Question
If AB = BA for any two sqaure matrices, prove by mathematical induction that (AB)n = AnBn.
So, P(1) is true.
Now, P(k) : (AB)k = AkBk $\text{k}\in\text{N}$ So, P(k) is true, whenever P(k + 1) is true. $\therefore$ P(k + 1) ⇒ (AB)k+1 = Ak+1.Bk+1 $\therefore$ P(k + 1 : AB)k+1 = Ak+1Bk+1 ⇒ Ak.Bk.AB ⇒ Ak.BkBA ⇒ AkBk+1A ⇒ AkA.Bk+1 ⇒ Ak+1Bk+1 ⇒ (A.B)k+1 = Ak+1Bk+1 So, P(k+1) is true for all $\text{n}\in\text{N},$ whenever P(k) is true. By mathematical induction (AB) = AnBn is true for all $\text{n}\in\text{N}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{And B}=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix},$ find (AB)-1.
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