Question

A and B are two matrices such that AB=A and BA=B then find k id (A+B)^2020=k(A+B)​

Answer 1

Answer:

A and B are two matrices such that AB=A and BA=B then find k if (A+B)^2020=k(A+B)​

Answer : k = $2^{2019} I$

Step-by-step explanation:

A and B are two matrices such that AB=A and BA=B

if AB = A

Multiplying both side by $A^{-1}$

$A^{-1}$AB = $A^{-1}$A

IB = I     [ $A^{-1}$A = I ]

B = I

again,

BA = B

Multiplying both side by $B^{-1}$

$B^{-1}$BA = $B^{-1}$B

IA = I

A = I

then,

$(A+B)^{2020}$ = k (A + B)

$(I + I)^{2020}$ = k ( I + I )

$(2I)^{2020}$ = k (2I)

$(2I)^{2019}$ = k

k = $2^{2019} I$

Therefore, the value of k will be $2^{2019} I$ .

Answer 2

Given as :

A and B are two matrices such that AB = A and BA = B

$(A + B)^{2020}$  = k (A + B)

To Find :

The value of k

Solution :

For two matrix A and B

Case I

 A B = A

multiplying both side by $A^{-1}$

So,  $A^{-1}$ ( A B ) = $A^{-1}$ A

Or,  [ $A^{-1}$ . A ] [ B ] = $A^{-1}$ A

∵      $A^{-1}$ A = I

So,     ( I ) ( B ) = I

Or,            B = $\dfrac{I}{I}$

∴               B = I               ..........1

Similarly

Case II

 B A = B

multiplying both side by $B^{-1}$

So,  $B^{-1}$ ( B A ) = $B^{-1}$ B

Or,  [ $B^{-1}$ . B ] [ A ] = $B^{-1}$ B

∵      $B^{-1}$ B = I

So,     ( I ) ( A ) = I

Or,            A = $\dfrac{I}{I}$

∴               A = I               ..........2

A/Q   $(A + B)^{2020}$  = k (A + B)

From eq 1 and eq2

      $(I + I)^{2020}$  = k (I + I)

Or,   $(2 I)^{2020}$   = k ( 2 I )

Or,    k = $\dfrac{(2I)^{2020} }{2I}$

Or,    k = $(2 I)^{2020 -1}$                      [ using base exponent formula ]

∴       k = $(2 I)^{2019}$

Hence, The value oh k is $(2 I)^{2019}$    Answer