MAT1503 ASSIGNMENT 2 2025
Question 1
(1.1)
𝑅𝐻𝑆 = 𝐴(𝐴 + 𝐵)−1 𝐵
𝑅𝐻𝑆 = (𝐴−1 )−1 (𝐴 + 𝐵)−1 𝐵
𝑅𝐻𝑆 = [(𝐴 + 𝐵)(𝐴−1 )]−1 𝐵 𝐸 −1 𝐹 −1 = [𝐹𝐸]−1
𝑅𝐻𝑆 = (𝐴𝐴−1 + 𝐵𝐴−1 )−1 𝐵
𝑅𝐻𝑆 = (𝐼 + 𝐵𝐴−1 )−1 𝐵
𝑅𝐻𝑆 = (𝐼 + 𝐵𝐴−1 )−1 (𝐵−1 )−1
𝑅𝐻𝑆 = [𝐵−1 (𝐼 + 𝐵𝐴−1 )]−1
𝑅𝐻𝑆 = [𝐵−1 𝐼 + 𝐵−1 𝐵𝐴−1 ]−1
𝑅𝐻𝑆 = [𝐵−1 + 𝐼𝐴−1 ]−1
𝑅𝐻𝑆 = (𝐵−1 + 𝐴−1 )−1
𝑅𝐻𝑆 = (𝐴−1 + 𝐵−1 )−1 Matrix addition is commutative
𝑅𝐻𝑆 = 𝐿𝐻𝑆
(1.2)
𝐴2 + 5𝐴 − 𝐼 = 0
𝐴2 + 5𝐴 = 𝐼
(𝐴2 + 5𝐴)𝐴−1 = 𝐼𝐴−1
𝐴2 𝐴−1 + 5𝐴𝐴−1 = 𝐴−1
𝐴𝐴𝐴−1 + 5𝐴𝐴−1 = 𝐴−1
𝐴 + 5𝐼 = 𝐴−1
𝐴−1 = 𝐴 + 5𝐼
,(1.3)
𝑃(𝑥) = 2𝑥 2 − 𝑥 + 1
1 −1
𝐴=[ ]
−2 3
1 −1 2 1 −1
𝑃(𝐴) = 2 [ ] −[ ] + 1𝐼2
−2 3 −2 3
1 −1 1 −1 1 −1 1 0
𝑃(𝐴) = 2 [ ][ ]−[ ]+[ ]
−2 3 −2 3 −2 3 0 1
1 + 2 −1 − 3 1 −1 1 0
𝑃(𝐴) = 2 [ ]−[ ]+[ ]
−2 − 6 2 + 9 −2 3 0 1
3 −4 1 −1 1 0
𝑃(𝐴) = 2 [ ]−[ ]+[ ]
−8 11 −2 3 0 1
6 −8 1 −1 1 0
𝑃(𝐴) = [ ]−[ ]+[ ]
−16 22 −2 3 0 1
6−1 −8 + 1 1 0
𝑃(𝐴) = [ ]+[ ]
−16 + 2 22 − 3 0 1
5 −7 1 0
𝑃(𝐴) = [ ]+[ ]
−14 19 0 1
5+1 −7 + 0
𝑃(𝐴) = [ ]
−14 + 0 19 + 1
6 −7
𝑃(𝐴) = [ ]
−14 20
(1.4)
𝑃(𝑥) = 𝑥 2 − 9
3 0 0
𝐴 = [0 −1 3 ]
0 −3 −1
3 0 0 2
𝑃(𝐴) = [0 −1 3 ] − 9𝐼3
0 −3 −1
3 0 0 3 0 0 1 0 0
𝑃(𝐴) = [0 −1 3 ] [0 −1 3 ] − 9 [0 1 0]
0 −3 −1 0 −3 −1 0 0 1
9+0+0 0−0−0 0+0−0 9 0 0
𝑃(𝐴) = [0 − 0 + 0 0 + 1 − 9 0 − 3 − 3] − [ 0 9 0]
0−0−0 0+3+3 0−9+1 0 0 9
9 0 0 9 0 0
𝑃(𝐴) = [0 −8 −6 ] − [ 0 9 0]
0 6 −8 0 0 9
, 9−9 0−0 0−0
𝑃(𝐴) = [0 − 0 −8 − 9 −6 − 0]
0 − 0 6 − 0 −8 − 9
0 0 0
𝑃(𝐴) = [0 −17 −6 ]
0 6 −17
Claim: 𝑄(𝑥) = 𝑥 + 3 and 𝑅(𝑥) = 𝑥 − 3
3 0 0 1 0 0 3 0 0 1 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 −1 3 ] + 3 [0 1 0]) ([0 −1 3 ] − 3 [0 1 0])
