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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Let \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) , show that \( (aI + bA)^n = a^nI+na^{n-1}bA, \) where I is the identity matrix of order zero \( n \in N. \) If $ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} $ prove that : $ \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 us consider LHS :
$(aI+bA)^n=\begin{bmatrix}a\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}\end{bmatrix}^n$
$\qquad\qquad=\begin{bmatrix}\begin{bmatrix}a & 0\\0 & a\end{bmatrix}+\begin{bmatrix}0 & b\\0 & 0\end{bmatrix}\end{bmatrix}^n$
$\qquad\qquad=\begin{bmatrix}a+0 & 0+b\\0+0 & a+0\end{bmatrix}^n$
$\qquad\qquad=\begin{bmatrix}a & b\\0 & a\end{bmatrix}^n$
Consider the RHS:
RHS$\Rightarrow a^nI+na^{n-1}bA$
$\Rightarrow a^n\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+na^{n-1}b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$
$=\begin{bmatrix}a^n & 0\\0 & a^n\end{bmatrix}+\begin{bmatrix}0 & na^{n-1}b\\0 & 0\end{bmatrix}$
$=\begin{bmatrix}a^n + 0 & 0+na^{n-1}b\\0+0 & a^n+0\end{bmatrix}$
$=\begin{bmatrix}a^n & na^{n-1}b\\0 & a^n\end{bmatrix}$
Now we have to prove that
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^n=\begin{bmatrix}a^n & na^{n-1}b\\0 & a^n\end{bmatrix}$------(1)
Step 2:
In order to prove that let us use the concept of mathematical induction.
Hence put P(n) where n=1,n=k.
Consider P(1)$\Rightarrow n=1$.
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^1=\begin{bmatrix}a^1 &1(a)^{1-1}b\\0 & a^1\end{bmatrix}$
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a &1(a)^{0}b\\0 & a^1\end{bmatrix}$
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a &b\\0 & a\end{bmatrix}$
Since $a^0=1$
Hence P(n) is true for n=1.
Now prove P(n) where n=k.
Substitute n=k in equation (1)
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^k=\begin{bmatrix}a^k &k(a)^{k-1}b\\0 & a^k\end{bmatrix}$
Multiply both side by $\begin{bmatrix}a & b\\0 & a\end{bmatrix}$
Consider the LHS :
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a & b\\0 & a\end{bmatrix}^{k+1}$
RHS :$\begin{bmatrix}a^k &ka^{k-1}b\\0 & a^k\end{bmatrix}\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a^{k+1} &ba^k+ka^kb\\0 & a^{k+1}\end{bmatrix}$
$\Rightarrow \begin{bmatrix}a^{k+1} & (k+1)a^kb\\0 &a^{k+1}\end{bmatrix}$
Hence we have P(n) is true for $n=k+1.$
Step 3:
$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} $
Let $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
For n=1.
LHS:$A^n=A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
$\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}\Rightarrow \begin{bmatrix} 3^0 & 3^0& 3^0 \\ 3^0 & 3^0 & 3^0\\ 3^0 & 3^0 & 3^0 \end{bmatrix}\Rightarrow\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
LHS=RHS
Hence P(n) is true for n=1
Step 4:
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
$A^k.A=\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
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 Apr 18, 2013 by sreemathi.v
 

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