logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Matrices
0 votes

If $A=\begin{bmatrix}0 & 1\\0 &0\end{bmatrix}$ prove that for all $n\in N.(aI+bA)^n=a^nI+na^{n-1}bA.$where I is the identity matrix of order 2.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Here we will use the principle of mathematical induction.
Let $P(n)=(aI+bA)^n=a^nI+na^{n-1}bA$
Where A=$\begin{bmatrix}0 &1\\0 & 0\end{bmatrix}$ and I=$\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}$
When n=1
LHS=$(aI+bA)^1$
$\qquad=aI+bA.$
RHS=$a^1I+na^0bA$
$\qquad=aI+1.a^0bA$
$\qquad=aI+bA$
$\qquad=LHS$
P(1) is true.
Let P(m) be true.
$\Rightarrow (aI+bA)^m=a^mI+ma^{m-1}bA$------(1)
To prove P(m+1) is true.
(i.e)$(aI+bA)^{m+1}=a^{m+1}I+(m+1)a^mbA.$
Multiply both sides of (1) by aI+bA we get
$(aI+bA)^{m+1}=(aI+A).(aI+bA)^m$
$\qquad\qquad\quad=(aI+bA).(a^mI+ma^{m-1}bA)$
$\qquad\qquad\quad=aI(a^mI+ma^{m-1}bA)+bA(a^m1+ma^{m-1}bA)$
$\qquad\qquad\quad=a^{m+1}+ma^mbA+a^mbA+ma^{m-1}b^2A^2 \quad( I.I=I,IA=A=AI)$
$\qquad\qquad\quad=a^{m+1}I+(m+1)a^mbA.\quad( A^2=0)$
answered Apr 4, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...