The 7th week
Quiz7
Exercise set 4.1
3 5
1. det ([
]) = 3x4 − 5x(−2) = 22
−2 4
−5 7
3. det ([
]) = (−5)x(−2) − 7x(−7) = 59
−7 −2
−2 1 4
7. det ([ 3 5 −7]) = 𝑎11 𝑎22 𝑎33 + 𝑎12 𝑎23 𝑎32 + 𝑎13 𝑎21 𝑎32 − 𝑎13 𝑎22 𝑎31 − 𝑎12 𝑎21 𝑎33 −
1 6 2
𝑎11 𝑎23 𝑎32 = −65
1 −2 3
25. A = [ 6
7 −1];
−3 1
4
a) det(A) = a11C11 + a12C12 + a13C13 = 1x29 + -2x(-21) + 3x27=152
d) det (A) = a12C12 + a22C22 + a32C32 = -2x(-21) + 7x13 + 1x19= 152
𝑎11
A = [𝑎21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13
𝑎23 ];
𝑎33
𝑎11
Theorem 4.2.1: A = [𝑎12
𝑎13
T
𝑎21
𝑎22
𝑎23
𝑎31
𝑎32 ]
𝑎33
det(A) = 𝑎11 𝐶11 + 𝑎21 𝐶21 + 𝑎31 𝐶31 = 𝑎11 (𝑎22 𝑎33 − 𝑎23 𝑎32 ) + 𝑎21 (−1)(𝑎12 𝑎33 − 𝑎13 𝑎32 ) +
𝑎31 (𝑎12 𝑎23 − 𝑎22 𝑎13 )
det(AT) = 𝑎11 (𝑎22 𝑎33 − 𝑎23 𝑎32 ) + 𝑎12 (−1)(𝑎21 𝑎33 − 𝑎31 𝑎23 ) + 𝑎13 (𝑎21 𝑎32 − 𝑎22 𝑎31 )
det(A) = det(AT) = 𝑎11 𝑎22 𝑎33 + 𝑎12 𝑎23 𝑎31 + 𝑎13 𝑎21 𝑎32 − 𝑎13 𝑎22 𝑎31 − 𝑎12 𝑎21 𝑎33 − 𝑎11 𝑎23 𝑎32
Theorem 4.2.2:
1/6
𝑎11 𝑘𝑎12 𝑎13
a) B = [𝑎21 𝑘𝑎22 𝑎23 ], det(B) = 𝑎11 𝐶11 + 𝑎21 𝐶21 + 𝑎31 𝐶31 = 𝑎11 (𝑘𝑎22 𝑎33 − 𝑎23 𝑘𝑎32 ) +
𝑎31 𝑘𝑎32 𝑎33
𝑎21 (−1)(𝑘𝑎12 𝑎33 − 𝑎13 𝑘𝑎32 ) + 𝑎31 (𝑘𝑎12 𝑎23 − 𝑘𝑎22 𝑎13 ) = k det(A)
𝑘𝑎11 𝑘𝑎12 𝑘𝑎13
𝑎22
𝑎23 ], det(B) = 𝑘𝑎11 𝐶11 + 𝑎21 𝐶21 + 𝑎31 𝐶31 = 𝑘𝑎11 (𝑎22 𝑎33 − 𝑎23 𝑎32 ) +
B = [ 𝑎21
𝑎31
𝑎32
𝑎33
𝑎21 (−1)(𝑘𝑎12 𝑎33 − 𝑘𝑎13 𝑎32 ) + 𝑎31 (𝑘𝑎12 𝑎23 − 𝑎22 𝑘𝑎13 ) = k det(A)
𝑎21 𝑎22 𝑎23
b) B = [𝑎11 𝑎12 𝑎13 ], interchange of row 1 and 2; det(B) = 𝑎21 𝐶11 + 𝑎11 𝐶21 + 𝑎31 𝐶31 =
𝑎31 𝑎32 𝑎33
𝑎21 (𝑎12 𝑎33 − 𝑎13 𝑎32 ) + 𝑎11 (−1)(𝑎22 𝑎33 −𝑎23 𝑎32 ) + 𝑎31 (𝑎22 𝑎13 −𝑎23 𝑎12 ) = - det(A)
