Download Machine Learning Essentials FOR BEGGINERS and more Exercises Data Analysis & Statistical Methods in PDF only on Docsity!
Solutions 11(a) 1
Complete solutions to Exercise 11(a)
- (a)
⎛ ⎞ ⎛ −^ ⎞ ⎛ +^ − ⎞ ⎛
⎝ −^ ⎠ ⎝ ⎠ ⎝ +^ − + ⎠ ⎝
A B ⎞⎟
(b) ( )
⎛ −^ ⎞ ⎛ ⎞ ⎛^ +^ − + ⎞ ⎛
B A
(Generally A + B = B + A ).
(c)
+ + = = ⎛^ ⎞^ =⎛
A A A A ⎞⎟
(d)
⎛ ⎞ ⎛ −^ ⎞ ⎛ ⎞ ⎛ − ⎞ ⎛
⎝ −^ ⎠ ⎝ ⎠ ⎝ − ⎠ ⎝ ⎠ ⎝
A B
(e) Cannot be evaluated because A and C are different size matrices. Similarly (f) cannot be evaluated.
(g)
⎛ ⎞⎛ − ⎞ ⎛^ ⎡⎣^ × −^ ⎤⎦+^ × ⎞ ⎛
⎝ −^ ⎠⎝ ⎠ ⎝ ⎡⎣^ × −^ ⎤⎦+ − × ⎠ ⎝−
AC
(h)
⎛ −^ ⎞⎛ −^ ⎞ ⎛−
BC
(i) ; cannot be evaluated because of the
different size matrices involved.
⎛ ⎞ ⎛^ −
⎝ −^ ⎠ ⎝−
A BC
(j)
− = ⎛^ ⎞^ − ⎛^ −^ ⎞^ = ⎛^ ⎞^ − ⎛^ − ⎞^ = ⎛^ − − ⎞=⎛
⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜^ − − − ⎟ ⎜−
AC BC^7
- Use (11.1) to find the determinants.
(a) ( ) ( )
det 1 7 5 3 8 5 7
⎛ ⎞ = × − × = −
det 1 7 3 5 8 3 7
⎛ ⎞ = × − × = −
(b) ( ) ( )
det 1 3 5 2 13 5 3
⎜ ⎟=^ − ×^ −^ ×^ = −
det 1 3 2 5 13 2 3
⎜ ⎟=^ − ×^ −^ ×^ = −
(c) , det
⎛ −^ −^ ⎞ ⎛ − ⎞
A B (^) A = det B =
The matrix A is transposed (rows→ columns) to give matrix B. The same numbers on each of the diagonals, so the determinant is the same, det A = det B.
- By (11.1)
det det det a b a c a b ad cb ad bc ad cb c d b d c d
- Use (11.1) to evaluate the determinants.
(a) (i) det A = ( 1 × 6 ) − ( 5 × 3 ) = − 9 (ii) det B = ( 3 × 5 ) − (− 1 × 7 ) = 22
(iii) det A × det B = − 9 × 22 = − 198
(11.1) det^
a b ad cb c d
Solutions 11(a) 2
(iv)
= ⎛^ ⎞⎛^ ⎞^ =⎛
AB ⎞⎟
det ( AB ) = ( 0 × 65 ) − ( 9 × 22 )= − 198
(b) (i) det^ A = − ×(^ 1 1.5^ )^ −^ (^10 ×^170 )= −1701.
(ii) det^ B =^ ⎡⎣^ −^30 × −(^ 1.9^ )^ ⎤⎦− −[^ 9.3^ ×^61 ]^ =624.
(iii) det^ A^ ×^ det^ B = −1701.5^ ×^ 624.3^ = −1062246.
(iv)
⎛ −^ ⎞⎛ −^ −^ ⎞ ⎛ −
⎝ ⎠⎝ −^ ⎠ ⎝−^ −
AB ⎞⎟
det ( AB ) = ⎡⎣ 640 × −( 1583.85 ) ⎤⎦ − ⎡⎣ ( −9.7 ) × −( 5008.5 )⎤⎦= −1062246.
