MAT1503 Assignment 4 2021

MAT1503 ASSIGNMENT 4 2021



Question 1


𝑥 𝑦 1
|𝑎1 𝑏1 1| = 0
𝑎2 𝑏2 1


𝑇𝑜 𝑎𝑑𝑑 𝑎𝑙𝑜𝑛𝑔 𝑟𝑜𝑤 1.
𝑏1 1 𝑎 1 𝑎 𝑏1
+𝑥 | |−𝑦| 1 | +1| 1 |=0
𝑏2 1 𝑎2 1 𝑎2 𝑏2



+𝑥[𝑏1 × 1 − 𝑏2 × 1] − 𝑦[𝑎1 × 1 − 𝑎2 × 1] + 1[𝑎1 × 𝑏2 − 𝑎2 × 𝑏1 ] = 0



+𝑥[𝑏1 − 𝑏2 ] − 𝑦[𝑎1 − 𝑎2 ] + 1[𝑎1 𝑏2 − 𝑎2 𝑏1 ] = 0


[𝑏1 − 𝑏2 ]𝑥 − [𝑎1 − 𝑎2 ]𝑦 + [𝑎1 𝑏2 − 𝑎2 𝑏1 ] = 0



−[𝑎1 − 𝑎2 ]𝑦 = −[𝑏1 − 𝑏2 ]𝑥 − [𝑎1 𝑏2 − 𝑎2 𝑏1 ]


[𝑏1 − 𝑏2 ] [𝑎1 𝑏2 − 𝑎2 𝑏1 ]
𝑦=− 𝑥−
−[𝑎1 − 𝑎2 ] −[𝑎1 − 𝑎2 ]


[𝑏1 − 𝑏2 ] [𝑎1 𝑏2 − 𝑎2 𝑏1 ]
𝑦= 𝑥+
[𝑎1 − 𝑎2 ] [𝑎1 − 𝑎2 ]

,Question 2



(2.1)
−4 2
𝐴=[ ]
3 −3


−4 × −1 2 × −1
−𝐴 = [ ]
3 × −1 −3 × −1


4 −2
−𝐴 = [ ]
−3 3


det(−𝐴) = 4 × 3 − (−3 × −2)



det(−𝐴) = 12 − (6)



det(−𝐴) = 12 − 6



det(−𝐴) = 6




−4 × −1 2 × −1 𝑇
−𝐴𝑇 = [ ]
3 × −1 −3 × −1


4 −2 𝑇
−𝐴𝑇 = [ ]
−3 3


4 −3
−𝐴𝑇 = [ ]
−2 3


det(−𝐴𝑇 ) = 4 × 3 − (−2 × −3)



det(−𝐴𝑇 ) = 12 − (6)



det(−𝐴𝑇 ) = 12 − 6

, det(−𝐴𝑇 ) = 6




𝑊𝑒 𝑛𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 det(−𝐴) = det(−𝐴𝑇 )



(2.2)


3 1 −2
𝐴 = [−5 3 −6]
−1 0 −4


3 1 −2
−𝐴 = −1 × [−5 3 −6]
−1 0 −4


−1 × 3 −1 × 1 −1 × −2
−𝐴 = [−1 × −5 −1 × 3 −1 × −6]
−1 × −1 −1 × 0 −1 × −4


−3 −1 2
−𝐴 = [ 5 −3 6]
1 0 4


−3 6 5 6 5 −3
det(−𝐴) = +(−3) | | − (−1) | | + (+2) | |
0 4 1 4 1 0


det(−𝐴) = −3[−3 × 4 − (0 × 6)] + 1[5 × 4 − (1 × 6)] + 2[5 × 0 − (1 × −3)]



det(−𝐴) = −3[12 − (0)] + 1[20 − (6)] + 2[0 − (−3)]



det(−𝐴) = −3[12 − 0] + 1[20 − 6] + 2[0 + 3]



det(−𝐴) = −3[12] + 1[14] + 2[3]



det(−𝐴) = −36 + 14 + 6



det(−𝐴) = −16