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Home  >>  CBSE XII  >>  Math  >>  Matrices
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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 2 and $n \in N$.

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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.
Step 1: Given
$I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$A=\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$
Let us consider LHS
$(aI+bA)^n=\begin{bmatrix}a\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}\end{bmatrix}^n$
$\;\;\;\;\qquad\;\;\;=\begin{bmatrix}\begin{bmatrix}a & 0\\0 & a\end{bmatrix}+\begin{bmatrix}0 & b\\0 & 0\end{bmatrix}\end{bmatrix}^n$
$\;\;\;\;\qquad\;\;\;=\begin{bmatrix}a+0 & 0+b\\0+0 & a+0\end{bmatrix}^n$
$\;\;\;\;\qquad\;\;\;=\begin{bmatrix}a & b\\0 & a\end{bmatrix}^n$
Consider RHS
RHS$\Rightarrow a^nI+na^{n-1}bA$
$\Rightarrow a^n\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+na^{n-1}b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$
$\Rightarrow\begin{bmatrix} a^n & 0\\0 & a^n\end{bmatrix}+\begin{bmatrix}0 & na^{n-1}b\\0 & 0\end{bmatrix}$
$\Rightarrow\begin{bmatrix} a^n+0 & 0+na^{n-1}b\\0+0 & a^n+0\end{bmatrix}$
$=\begin{bmatrix} a^n & na^{n-1}b\\0 & a^n\end{bmatrix}$
Step :2 Now we have to prove that
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a^n & na^{n-1}b\\0 & a^n\end{bmatrix}$------(1)
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
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^1=\begin{bmatrix}a^1 & 1a^{1-1}b\\0 & a^1\end{bmatrix}$
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^1=\begin{bmatrix}a & 1a^0b\\0 & a^1\end{bmatrix}$
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^1=\begin{bmatrix}a & b\\0 & a\end{bmatrix}$[since $a^0=1$]
Hence P(n) is true for n=1 .Now prove p(n) where n=k substitute n=k in equ(1)
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^k=\begin{bmatrix}a^k & ka^{k-1}b\\0 & a^k\end{bmatrix}$
Multiply both the side by $\begin{bmatrix}a & b\\0 & a\end{bmatrix}$
Step 3: Consider the LHS
$\begin{bmatrix}a & b\\0 & a\end{bmatrix}^k\begin{bmatrix}a & b\\0 & a\end{bmatrix}=\begin{bmatrix}a & b\\0 & a\end{bmatrix}^{k+1}$
RHS:
$\begin{bmatrix}a^k & ka^{k-1}b\\0 & a^k\end{bmatrix}^k\begin{bmatrix}a & b\\0 & a\end{bmatrix}$
$=\begin{bmatrix}a^{k +1}& ba^k+ka^kb\\0 & a^{k+1}\end{bmatrix}$
$=\begin{bmatrix}a^{k +1}& (k+1)a^kb\\0 & a^{k+1}\end{bmatrix}$
Hence we have P(n) is true for n=k+1.
[Then by principle of mathematical induction p(n) is true for all positive integral value of n.]
answered Mar 18, 2013 by sharmaaparna1
edited Mar 19, 2013 by sreemathi.v
 

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