2 0 0 1 y 3
The C.F. is 2 y12 y2 2 y32 rank of Q.F.is = no. of terms in C.F=3 index of Q.F. = no. of positive terms in C.F. = 1 signature of Q.F. = ( no. of positive terms) – (no. of negative terms) = 1-2 = -1 Nature of the Q.F. is indefinite.
Problem 14. Reduce the quadratic form 8 x12 7 x22 3 x32 12 x1 x2 4 x1 x3 8 x2 x3 to the canonical form by an orthogonal transformation. Find also the rank, index, signature and the nature of the quadratic form.
17
Matrices
Solution:
The matrix of the quadratic form is
8 6 2 A 6 7 4 2 4 3
The eigen values of this matrix are 0, 3 and 15 and the corresponding eigen vectors are 2 1 2
2
X1 2 ,
X2
1
1 , 2
X 3 2 , which are mutually orthogonal.
2/3 1/3 2/3 The normalized modal matrix is N 2/3 1/3 2/3 2/3 2/3 1/3 0 0 0 T and N AN = D 0 3 0 0 0 15
Now the orthogonal transformation X = NY will reduce the given quadratic form to the canonical form 0y12 3y 22 15y32 . Also rank = 2, index = 2, signature = 2. The quadratic form is positive semi definite.
Problem 15. Find the orthogonal transformation which reduces the quadratic form 2 x12 2 x22 2 x32 2 x1 x 2 2 x 2 x3 2 x1 x3 into the canonical form. Determine the rank, index, signature and the nature of the quadratic form.
Solution:
2 1 1 The matrix of the quadratic form is A 1 2 1 1 1 2 The characteristic equation of A is 3
2 -
-1
-1
2 -
1
-1
1 -1 0 2 -
2
Expanding - 6 + 9 - 4 = 0 = 1 is a root 3 2 Dividing - 6 + 9 - 4 by -1, 1 6 9 4 0
1
1 5
5
4
4
0 2
The remaining roots are given by -5 + 4 = 0 2 - 5 + 4 = ( - 1) ( - 4) = 0 i.e., = 1, 4
18
Matrices
The
eigen values of A are = 4, 1, 1
Case 1 =
4
1 x 1 0 2 - 4 - 1 x 0 The eigen vectors are given by - 1 2 - 4 - 1 2 1 - 1 2 - 4 x 3 0 2 1 1 1 - 1 - 2 1 2 1 ~ 0 - 3 - 3 1 1 2 0 0 0 x1 - x2 - 2x3 = 0 -3x2 - 3x3 = 0 A solution is x3 = 1, x2 = -1, x1 = 1.
1 The corresponding eigen vector is X1 = 1 1 Case 2 = 1 1 x1 0 2 - 1 - 1 The eigen vectors are given by - 1 2 - 1 - 1 x 2 0 1 - 1 2 - 1 x3 0
1 - 1 1 1 1 1 - 1 1 - 1 ~ 0 0 0 1 - 1 1 0 0 0 x1- x2 + x3 = 0 Put x3 = 0. We get x1 = x2 = 1. Let x1 = x2 = 1
1 The eigen vector corresponding to = 1 is X 2 = 1 0 X1 and X2 are orthogonal as X 1T X 2 = 10 + (-1) 1 + 11 = 0. a To find another vector X3 = b corresponding to =1 such that it is orthogonal to both c X1 and X2 and satisfies x1- x2 + x3 = 0 i.e., X1.X3 = 0, X2.X3 = 0 and a – b + c = 0 i.e., 1.a -1.b + 1.c = 0, 1.a + 1.b + 0.c = 0 and a – b + c = 0. i.e., a – b + c = 0 and a+b=0 i.e., a = -b and c = 2b Put b =1, so that a = -1, c = 2
19
Matrices
1 X3 1 2
1 1 1 The modal matrix is 1 1 1 1 0 2 1/ 3 1/ 2 1/ 6 Hence the normalized modal matrix is N 1 / 3 1 / 2 1 / 6 1/ 3 0 2 / 6 The required orthogonal transformation is X = NY will reduce the given quadratic form to the canonical form. C.F= 4 y12 y 22 y32 Rank of the quadratic form = 3, index = 3, signature = 3. The quadratic form is positive definite.