Show that if A and B are square matrices such that $A B = B A$ $AB = BA$ then: ${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$ $(A+B)^2=A^2+2AB+B^2$ ?

Question

12299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T10:19:34

Kindly refer to the Explanation.

Since $A and B$ $A and B$ are square matrices, all the multiplications

reqd. in the Question are defined.

Now, ${(A + B)}^{2} = (A + B) \cdot (A + B)$ $(A+B)^2=(color(red)A+color(red)B)*(color(green)A+color(green)B)$ ,

$= A (A + B) + B (A + B)$ $=color(red)A(color(green)(A+B))+color(red)B(color(green)(A+B))$ ,

$= A \cdot A + A \cdot B + B \cdot A + B \cdot B$ $=A*A+A*B+B*A+B*B$ ,

$=A^2+A*B+A*B+B^2.......[because, A*B=B*A," Given]"$ ,

$=A^2+2A*B+B^2, i.e.,$

$=A^2+2AB+B^2$ , as desired!

Show that if A and B are square matrices such that AB = BAAB=BAAB = BA then: (A+B)^2=A^2+2AB+B^2(A+B)2=A2+2AB+B2(A+B)^2=A^2+2AB+B^2?