if I= 1 0 E= 0 1 0 1 1 0 and show that aI+bE while cube= a cube I+3a square bE
Answers
Answered by
0
Answer:
I=(
1
0
0
1
)E=(
0
0
1
0
)
aI+bE=(
a
0
0
a
)+(
0
0
b
0
)
=(
a
0
b
a
)
⇒(aI+bE)
3
=(
a
0
b
a
)(
a
0
b
a
)(
a
0
b
a
)
=(
a
3
0
3a
2
b
a
3
)
=(
a
3
0
0
a
3
)+(
0
0
3a
2
b
0
)
=a
3
(
1
0
0
1
)+3a
2
b(
1
0
0
1
)
(aI+bE)
3
=a
3
I+3a
2
bE
Step-by-step explanation:
Similar questions