Question

If A is a square matrix such that A2=A, then write the value of 7A−(I+A)3 where I is an identity matrix.​

Answer 1

$\textbf{Given:}$

$A^2=A$

$\text{Consider,}$

$7A-(I+A)^3$

$\text{Using}$

$\boxed{\bf(a+b)^3=a^3+b^3+3a^2b+3ab^2}$

$=7A-(I^3+A^3+3I^2A+3IA^2)$

$\text{For any identity matrix I, I^n=I }$

$=7A-(I+(A^2)A+3IA+3IA)$

$=7A-(I+(A)A+3A+3A)$

$=7A-(I+A^2+6A)$

$=7A-(I+A+6A)$

$=7A-(I+7A)$

$=7A-I-7A$

$=-I$

Answer 2

Answer:

Step-by-step explanation:

A is a square matrix such that A² = A and I is the identity matrix.

To find---> Value of 7A - ( I + A )³

Solution---> We know that if we multiply identity matrix to any other matrix we get same matrix

i. e.

I P = P , where I is identity matrix and P is any matrix .

And , I I = I

Now , we know that,

( a + b )³ = a³ + b³ + 3ab ( a + b )

Now , returning to original problem ,

7A - ( I + A )³

= 7A - { ( I )³ + ( A )³ + 3 I A ( I + A ) }

= 7A - ( I³ + A³ + 3 I² A + 3 I A² )

= 7A - ( I² I + A² A + 3 I² A + 3 I A² )

Putting I² = I , A² = A and

= 7A - ( I I + A A + 3 I A + 3 I A )

Putting I A = A , we get,

= 7 A - ( I + A² + 3 A + 3 A )

Putting A² = A , we get,

= 7A - ( I + A + 6A )

= 7 A - ( I + 7A )

= 7 A - I - 7A

= - I