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Transpose of a Matrix

If A=[ 1 ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦$ , prove that $A^{n} = ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦, n \in N$

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Solution

It is given that $A = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦$
To show: $P (n) : A^{n} = ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦, n \in N$
We shall prove that the result by using the principle of mathematical induction .
For $n = 1$ , we have:
$P (1) : ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3^{0} 3^{0} 3^{0} \end{matrix} \begin{matrix} 3^{0} 3^{0} 3^{0} \end{matrix} \begin{matrix} 3^{0} 3^{0} 3^{0} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = A$
Therefore, the result is true for $n = 1$ .
Let the result be true for $n = k$ .
That
is $P (k) : A^{k} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} ⎤ ⎥ ⎥ ⎦$
Now, we prove that the result is true for $n = k + 1$ .
Now, $A^{k + 1} = A . A^{k}$
${⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦}^{k} ⎡ ⎢ ⎢ ⎣ \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} \begin{matrix} 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{matrix} ⎤ ⎥ ⎥ ⎦$
$= ⎡ ⎢ ⎢ ⎣ \begin{matrix} 3. 3^{k - 1} 3. 3^{k - 1} 3. 3^{k - 1} \end{matrix} \begin{matrix} 3. 3^{k - 1} 3. 3^{k - 1} 3. 3^{k - 1} \end{matrix} \begin{matrix} 3. 3^{k - 1} 3. 3^{k - 1} 3. 3^{k - 1} \end{matrix} ⎤ ⎥ ⎥ ⎦$
$= ⎡ ⎢ ⎢ ⎣ \begin{matrix} 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \end{matrix} \begin{matrix} 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \end{matrix} \begin{matrix} 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \end{matrix} ⎤ ⎥ ⎥ ⎦$
Therefore, the result is true for $n = k + 1$ .
Thus by the principle of mathematical induction , we have:
$A^{n} = ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} \begin{matrix} 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦, n \in N$

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