Let A= [ beginmatrix2&0&0 0&2&0 0&0&2 endmatrix] , then A3-6A2+12A-8I =

Question

Let A= [ beginmatrix2&0&0 0&2&0 0&0&2 endmatrix] , then A3-6A2+12A-8I = (A)  [ beginmatrix0&0&0 0&0&0 0&0&0 endmatrix]  (B)  [ beginmatrix1&0&0 0&1&0 0&0&1 endmatrix]  (C)  [ beginmatrix3&0&0 0&3&0 0&0&3 endmatrix]  (D)  [ beginmatrix2&0&0 0&2&0 0&0&0 endmatrix]

Tardigrade Admin · Accepted Answer

We have, A = [2&0&0 0&2&0 0&0&2] = 2[1&0&0 0&1&0 0&0&1] = 2I A2 2I ⋅ 2I = 22 I Similarly A3 = 23I Now, A3 - 6A2 + 12A - 8I 8I - 6(4I) + 12 (2I) - 8I = 0 =[0&0&0 0&0&0 0&0&0]

Q. Let $A = ⎣ ⎡ 200020002 ⎦ ⎤$ , then $A^{3} - 6 A^{2} + 12 A - 8 I$ =

$⎣ ⎡ 000000000 ⎦ ⎤$

$⎣ ⎡ 100010001 ⎦ ⎤$

$⎣ ⎡ 300030003 ⎦ ⎤$

$⎣ ⎡ 200020000 ⎦ ⎤$

We have,
$A = ⎣ ⎡ 200020002 ⎦ ⎤ = 2 ⎣ ⎡ 100010001 ⎦ ⎤ = 2 I$
$A^{2} 2 I \cdot 2 I = 2^{2} I$
Similarly $A^{3} = 2^{3} I$
Now, $A^{3} - 6 A^{2} + 12 A - 8 I$
$8 I - 6 (4 I) + 12 (2 I) - 8 I = 0$
$= ⎣ ⎡ 000000000 ⎦ ⎤$

Q. Let A=⎣⎡​200​020​002​⎦⎤​ , then A3−6A2+12A−8I =

⎣⎡​000​000​000​⎦⎤​

⎣⎡​100​010​001​⎦⎤​

⎣⎡​300​030​003​⎦⎤​

⎣⎡​200​020​000​⎦⎤​