0 −3 −1 0 0 1 0 −3 −1 0 0 1
3 0 0 3 0 0 3 0 0 3 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 −1 3 ] + [0 3 0]) ([0 −1 3 ] − [0 3 0])
0 −3 −1 0 0 3 0 −3 −1 0 0 3
6 0 0 0 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 2 3]) ([0 −4 3 ])
0 −3 2 0 −3 −4
0 0 0
𝑄(𝐴)𝑅(𝐴) = [0 −17 −6 ]
0 6 −17
∴ 𝑃(𝐴) = 𝑄(𝐴)𝑅(𝐴)
Question 2
(2.1)
1 𝑥 1 𝑥
(𝑒 + 𝑒 −𝑥 ) (𝑒 − 𝑒 −𝑥 )
𝑀=[2 2 ]
1 𝑥 1 𝑥
(𝑒 − 𝑒 −𝑥 ) −𝑥
(𝑒 + 𝑒 )
2 2
1 1 1 1
det 𝑀 = (𝑒 𝑥 + 𝑒 −𝑥 ) × (𝑒 𝑥 + 𝑒 −𝑥 ) − (𝑒 𝑥 − 𝑒 −𝑥 ) × (𝑒 𝑥 − 𝑒 −𝑥 )
2 2 2 2
1 1
det 𝑀 = (𝑒 𝑥 + 𝑒 −𝑥 )2 − (𝑒 𝑥 − 𝑒 −𝑥 )2
4 4
1 1
det 𝑀 = (𝑒 2𝑥 + 2 + 𝑒 −2𝑥 ) − (𝑒 2𝑥 − 2 + 𝑒 −2𝑥 )
4 4
1 1 1 1 1 1
det 𝑀 = 𝑒 2𝑥 + + 𝑒 −2𝑥 − 𝑒 2𝑥 + − 𝑒 −2𝑥
4 2 4 4 2 4
det 𝑀 = 1
1
𝑀−1 = [matrix of cofactors]𝑇
det 𝑀
Question 1
(1.1)
𝑅𝐻𝑆 = 𝐴(𝐴 + 𝐵)−1 𝐵
𝑅𝐻𝑆 = (𝐴−1 )−1 (𝐴 + 𝐵)−1 𝐵
𝑅𝐻𝑆 = [(𝐴 + 𝐵)(𝐴−1 )]−1 𝐵 𝐸 −1 𝐹 −1 = [𝐹𝐸]−1
𝑅𝐻𝑆 = (𝐴𝐴−1 + 𝐵𝐴−1 )−1 𝐵
𝑅𝐻𝑆 = (𝐼 + 𝐵𝐴−1 )−1 𝐵
𝑅𝐻𝑆 = (𝐼 + 𝐵𝐴−1 )−1 (𝐵−1 )−1
𝑅𝐻𝑆 = [𝐵−1 (𝐼 + 𝐵𝐴−1 )]−1
𝑅𝐻𝑆 = [𝐵−1 𝐼 + 𝐵−1 𝐵𝐴−1 ]−1
𝑅𝐻𝑆 = [𝐵−1 + 𝐼𝐴−1 ]−1
𝑅𝐻𝑆 = (𝐵−1 + 𝐴−1 )−1
𝑅𝐻𝑆 = (𝐴−1 + 𝐵−1 )−1 Matrix addition is commutative
𝑅𝐻𝑆 = 𝐿𝐻𝑆
(1.2)
𝐴2 + 5𝐴 − 𝐼 = 0
𝐴2 + 5𝐴 = 𝐼
(𝐴2 + 5𝐴)𝐴−1 = 𝐼𝐴−1
𝐴2 𝐴−1 + 5𝐴𝐴−1 = 𝐴−1
𝐴𝐴𝐴−1 + 5𝐴𝐴−1 = 𝐴−1
𝐴 + 5𝐼 = 𝐴−1
𝐴−1 = 𝐴 + 5𝐼
,(1.3)
𝑃(𝑥) = 2𝑥 2 − 𝑥 + 1
1 −1
𝐴=[ ]
−2 3
1 −1 2 1 −1
𝑃(𝐴) = 2 [ ] −[ ] + 1𝐼2
−2 3 −2 3
1 −1 1 −1 1 −1 1 0
𝑃(𝐴) = 2 [ ][ ]−[ ]+[ ]
−2 3 −2 3 −2 3 0 1
1 + 2 −1 − 3 1 −1 1 0
𝑃(𝐴) = 2 [ ]−[ ]+[ ]
−2 − 6 2 + 9 −2 3 0 1
3 −4 1 −1 1 0
