If $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\1&{ - 1}\end{array}} \right]$, then prove that ${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right],$where n is any positive integer.

.:

We have, $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\1&{ - 1}\end{array}} \right]$and ${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right]$

….(i)
For n = 1,
${A^1} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2 \times 1}&{ - 4 \times 1}\\1&{1 - 2 \times 1}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\1&{ - 1}\end{array}} \right] = A$

So, (i) is true for n = 1.

Assume that (i) is true for n = k i.e.,

${A^k} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right]$

Also, ${A^{k + 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2(k + 1)}&{ - 4(k + 1)}\\{k + 1}&{1 - 2(k + 1)}\end{array}} \right]$for $n = k + 1$.

$\Rightarrow$ ${A^{k + 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2k + 3}&{ - 4k - 4}\\{k + 1}&{ - 2k - 1}\end{array}} \right]$

Now, ${A^{k + 1}} = A \cdot {A^k} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\1&{ - 1}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2k}&{ - 4k}\\k&{1 - 2k}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3 + 6k - 4k}&{ - 12k - 4 + 8k}\\{1 + k - k}&{ - 4k - 1 + 2k}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2k + 3}&{ - 4k - 4}\\{k + 1}&{ - 2k - 1}\end{array}} \right]$ $= {A^{k + 1}}$

So, (i) is true for n = k + 1.

Hence, by mathematical induction ${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 2n}&{ - 4n}\\n&{1 - 2n}\end{array}} \right]$is true.