CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Open in App

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

Suggest Corrections

thumbs-up

0

Similar questions

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

A^{n} = ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦, n \in N

lim n \to \infty \frac{(2 n - 1) (3 n + 5)}{(n - 1) (3 n + 1) (3^{n} + 2^{n})} =

lim n \to \infty \frac{(2 n - 1) (3 n + 5)}{(n - 1) (3 n + 1)} =

\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}

Q. Prove that the

12^{t h}

power of any number is of the form

13 n

13 n + 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Transpose of a Matrix

MATHEMATICS

Watch in App

explore_more_icon

Explore more

NCERT - Standard XII

Transpose of a Matrix

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon