STA3710 ASSIGNMENT 1 2023

STA3710
ASSIGNMENT 1
2023

,
, Solution:



1.1).



1 0 0 1 1
A=( ) ,B = ( ) ,c = − and k = 5
0 2 1 0 4


1.1.1).


1 0 0 1 0 1
AB = ( )( )=( )
0 2 1 0 2 0


0 1 1 0 0 2
BA = ( )( )=( )
1 0 0 2 1 0


We can see that AB ≠ BA is true



1.1.2).


1
(c + k)A = (− + 5) (1 0)
4 0 2
19 1 0
= ( )
4 0 2
19⁄ 0
=( 4 )
0 19⁄2



1 1 0 1 0
cA + kA = − ( ) +5( )
4 0 2 0 2
− 1⁄4 0 5 0
=( )+( )
0 1
− ⁄2 0 10

19⁄ 0
=( 4 )
0 19⁄
2


It shows (c + k)A = cA + kA