(c) (i) det^ A =^ ⎡⎣^5 × −(^ 5.6^ )^ ⎤⎦^ −^ ⎡⎣^ 2.2^ × −(^3 )^ ⎤⎦= −21.
(ii) det^ B =^ ⎡⎣^ −7.1^ × −(^ 12.2^ )^ ⎤⎦^ −^ ⎡⎣^ −3.5^ × −(^ 2.1^ )^ ⎤⎦=79.
(iii) det^ A^ ×^ det^ B = −21.4^ ×^ 79.27^ = −1696.
(iv)
⎛ −^ ⎞⎛ −^ −^ ⎞ ⎛−
⎝ −^ ⎠⎝ −^ − ⎠ ⎝
AB ⎞⎟
det ( AB ) = ( − 25 × 63.7) − ( 3.98 × 26.1) = −1696.
All the results satisfy det A det B = det ( AB ).
- In each case det A = 1 so we use (11.3). (a) Exchanging numbers 3 and 9 and placing a negative sign in front of the other numbers gives: 1 3 2 13 9
− = ⎛⎜^ − ⎞⎟
A
(b) 1
− = ⎛⎜^ − ⎞⎟
A
(c) In this case A is the identity matrix, and so.
− = = ⎛⎜^ ⎞⎟
I I
- We first evaluate the determinant.
(a) By (11.1), ( ) ( ). So using (11.4)
det 5 4 5 1 3 4 7 3 1
⎛ ⎞ = × − × = −
1 1 1 4 1 7^ 4 7
− = − ⎛^ −^ ⎞^ =⎛−
A ⎜⎝ − ⎟⎠ ⎜⎝ −
(b) We have ( ) ( ). Thus applying (11.4)
det 3 6 3 8 7 6 24 42 18 7 8
⎛ ⎞ = × − × = − = −
1 1 3 6 1 8 6 8 18^ 6 18^ 4 9^ 1 3 7 8 18 7 3 7 18 3 18 7 18 1 6
− − (^) = ⎛^ ⎞^ = − ⎛^ −^ ⎞^ = ⎛^ −^ ⎞^ =⎛− A (^) ⎜⎝ ⎟⎠ ⎜⎝ (^) − ⎟⎠ ⎜⎝ (^) − ⎟⎠ ⎜⎝
(c) det A = ( 7 × 2 )− ( 14 × 1 ) = 0. Since det A = 0 , so A cannot have an inverse.
(11.1) det A = ( a × d )− ( b × c )
a b^11 d b c d ad bc c a
⎛ ⎞ − ⎛^ −
Solutions 11(b) 2
- In each case det A = 1 so we use (11.3). (a) Exchanging numbers 3 and 9 and placing a negative sign in front of the other numbers gives: 1 3 2 13 9
− = ⎛^ − ⎞
A ⎜⎝ − ⎟⎠
(b) 1
− = ⎛^ − ⎞
A ⎜⎝ − ⎟⎠
(c) In this case A is the identity matrix, and so.
− = = ⎛^ ⎞
I I ⎜⎝ ⎟⎠
- We first evaluate the determinant.
(a) By (11.1), ( ) ( ). So using (11.4)
det 5 1 3 4 7 3 1
⎛ ⎞ = × − × = −
1 1 1 4 1 7^ 4 7
− = − ⎛⎜^ −^ ⎞⎟^ =⎛⎜−
⎝ −^ ⎠ ⎝ −
A ⎞⎟
(b) We have ( ) ( ). Thus applying (11.4)
det 3 8 7 6 24 42 18 7 8
⎛ ⎞ = × − × = − = −
1 1 3 6 1 8 6 8 18^ 6 18^ 4 9^ 1 3 7 8 18 7 3 7 18 3 18 7 18 1 6
− − (^) = ⎛^ ⎞^ = − ⎛^ −^ ⎞^ = ⎛^ −^ ⎞^ =⎛− ⎜ ⎟ ⎜ (^) − ⎟ ⎜ (^) − ⎟ ⎜ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝
A ⎞⎟
(c) det A = ( 7 × 2 )− ( 14 × 1 ) = 0. Since det A = 0 , so A cannot have an inverse.
