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Home  >>  CBSE XII  >>  Math  >>  Matrices
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if $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ prove that $A^n = \begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} , n \in N$.

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Toolbox:
  • The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$ j $\leq$ n.
  • We use the principle of mathematical induction, where we need to prove P(n) is true for n=1, n=k, n=k+1
Step 1: Let P(n)
Given
$A^n = \begin{bmatrix} 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{bmatrix} $
Given $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
For n=1.
Step 2: LHS:$A^n=A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
RHS:Substitute n=1
$\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}\Rightarrow \begin{bmatrix} 3^0 & 3^0& 3^0 \\ 3^0 & 3^0 & 3^0\\ 3^0 & 3^0 & 3^0 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
LHS=RHS
Hence P(n) is true for n=1
Now consider P(n)$\Rightarrow n=k$
Hence now prove P(n) to be true for n=k
$A^k=\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix} $ [Relplacing n=k]
Multiply A on both sides
$A^k.A=\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
Step 3: LHS:$A^k.A\Rightarrow A^{k+1}$
RHS:
$\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 3^{k-1}+3^{k-1}+3^{k-1} & 3^{k-1}+3^{k-1} +3^{k-1}& 3^{k-1} +3^{k-1}+3^{k-1}\\ 3^{k-1} +3^{k-1}+3^{k-1}& 3^{k-1}+3^{k-1}+3^{k-1} & 3^{k-1}+3^{k-1}+3^{k-1} \\ 3^{k-1} +3^{k-1}+3^{k-1}& 3^{k-1}+3^{k-1}+3^{k-1} & 3^{k-1}+3^{k-1}+3^{k-1} \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 3.3^{k-1} &3. 3^{k-1} &3. 3^{k-1} \\3. 3^{k-1} &3. 3^{k-1} & 3.3^{k-1} \\ 3.3^{k-1} &3. 3^{k-1} & 3.3^{k-1} \end{bmatrix}$
$\Rightarrow\begin{bmatrix}3^k & 3^k&3^k\\3^k & 3^k&3^k\\3^k & 3^k&3^k\end{bmatrix}$
$\Rightarrow \begin{bmatrix} 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \end{bmatrix}$
P(n) is true for n=k+1.
By principle of mathematical induction that P(n) is true for all $n\in N$

 

answered Mar 3, 2013 by sharmaaparna1
edited Mar 20, 2013 by sharmaaparna1
 

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