Question

if AandB are matrices of the same order and AB =BA, then prove that (i) A²-B²=(A-B) (A+B) (ii) A²+2AB+B²=(A+B) ² (iii) A²-2AB+B²=(A-B) ²​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

If A and B are square matrices of same order and AB = BA

Prove that,

(i) A²-B²=(A-B) (A+B)

(ii) A²+2AB+B²=(A+B) ²

(iii) A²-2AB+B²=(A-B) ²

$\large\underline{\sf{Solution-}}$

Given that,

A and B are square matrices of same order such that AB = BA

Consider,

$\rm :\longmapsto\:(A + B)(A - B)$

$\rm \:  =  \: {A}^{2} - AB + BA - {B}^{2}$

$\rm \:  =  \: {A}^{2} - AB + AB - {B}^{2}$

$\red{\bigg \{ \because \: AB = BA\bigg \}}$

$\rm \:  =  \: {A}^{2} - {B}^{2}$

Hence, Proved

Consider

$\rm :\longmapsto\: {(A + B)}^{2}$

$\rm \:  =  \: (A + B)(A + B)$

$\rm \:  =  \: {A}^{2} + AB + BA + {B}^{2}$

$\rm \:  =  \: {A}^{2} + AB + AB + {B}^{2}$

$\red{\bigg \{ \because \: AB = BA\bigg \}}$

$\rm \:  =  \: {A}^{2} + 2AB + {B}^{2}$

Hence, Proved

Consider

$\rm :\longmapsto\: {(A - B)}^{2}$

$\rm \:  =  \: (A - B)(A - B)$

$\rm \:  =  \: {A}^{2} - AB - BA + {B}^{2}$

$\rm \:  =  \: {A}^{2} - AB - AB + {B}^{2}$

$\red{\bigg \{ \because \: AB = BA\bigg \}}$

$\rm \:  =  \: {A}^{2} - 2AB + {B}^{2}$

Hence, Proved