- Putting the equations into matrix form gives 1 2
i i
We need to find the inverse of (^) ⎟. By (11.4)
=^ ⎛^ ⎞
A
⎛ ⎞^ − ⎛ −^ ⎞ ⎛ − ⎞
⎜ ⎟ =^ ⎜ ⎟ = ⎜ ⎟
⎝ ⎠ ×^ −^ ×^ ⎝ −^ ⎠ − ⎝ − ⎠
Using (11.5) we have 1 2
i i
⎛ ⎞ ⎛ −^ ⎞⎛ ⎞ ⎛ ⎞ ⎛−
⎜ ⎟ = −^ ⎜ ⎟⎜ ⎟ = −^ ⎜ ⎟ =⎜
⎝ ⎠ ⎝ −^ ⎠⎝ ⎠ ⎝ − ⎠ ⎝
i 1 (^) = −1.25 A and i 2 =2.5 A
(11.1) det A = ( a × d )− ( b × c )
1 a b (^) 1 d b c d (^) ad bc c a
(11.5) u = A −^1 b
Solutions 11(b) 3
- These equations can be written as 30 i 1 − 10 i 2 = 12 − 10 i 1 + 35 i 2 = 5 In matrix form we have 1 2
i i
⎛ −^ ⎞ ⎛ ⎞ ⎛
Let A =
⎠. Then by (11.4)
−^ −
⎜⎝ − ⎟⎠ × − × ⎜⎝ ⎟⎠ ⎜⎝
Using (11.5) gives 1 2
i i
1 2
(^470 47) A and 270 27 A 950 95 950 95
i = = i = =
- Similar to solutions 6 and 7. (a) x = 2 and y = 4 (b) x = −1 and y = 1
(c) x = 1 4 and y = − 1 3
(11.1) det A = ( a × d )− ( b × c )
a b^11 d b c d ad bc c a
⎛ ⎞ − ⎛^ −
(11.5) u = A −^1 b
Solutions 11(c) 2
- We have
N (^ )^ (^ (^ ))^ (^ )^ (^ )
[ ]
by (11.1) 2 2 2
det 5 7 1det 0 3det 9 1 3 9 3 9 1 1 9 7 3 5 9 3
63 3 45 3 63 135 9 10 198
t t t t t t t t
t t t t t t t t
⎜ − ⎟= ⎛^ ⎞^ − + ⎛^ ⎞
⎜ ⎟ ⎜^ − ⎟^ ⎜^ ⎟
⎜ − ⎟ ⎝^ ⎠^ ⎝^ ⎠
= ⎡⎣^ − − × − ⎤⎦+ ⎡⎣ × − × ⎤⎦
Since the determinant is zero we have t^2 − 10 t + 198 = 0 How do we solve this quadratic equation? Use (1.16) with a = 1, b = − 10 and c = 198
t j (^) j
± − − × ×
= ±^ −^ = ± = ±
Thus t^ =^5 +^ j^ 173, 5^ − j^173
- We need to find the cofactors of each element of the matrix. Cofactor of 1 is
det 3 0 1 7 7 1 0
⎛ ⎞ = × − − × =
Cofactor of 0 is
det 2 0 6 7 42 6 0
− ⎜ ⎟ = − ⎡⎣ − × − × ⎤⎦
Cofactor of 5 is
( ( )) ( )
det 2 1 6 3 16 6 1
⎜ ⎟ =^ ⎣ − × −^ −^ ×^ ⎦= −
Cofactor of − 2 is
det 0 0 1 5 5 1 0
− ⎛^ ⎞= − ⎡ × − − × ⎤
Cofactor of 3 is
det 1 0 6 5 30 6 0
⎛ ⎞ = ⎡ × − × ⎤= −
