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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
If $\text{A} = \begin{bmatrix}3&-4\\1&-1\end{bmatrix},$then prove that $\text{A}'' = \begin{bmatrix}1 + 2n & -4n \\n & 1 - 2n \end{bmatrix}$ where n is any positive integer.
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Answer

Given: $\text{A}^{"}=\begin{bmatrix}1+2n&-4n\\n&1-2n\end{bmatrix}\ \ \ \therefore\ \text{A}^{n}=\begin{bmatrix}1+2n&-4n\\n&1-2n\end{bmatrix}$
$\Rightarrow \text{A}^{1}=\begin{bmatrix}1+2&-4\\1&1-2\end{bmatrix}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$which is true for n =1.
Now, $\text{A}^{k}=\begin{bmatrix}1+2k&-4k\\k&1-2k\end{bmatrix}...\text{(i)}$
Again $\text{A}^{k + 1}=\begin{bmatrix}1 + 2(k + 1)&-4(k + 1)\$k + 1)&1-2(k + 1)\end{bmatrix}...\text{(ii)}$
$\Rightarrow\text{A}^{k}.\text{A}=\begin{bmatrix}1+2(k+1)&-4(k+1)\$k+1)&1-2(k+1)\end{bmatrix}\begin{bmatrix}3&-4\\1&1\end{bmatrix}$ [From eq.(i)]
$\Rightarrow\text{ A}^{k}.\text{A}=\begin{bmatrix}3+6k-4k&-4-8k+4k\\3k+1-2k&-4k-1+2k\end{bmatrix}=\begin{bmatrix}3+2k&-4-4k\\1+k&-1-2k\end{bmatrix}$
$\Rightarrow \text{A}^{k}.\text{A=}\begin{bmatrix}1+2(k+1)&-4(k+1)\$k+1)&1-2(k+1)\end{bmatrix}$
Therefore, the result is true for n = k + 1.
Hence, by the principal of mathematical induction, the result is true for all positive integers n.

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