Matrix A is such that A2=2A-I, where I is the identity matrix, then for n ge 2, An is equal to

Question

Matrix A is such that A2=2A-I, where I is the identity matrix, then for n ge 2, An is equal to (A)  2n-1A-(n-1)I  (B)  2n-1A-I  (C)  nA-(n-1)I  (D)  nA-I

Tardigrade Admin · Accepted Answer

Given, A2=2A-I Now, A3=A2.A =(2A-I)A =2A2-A =2(2A-I)-A =3A-2I Now, A4=A3.A =(3A-2I).A =3A2-2A =3(2A-I)-2A =4A-3I Similarly, An=nA-(n-1)I

Q. Matrix $A$ is such that $A^{2} = 2 A - I,$ where $I$ is the identity matrix, then for $n \geq 2, A^{n}$ is equal to

$2^{n - 1} A - (n - 1) I$

$2^{n - 1} A - I$

$n A - (n - 1) I$

$n A - I$

Given, $A^{2} = 2 A - I$
Now, $A^{3} = A^{2} . A$
$= (2 A - I) A$ $= 2 A^{2} - A$
$= 2 (2 A - I) - A$ $= 3 A - 2 I$
Now, $A^{4} = A^{3} . A$
$= (3 A - 2 I) . A$ $= 3 A^{2} - 2 A$
$= 3 (2 A - I) - 2 A$ $= 4 A - 3 I$
Similarly, $A^{n} = n A - (n - 1) I$

Q. Matrix A is such that A2=2A−I, where I is the identity matrix, then for n≥2,An is equal to

2n−1A−(n−1)I

2n−1A−I

nA−(n−1)I

nA−I