𝑃(𝐴) = 2 [ ]−[ ]+[ ]
−8 11 −2 3 0 1
6 −8 1 −1 1 0
𝑃(𝐴) = [ ]−[ ]+[ ]
−16 22 −2 3 0 1
6−1 −8 + 1 1 0
𝑃(𝐴) = [ ]+[ ]
−16 + 2 22 − 3 0 1
5 −7 1 0
𝑃(𝐴) = [ ]+[ ]
−14 19 0 1
5+1 −7 + 0
𝑃(𝐴) = [ ]
−14 + 0 19 + 1
6 −7
𝑃(𝐴) = [ ]
−14 20
(1.4)
𝑃(𝑥) = 𝑥 2 − 9
3 0 0
𝐴 = [0 −1 3 ]
0 −3 −1
3 0 0 2
𝑃(𝐴) = [0 −1 3 ] − 9𝐼3
0 −3 −1
3 0 0 3 0 0 1 0 0
𝑃(𝐴) = [0 −1 3 ] [0 −1 3 ] − 9 [0 1 0]
0 −3 −1 0 −3 −1 0 0 1
9+0+0 0−0−0 0+0−0 9 0 0
𝑃(𝐴) = [0 − 0 + 0 0 + 1 − 9 0 − 3 − 3] − [ 0 9 0]
0−0−0 0+3+3 0−9+1 0 0 9
9 0 0 9 0 0
𝑃(𝐴) = [0 −8 −6 ] − [ 0 9 0]
0 6 −8 0 0 9
, 9−9 0−0 0−0
𝑃(𝐴) = [0 − 0 −8 − 9 −6 − 0]
0 − 0 6 − 0 −8 − 9
0 0 0
𝑃(𝐴) = [0 −17 −6 ]
0 6 −17
Claim: 𝑄(𝑥) = 𝑥 + 3 and 𝑅(𝑥) = 𝑥 − 3
3 0 0 1 0 0 3 0 0 1 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 −1 3 ] + 3 [0 1 0]) ([0 −1 3 ] − 3 [0 1 0])
0 −3 −1 0 0 1 0 −3 −1 0 0 1
3 0 0 3 0 0 3 0 0 3 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 −1 3 ] + [0 3 0]) ([0 −1 3 ] − [0 3 0])
0 −3 −1 0 0 3 0 −3 −1 0 0 3
6 0 0 0 0 0
𝑄(𝐴)𝑅(𝐴) = ([0 2 3]) ([0 −4 3 ])
0 −3 2 0 −3 −4
0 0 0
𝑄(𝐴)𝑅(𝐴) = [0 −17 −6 ]
0 6 −17
∴ 𝑃(𝐴) = 𝑄(𝐴)𝑅(𝐴)
Question 2
(2.1)
1 𝑥 1 𝑥
(𝑒 + 𝑒 −𝑥 ) (𝑒 − 𝑒 −𝑥 )
𝑀=[2 2 ]
1 𝑥 1 𝑥
(𝑒 − 𝑒 −𝑥 ) −𝑥
(𝑒 + 𝑒 )
2 2
1 1 1 1
det 𝑀 = (𝑒 𝑥 + 𝑒 −𝑥 ) × (𝑒 𝑥 + 𝑒 −𝑥 ) − (𝑒 𝑥 − 𝑒 −𝑥 ) × (𝑒 𝑥 − 𝑒 −𝑥 )
2 2 2 2
1 1
det 𝑀 = (𝑒 𝑥 + 𝑒 −𝑥 )2 − (𝑒 𝑥 − 𝑒 −𝑥 )2
4 4
1 1
det 𝑀 = (𝑒 2𝑥 + 2 + 𝑒 −2𝑥 ) − (𝑒 2𝑥 − 2 + 𝑒 −2𝑥 )
4 4
1 1 1 1 1 1
det 𝑀 = 𝑒 2𝑥 + + 𝑒 −2𝑥 − 𝑒 2𝑥 + − 𝑒 −2𝑥
4 2 4 4 2 4
det 𝑀 = 1
1
𝑀−1 = [matrix of cofactors]𝑇
det 𝑀