𝑎11 𝑎13 𝑎12
𝑎
B = [ 21 𝑎23 𝑎22 ], interchange of column 2 and 3; det(B) = 𝑎11 𝐶11 + 𝑎21 𝐶21 + 𝑎31 𝐶31 =
𝑎31 𝑎33 𝑎32
𝑎11 (𝑎23 𝑎32 − 𝑎22 𝑎33 ) + 𝑎21 (−1)(𝑎13 𝑎32 − 𝑎12 𝑎33 ) + 𝑎31 (𝑎22 𝑎13 −𝑎12 𝑎23 )= - det(A)
𝑎11
𝑎12
𝑎13
c) B = [𝑎21 + 𝑘𝑎11 𝑎22 + 𝑘𝑎12 𝑎23 + 𝑘𝑎13 ] , det(B) = 𝑎11 𝐶11 + (𝑎21 + 𝑘𝑎11 )𝐶21 + 𝑎31 𝐶31 =
𝑎31
𝑎32
𝑎33
𝑎11 ((𝑎22 + 𝑘𝑎12 )𝑎33 − (𝑎23 + 𝑘𝑎13 )𝑎32 ) + (𝑎21 + 𝑘𝑎11 )(−1)(𝑎12 𝑎33 − 𝑎13 𝑎32 ) +
𝑎31 (𝑎12 (𝑎23 + 𝑘𝑎13 ) − 𝑎13 (𝑎22 + 𝑘𝑎12 )) = det (A)
Quiz in Exercise set 4.2
3. a) det (A) = 3x(1/3)x(-2)(2)= -4
b) det (A) = 0 Since the1st column times 3 equals the 3rd column.
c) det(A) = 3x5x(-2)=-30
HW7
Exercise set 4.1
𝜆−2
1
13. det ([
]) = (𝜆 − 2)(𝜆 + 4) − 1(−5) = (𝜆 − 1)(𝜆 + 3) = 0
−5 𝜆 + 4
=> 𝜆 = 1, −3
2/6
𝜆−4 0
14. det([ 0
𝜆
0
3
0
2 ]) = (𝜆 − 4)𝜆(𝜆 − 1) + 0 + 0 − (𝜆 − 4)6 = 0
𝜆−1
=> (𝜆 − 4)(𝜆(𝜆 − 1) − 6) = 0 => (𝜆 − 4)((𝜆 + 2)(𝜆 − 3)) = 0
=> 𝜆 = 4, −2, 3
1
19. a) det ([0
0
0 0
−1 0]) = 𝑎11 𝑎22 𝑎33 = −1
0 1
0
1
b) det ([
0
1
0
2
4
2
0
0
3
4
1
0
c) det ([
0
0
2
1
0
0
7 −3
−4 1
]) = 𝑎11 𝑎22 𝑎33 = 1x1x2x3 = 6
2
7
0
3
0
0
]) = 𝑎11 𝑎22 𝑎33 = 0
0
8
1 −2 3
21. A = [ 6
7 −1]; a) M11 = 7x4-(-1)x1 = 29, M12 = 6x4-(-1)x(-3)=21, M13 = 6x1-7x(-3)=27,
−3 1
4
M21 = -2x4-3x1=-11, M22 = 1x4-3x(-3)=13, M23 = 1x1-(-2)x(-3)=-5,
M31 = -2x(-1)-3x7=-19, M32 = 1x(-1)-3x6=-19, M33 = 1x7-(-2)x6=19,
b) C11 = 1x29=29, C12 = (-1)x21=-21, C13 = 1x27 = 27,
C21 = (-1)x(-11)=11, C22 =1x(13)=13, C23 = (-1)x(-5)= 5,
C31 = 1x(-19)=-19, C32 = (-1)x(-19)=19, C33 = 1x19=19.