Cofactor of 7 is
( ( )) ( )
det 1 1 6 0 1 6 1
− ⎛^ ⎞^ = − ⎡^ × − − × ⎤
(1.16) x = −^ b^ ±^ b
(^2) − 4 ac 2 a
(11.1) det^ a b (^) ad cb c d
Solutions 11(c) 3
Cofactor of 6 is
det 0 5 0 7 3 5 15 3 7
⎛ ⎞ = ⎡ × − × ⎤= −
Cofactor of − 1 is
det 1 7 2 5 17 2 7
− ⎛^ ⎞= − ⎡ × − − × ⎤
Cofactor of 0 is
det 1 3 2 0 3 2 3
⎛ ⎞ = ⎡ × − − × ⎤=
Collecting the cofactors gives the cofactor matrix: 7 42 16 5 30 1 15 17 3
= ⎜^ − − ⎟
⎝ −^ − ⎠
C
Transposing this matrix (interchanging rows and columns) gives 7 5 1 42 30 17 16 1 3
T
⎛ −^ − ⎞
= ⎜^ − − ⎟
C
- Writing the equations in matrix form yields 1 2 3
i i i
Let (^) ⎟. What do we need to find?
A
The inverse matrix, A −^1. The currents can be obtained from 1 2 1 3
i i i
−
A
To find A −^1 we have to evaluate the determinant and the adjoint of A.
( ) (^) ( ( )) ( ) ( ) ( ) ( )
det 2 1 7 3det 5 det 3det 4 5 10 5 10 4 10 4 5 3 1 5 4 7 5 2 5 10 7 3 2 4 10 1 147
⎜ ⎟ ⎛^ −^ ⎞^ ⎛^ − ⎞^ ⎛^ ⎞
⎜ −^ ⎟ =^ ⎜^ ⎟^ − −^ ⎜^ ⎟^ + ⎜^ ⎟
⎜ ⎟ ⎝^ ⎠^ ⎝^ −^ ⎠^ ⎝^ − ⎠
= ⎡⎣^ × − × − ⎤⎦+ ⎡⎣ × − × ⎤⎦ + ⎡⎣ × − − × ⎤
Next we find adj , which is the cofactor matrix transposed. The cofactor matrix can be obtained using the method described in solution 5. Thus
A
= ⎜^ ⎟
C
Solutions 11(c) 5
(c) Expand along the first column since it contains 2 zeros:
det 0 3 7 1 det 3 7 0 det 5 1 0 det^5 0 2 9 2 9 2 9 3 7 1 3 9 2 7 0 0 13
⎜ ⎟ = ⎡^ ⎛^ ⎞^ ⎤^ − ⎡^ ⎛^ ⎞^ ⎤^ + ⎡ ⎛ ⎞⎤
⎜ ⎟ ⎢^ ⎜^ ⎟^ ⎥^ ⎢^ ⎜^ ⎟^ ⎥^ ⎢ ⎜
⎜ ⎟ ⎣^ ⎦^ ⎣^ ⎦^ ⎣ ⎠
= ⎡⎣ × − × ⎤⎦− + =
(d) Expanding along the second column
det 13 0 2 5det 0 det 0 det 11 3 11 3 13 2 11 0 3 5 13 3 11 2 85
⎜ ⎟ = − ⎛^ ⎞^ + ⎛^ ⎞^ − ⎛^ ⎞
⎜ ⎟ ⎜^ ⎟^ ⎜^ ⎟^ ⎜^ ⎟
⎜ ⎟ ⎝^ ⎠^ ⎝^ ⎠^ ⎝^ ⎠
= − ⎡⎣ × − × ⎤⎦= −
- The following shows the MAPLE output but you could use a graphical calculator.
