[MOVED] Matrices question: If A^3 = I, then simplify...

[mathfun]

IF A^3 = I

Simplify A^2(A+I)^2

the answer that i got was

2A^2 + A + I

the CORRECT answer is A^2 + A + 2I

Can somebody please explain?
Thanks in advance.

 

[jonboy]

Show us/tell us how you got 2A^2 + A + I so we can see what you did wrong.

 

[mathfun]

oh ok. btw. i forgot to mention that I is identity matrix.

here is my work.

A^2(A+I)^2
= (A^3 + A^2I)^2
= (I + A^2)^2
= (I + A^2)(I + A^2)
= I^2 + A^2 + A^2 + A^4
= I + 2A^2 + A^4
= A(2A + A^3) + I
= A(2A + I) + I
= 2A^2 + A + I

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
A^2 \left( {A + I} \right)^2 & = & A^2 \left( {A + I} \right)\left( {A + I} \right) \\
& = & A^2 \left( {A^2 + IA + AI + I^2 } \right) \\
& = & A^2 \left( {A^2 + AI + AI + I} \right) \\
& = & A^2 \left( {A^2 + 2AI + I} \right) \\
& = & \left( {A^4 + 2A^3 I + A^2 } \right) \\
& = & \left( {A^3 A + 2II + A^2 } \right) \\
& = & \left( {A^2 + A + 2I} \right) \\
\end{array}\)

 

[mathfun]

for the fourth step,
A^2(A^2 + 2AI + I)

how come you can't take away the I in 2AI?

the last step..
how come u can take away ONE I only?

btw, ur signature sounds interesting. i think i am the kind that doesn't understand binary #'s. actually, i haven't really heard of it. do u mind explain how there are only 10 kinds? i think that its really cool.

 

[pka]

for the fourth step,
A^2(A^2 + 2AI + I) how come you can't take away the I in 2AI?

One can do that: 2AI=2A and A<SUP>2</SUP>(2A)=2A<SUP>3</SUP>

the last step..how come u can take away ONE I only?

I times I is I. II=I

 

[pka]

In the binary numbers 10 is the same as two in the decimal number system.