if A = [1&1 0&2], then A4 - 24(A-I) =

Question

if A = [1&1 0&2], then A4 - 24(A-I) = (A) I-A (B) 2I - A (C) I + A (D) A - 2I

Tardigrade Admin · Accepted Answer

We have A = [1&1 0&2] A2 = [1&1 0&2][1&1 0&2] = [1&3 0&4] A3 = [1&7 0&4] [1&1 0&2][1&7 0&8] A4 = [1&7 0&8] [1&1 0&2] = [1&15 0&16] Now A4 - 24 ( A - I )=[1&15 0&16] -16[0&1 0&1] = [1&15 0&16] -[1&16 0&16] = [1&-1 0&0] Also 2I - A = [2&0 0&2]-[1&1 0&2] =[1&-1 0&0] ∴ A4 - 24 ( A-I ) = 2I - A

Q. if $A = [1012]$ , then $A^{4} - 2^{4} (A - I) =$

$I - A$

$2 I - A$

$I + A$

$A - 2 I$

We have $A = [1012]$
$A^{2} = [1012] [1012] = [1034]$
$A^{3} = [1074] [1012] [1078]$
$A^{4} = [1078] [1012] = [101516]$
Now $A^{4} - 2^{4} (A - I) = [101516] - 16 [0011]$
$= [101516] - [101616] = [10 - 1 0]$
Also $2 I - A = [2002] - [1012] = [10 - 1 0]$
$∴ A^{4} - 2^{4} (A - I) = 2 I - A$

Q. if A=[10​12​], then A4−24(A−I)=

I−A

2I−A

I+A

A−2I

Q. if $A = [1012]$ , then $A^{4} - 2^{4} (A - I) =$

$I - A$

$2 I - A$

$I + A$

$A - 2 I$