A:=<<1 | 2 | 3 | 4> , <5 | 6 | 7 | 8> , <9 | 10 | 11 | 12> , <13 | 14 | 15 | 16>>;
A :=
with(linalg): Warning, the protected names norm and trace have been redefined and unprotected
det(A); 0
B:=matrix([[-1.1, 4.23, 2.67, 7.45,9.62], [19.61, 6.40, 3.12, 11.89,2.36], [-17.5, -9.73, 5.23, 8.54,2.51], [6.19, 2.91, 17.64, 8.93,8.98],[3.98,11.84,4.78,9.85,3.22]]);
B :=
det(B); -509092.
Solutions 11(d) 1
Complete solutions of Exercise 11(d)
- (a) In matrix form we have 1 2 3 3 2 1 1 11 3 2 1 5
x y z
The augmented matrix can be written
1 2 3
R
R
R
We execute to get 0 in place of 2 in row 2. Similarly to achieve a 0 in place of 3 in row 3:
R 2 − 2 R 1 R 3 − 3 R 1
1 2 * 2 1 3 * 3 1
R
R R R
R R R
= − ⎜^ − ⎟
= − ⎜⎝^ − − ⎟⎠
1 2 *
3 3 2
R
R
R R R
From the last row, R 3 **, we have 6 z = − 18 which gives z = − 3 From R 2 *we have
5 5 3 5 gives 4
y z y y
− + ⎡⎣ × − ⎤⎦= = −
From first row, R 1 , we have x + 2 y − 3 z = 3 Substituting y = − 4 and z = − 3 gives
x + ⎡⎣ 2 × − ( 4 ) ⎤⎦ − ⎡⎣ 3 × −( 3 ) ⎤⎦ = 3 gives x = 2
Thus x = 2, y = −4, z = − 3. Similarly (b) x = 1, y = 2, z = 3 (c) x = 1 2, y = 1 4, z = 1 8
- The augmented matrix is 9 × 103 − 3 × 103 − 3 × 103 13 × 103
Divide each row by 10 3 gives R 1 R 2
10 × 10 −^3
Interchanging row 1 and row 2 R 1 ' R 2 '
10 × 10 −^3
To get 0 in place of 9 we execute R 2 ' + 3 R 1 '
Solutions 11(d) 3
1 2 2 1 3
R
R R R g R g
= + ⎜^ ⎟
Getting 0 in place of − 2 1 2 3 3 2
R
R g R R R g
= + ⎜⎝^ ⎟⎠
From the last row we have
3 4 gives 4 3
T = g T = g
From R 2 'and substituting the above we have
2 2
(^4) which gives 4 3 3 x ^ + g = g x^ = − g + g = −^ g 3
From the first row, R 1 , we have (^1 ) x ^ =^ g .
- The augmented matrix is 1 2 3 1 2 2 1 3 1 2 3 3 2
' 2 '^0 0 5 3^4
R
R g R g R R R R g R g R R g R R R g
= + ⎜^ ⎟
By R 3 ' we have (^5 4) which gives 1 3 5 T = g T = 2 g
Similarly from R 2 ' and R 1 we have
1 5 , (^25) x (^) = −^ g^ x = g
- The augmented matrix is 3 3 3 3 3 3 3
⎛ × − × ⎞
⎜ − × × − × ⎟
⎜⎝ − × × − 5 ⎟⎠
Dividing each row by 10 3 gives
Solutions 11(d) 4
1 2 3 1 3 2 3 3
I I I
R
R
R
−
−
⎛ − × ⎞
⎜⎝ − − × ⎟⎠
Interchanging columns I 1 and I 3 : 3 2 1 3 1 2 3 3
I I I
R
R
R
−
−
⎛ − × ⎞
⎜⎝ − − × ⎟⎠
Divide row 2, R 2 , by − 10 gives
1 3 2 3 3
R
R
R
−
−
⎛ − × ⎞
⎜⎝ − − × ⎟⎠
To get 0 in place of 23 we need to execute R 3 − 23 R 2 ' 3 1 2 3 3
R
R
R
−
−
⎛ − × ⎞
⎜⎝ − − × ⎟⎠
To achieve 0 in place of 31.4 we have to implement R 3 ' + 31. 3
R 1
1 3 2 3 3
R
R
R
−
−
⎛ − × ⎞
⎜⎝ × ⎟⎠
Interchanging R 1 and R 2 ' gives 0's in the required position 3 2 1
3 3
I I I
− −
⎜ − × ⎟
⎜⎝ × ⎟⎠
From the last row we have 3 3 3 1 1 34.967 89.667 10 which gives 89.667^10 2.564332 10
I I
− ×^ − −
= × = = ×
Substituting I 1 = 2.564332 × 10 −^3 into the penultimate row gives
2 3 3 3 3 5 2
I
I
− − − − −
− + × × = ×
× − × ×
×
From the first row we have I 3 − 1.8 I 2 + 0.3 I 1 = 0 () Substituting I 1 = 2.564332 × 10 −^3 and I 2 = 8.5776 × 10 −^5 into () yields
Solutions 11(e) 1
Complete solutions to Exercise 11(e)
- Similar to EXAMPLE 23. Let A be the matrix of coefficients: 1 3 5 7 1 1 2 3 8
= ⎜^ − ⎟
A
We need to find the determinant of this matrix:
by (11.1)^ N^ (^ )^ (^ )^ (^ )
det 1det 3det 5det 3 8 2 8 2 3 8 3 3 56 2 5 21 2 294
⎛ −^ ⎞ ⎛ ⎞ ⎛ −
⎝ −^ ⎠ ⎝ − ⎠ ⎝
A ⎞⎟
Since det A ≠ 0 so by (11.16) we only have the trivial solution x = 0, y = 0, z = 0
- Solving the equations by the inverse matrix method gives x =1, y = 1. To check if this solution is unique we test whether the following matrix has a non zero determinant: 2 3 4 5
=^ ⎛^ ⎞
A ⎜⎝ ⎟⎠
det 2 3 2 5 4 3 2 0 4 5
⎛ ⎞ = × − × = − ≠
By (11.13) x = 1, y = 1 is the unique solution of the given equations.
- We can write the given equations as Au = b where 2 1 0 , , 4 2 0
x y
= ⎛^ ⎞^ = ⎛^ ⎞^ =⎛
A u b ⎞⎟ ⎠ We can find the determinant of A by using (11.1):
det 2 2 4 1 0 4 2
⎛ ⎞ = × − × =
By (11.15) there are an infinite number of solutions for 2 0 ( 4 2 0 (**)
x y x y
From (*) we have y = − 2 x The general solution is x = a , y = − 2 a where a^ is any real number.
- For the solution to be unique, det A ≠ 0 where A is the matrix of coefficients.
For question 1(a), let
A ⎟⎟ and so the determinant is given by
det 2 1 1 1det 2 det 3det 30 2 1 3 1 3 2 3 2 1
⎜ ⎟ ⎛^ −^ −^ ⎞^ ⎛^ −^ ⎞^ ⎛^ − ⎞
⎜ −^ −^ ⎟ =^ ⎜^ ⎟^ −^ ⎜^ ⎟^ −^ ⎜^ ⎟= −
⎜ ⎟ ⎝^ ⎠^ ⎝^ ⎠^ ⎝^ ⎠
Since det A = − 30 ≠ 0 the solution to the given equations is unique.
Solutions 11(e) 2
Similarly for (b) det A = 8 (c)det A = − 8 Therefore the solutions are unique.
For question 4;. The solution is unique.
1 0 1 , det 3 0 2 1
= ⎜^ − ⎟ =
A A
For question 5; A. Solution is unique.
3 0 1 , det 5 0 2 1
= ⎜^ − ⎟ =
A
- (a) For a non trivial solution we need
det ( A − λ I )= 0
Using the given A we have
det det 9 1 0 1 1 1 0 det 9 1 0 1 1 det 9 1 1 1 9 1 1 9 1 2 9 2
⎣⎝ − ⎠^ ⎝^ ⎠⎦
⎣⎝ − ⎠^ ⎝^ ⎠⎦
A I
Solving the quadratic equation by putting it to zero gives
4 2 0 which gives 4, 2
The values of λ for a non trivial solution are λ = −4, λ = 2.
