Question

Let $A$ and $B$ be two square matrices of order $3$ such that $AB=A$ and $BA=B.$ If $(A+B{)}^{10}=k(A+B),$ then the value of $k$ is

Answer 1

Given : $AB=A$ and $BA=B$
$ABA=A^2\Rightarrow AB=A^2\Rightarrow A^2=A$
Similarly, $B^2=B$
Now, $(A+B{)}^2=(A+B\left)\right(A+B)$
$=A^2+B^2+AB+BA$
$=A+B+A+B$
$=2(A+B)$

and $(A+B{)}^3=(A+B{)}^2(A+B)$
$=2(A+B)(A+B)$
$=2(A^2+B^2+AB+BA)$
$=2(A+B+A+B)$
$=4(A+B)$
Hence, $(A+B{)}^n=2^{n-1}(A+B)$
For $n=10,$ we have
$(A+B{)}^{10}=2^9(A+B)$
Hence, $k=512$