CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$A = \begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}$$, then $$A^{n}$$ (where $$n\epsilon N$$) equals.


A
[1na01]
loader
B
[1n2a01]
loader
C
[1na00]
loader
D
[nna0n]
loader

Solution

The correct option is A $$\begin{bmatrix}1 & na\\ 0 & 1\end{bmatrix}$$
Given,
$$A = \begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}$$.

Now,
$$A^2 = \begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}$$$$\begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}$$

or, $$A^2 = \begin{bmatrix}1 & 2a\\ 0 & 1\end{bmatrix}$$.

Again,

$$A^3 = \begin{bmatrix}1 & 2a\\ 0 & 1\end{bmatrix}$$$$  \begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}$$

or, $$A^3 = \begin{bmatrix}1 & 3a\\ 0 & 1\end{bmatrix}$$.

Proceeding in this way using mathematical induction we've,
$$A^n = \begin{bmatrix}1 & na\\ 0 & 1\end{bmatrix}$$.

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image