25. a) det(A) = a11C11 + a12C12 + a13C13 = 1x29 + -2x(-21) + 3x27=152
d) det (A) = a12C12 + a22C22 + a32C32 = -2x(-21) + 7x13 + 1x19= 152
sin(𝜃)
𝑐𝑜𝑠(𝜃)
0
−𝑐𝑜𝑠(𝜃)
𝑠𝑖𝑛(𝜃)
0];
33. 𝐴 = [
𝑠𝑖𝑛(𝜃) − 𝑐𝑜𝑠(𝜃) 𝑠𝑖𝑛(𝜃) + 𝑐𝑜𝑠(𝜃) 1
3/6
det(A) = 1x(𝑠𝑖𝑛(𝜃) sin(𝜃) − 𝑐𝑜𝑠(𝜃)(−𝑐𝑜𝑠(𝜃))) = 𝑠𝑖𝑛2 (𝜃) + 𝑐𝑜𝑠 2 (𝜃) = 1
Exercise set 4.2
𝑎
𝑑
5. det([
𝑔
𝑏
𝑒
ℎ
𝑐
𝑑
𝑓 ]) = 6; a) way 1: det ([𝑔
𝑖
𝑎
𝑑
way 2: det ([𝑔
𝑎
𝑓
0 1 0 𝑎
𝑖 ]) = det ([0 0 1] [𝑑
1 0 0 𝑔
𝑐
𝑒
ℎ
𝑏
𝑎
𝑓
2
𝑑
𝑖 ]) = (-1) det([
𝑔
𝑐
𝑒
ℎ
𝑏
𝑏
𝑒
ℎ
3𝑎 3𝑏 3𝑐
𝑎
b) way 1: det([−𝑑 −𝑒 −𝑓 ]) = 3det ([−𝑑
4𝑔 4ℎ 4𝑖
4𝑔
𝑎 𝑏 𝑐
𝑑
𝑒 𝑓]) = −12x6 = −72
−12 det ([
𝑔 ℎ 𝑖
3𝑎
way 2: det([−𝑑
4𝑔
3𝑏
−𝑒
4ℎ
𝑎+𝑔
c) way 1: det([ 𝑑
𝑔
𝑎+𝑔
way 2: det([ 𝑑
𝑔
3
11. 𝐴 = [ 0
−2
𝑏
−𝑒
4ℎ
𝑏+ℎ
𝑒
ℎ
𝑐+𝑖
𝑎
𝑓 ]) = det([𝑑
𝑔
𝑖
𝑐+𝑖
1
𝑓 ]) = det ([0
𝑖
0
𝑐
𝑓]) = 6
𝑖
𝑐
𝑓]) = 1x6 = 6
𝑖
𝑐
𝑎
−𝑓]) = −3det ([ 𝑑
4𝑖
4𝑔
3𝑐
3 0 0 𝑎
−𝑓 ]) = det ([0 −1 0] [𝑑
4𝑖
0 0 4 𝑔
𝑏+ℎ
𝑒
ℎ
𝑏
𝑒
ℎ
𝑏
𝑒
ℎ
𝑏
𝑒
ℎ
𝑐
𝑓 ]) = -12 x 6 = -72
𝑖
𝑐
𝑓]) = 6
𝑖
0 1 𝑎
1 0] [𝑑
0 1 𝑔
6 −9
0 −2];
1 5
4/6
𝑏
𝑒
ℎ
𝑐
𝑓]) = 1 x 6 = 6
𝑖
𝑏
𝑒
4ℎ
𝑐
𝑓 ]) =
4𝑖
1
way 1: det(𝐵) = 3 det(𝐴) , 𝑑𝑒𝑡(𝐶) = 𝑑𝑒𝑡(𝐵), det(𝐷) = − det(𝐶) , det(𝐷) = −10 => det(𝐴) = 30
3 0
way2: det ([0 1
0 0
0 1 0 0 1 0
0] [ 0 1 0] [0 0
1 −2 0 1 0 1
0 1
1] [0
0 0
2 −3
5 −1]) =(3x1x1)x(1x1x1)x(-1)x(1x5x(-2))= 30
0 −2
27. det(A) = 7; a) det(3A) = 33det(A) = (3x3x3)x7=189
1
1
b) det(A-1) = det(𝐴) = 7
1
8
c) det( (2A-1) = 23det(A-1) = 2x2x2x7= 7
d) det( (2A)-1 ) =
1
det(2𝐴)
1
1
7
2x2x2
= x
=
1
56
1 −2 3
1
5 −9 6
3
31. 𝐴 = [
];
−1 2 −6 −2
2
8
6
1
5/6
1
0
=> U =
0
[0
−2 3
1
1
0
1 −9 −2
5
1
1 , L=[
0
1
−1
0
3
2 12
0
0
1]
0
0
−3
108
0
0
]
0
−13
det(A) = det(LU)=(1x1x(-3)x(-13))x1=39
D7. a) det(I+A) = I + det(A); F, scalar ≠ matrix + scalar
b) det(A4) = (det(A))4; T, det(A4) =det(A) det(A) det(A) det(A) = (det(A))4
c) det(3A) = 3det(A); F, det(3A) = 3ndet(A), n depends on size of matrix A.
d) If det(A) = 0, then Ax=0 has infinitely many solutions.
T, det(A) = 0 means matrix A is not invertible, and so Ax=0 has infinitely many solutions.
6/6
