Question

Proof this theorem
If A and B are square matrices of the same order then-
(A+B)^2=A^2+AB+BA+B^2.
Also when AB=BA then (A+B)^2=A^2+2AB+B^2​

Answer 1

$\huge \mathtt { \purple{SOLUTION:-}}$

Let A and B be n-rowed square matrices.

Then, clearly,(A+B) is a square matrix of order n.

So,$\large(a + b) ^{2}$is defined

Now,

$(a + b) ^{2} = (a + b).(a + b)$

$\implies {a.(a + b) + b.(a + b)}$

$\implies{aa + ab + ba + bb}$

$\implies{ {a}^{2} + ab + ba + {b}^{2} }$

Hence,

$\implies{ ( {a + b)}^{2} = ( {a}^{2} + ab + ba + {b}^{2}) }$

Particular case-

When AB=BA

In this case,we have

$\implies(a + b) ^{2} = ( {a}^{2} + ab + ab + {b}^{2} )$

$\implies{ ({ a}^{2} + 2ab + {b}^{2} )}$

[Hence BA= AB]

$\large{ \mathtt{ \pink{Hence \: proved}}}$

Answer 2

${ \bold{ \boxed{ \boxed{ \red{ \mathtt{ \bigstar \: solution \: \bigstar}}}}}}$

${ \bold { \orange{ \mathtt{Let \: \: A \: \: and \: \: B \: \: be \: \: n-rowed \: \: square \: \: matrices.}}}} \\ \\ </p><p>{ \bold { \orange{ \mathtt{</p><p>Then, \: \: clearly \: \: ,(A+B) \: \: is \: \: a \: \: square \: \: matrix \: \: of \: \: order \: \: n.}}}} \\ \\ { \bold { \orange{ \mathtt{so \: \: A + B \: \: defined \: \: as - }}}} \\ \\ { \bold{ \orange{ \mathtt{ \implies{(A + B) }^{2} = (A + B)(A + B)}}}} \\ \\ { \bold{ \orange{ \mathtt{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {A}^{2} + AB + BA + {B}^{2} }}}} \\ \\ { \bold{ \blue{ \mathtt{but \:AB= BA \:, \: so \: \: that - }}}} \\ \\ { \bold{ \boxed{ \orange{ \mathtt{ {(A+B)}^{2} = {A}^{2} + 2 AB + {B}^{2} }}}}} \\ \\ { \bold{ \boxed{ \red { \huge{ \mathtt{ Hence \: \: Proved }}}}}}$