If *A* and *B* are square matrices of the same order such that *AB* = *BA*, then prove by induction that AB" = B"A. Further, prove that (AB)" = A"B" for all *n* ∈ **N**

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#### Solution

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

To prove P(n) : AB" = B"A, `n in N`

For *n* = 1, we have:

Therefore, the result is true for *n* = 1.

Let the result be true for *n* = *k*.

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have AB" = B"A, `n in N`

Now, we prove that (AB)" = A"B" for all *n* ∈ **N**

For *n* = 1, we have:

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have (AB)" = A"B", for all natural numbers.

Concept: Types of Matrices

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