(b) Similar to (a): 1 3 0 1 3 6 5 0 6 5
⎝ −^ ⎠ ⎝ ⎠ ⎝−
A I ⎞⎟
For a non trivial solution we need the determinant of this to be zero.
det 1 3 1 5 18 5 5 18 6 23 6 5
Putting the quadratic to zero and solving for λ.
λ^2 − 6 λ + 23 = 0 Using the quadratic formula, (1.16), with a = 1, b = − 6 and c = 23 gives
λ j
± − − × × ± − ± − ×
The values of λ are 3 + j 14 , 3 − j 14.
x b^ b^ ac a
=^ −^ ±^ −
Solutions 11(f) 2
x y x y
⎝ −^ − −^ ⎠⎝^ ⎠^ ⎝^ ⎠
Multiplying out the first row yields 11 3 0 3 11
x y x y
If y^ =1 then x = − 3 11, thus
⎛^ −
v ⎞⎟ or using smallest integers gives
(b) Let
x y
=^ ⎛^ ⎞
u (^) be the eigenvector for λ = 1 : 5 1 2 0 4 1 1 0 4 2 0 4 2 0
x y x y
⎛ −^ − ⎞⎛ ⎞
Multiplying out the matrix 4 2 4 2
x y x y
Solving these gives x = 1, y = 2. Thus is an eigenvector for
=^ ⎛^ ⎞
u λ = 1.
Let v = x y
⎠ be the eigenvector for^ λ^ =^3 : 5 3 2 0 4 1 3 0 2 2 0 4 4 0
x y x y
⎛ −^ − ⎞⎛ ⎞ =
Multiplying gives 2 2 4 4
x y x y
Solving these gives x = y = 1. An eigenvector corresponding to λ = 3 is.
(c) Let
x y
=^ ⎛^ ⎞
u (^) be the eigenvector for λ = − 3.
x y x y
⎝ − −^ ⎠⎝^ ⎠^ ⎝^ ⎠
Solutions 11(f) 3
Solving gives x = −2, y = 1. An eigenvector is
⎛^ −
⎟ corresponding to^ λ^ = −^3.
Let
x y
=^ ⎛^ ⎞
v (^) be an eigenvector for λ = 3 : 1 3 4 0 2 1 3 0 4 4 0 2 2 0
x y x y
Hence x = y = 1. The eigenvector corresponds to
λ = 3.
3. (a) Using det ( A − λ I ) = 0 :
( )( ) ( 2 )^2
det 1 1 2 1 1 1 2 1 1 1 2 1 2 1 2
λ^ λ λ λ λ λ λ λ λ λ
− = ⎛^ − − ⎞
= − − − − − ×
A I
Putting this quadratic to zero yields 2 2
1 which gives 1 j
λ λ λ
The system poles are λ 1 = j , λ 2 = − j
(b) Substituting the given matrix into det ( A − λ I ) gives
2 2 2 2
det 1 3 4 2 4 3 1 3 8 3 3 8 3 2 8 3 2 8 2 5
⎜ ⎟=^ −^ −^ −^ − − ×
⎝ −^ −^ − ⎠
Putting the resulting quadratic to zero λ^2 + 2 λ + 5 = 0. How do we solve this quadratic? Using the quadratic formula (1.16) with a = 1, b = 2 and c = 5 gives
λ = −^ ±^ −^ × ×^ = −^ ±^ −^ = −^ ±^ j^ = − ± j
The system poles are λ 1 = − 1 + j 2, λ 2 = − 1 − j 2. (c) The system poles are given by the eigenvalues of the matrix.
(1.16) x = −^ b^ ±^ b
(^2) − 4 ac 2 a
