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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ for all positive integers n.
✓

Answer

Given,
$\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
To prove $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ we will use the principle of mathematical induction.
Step 1: Put n - 1
$\text{A}^1=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
So,
$A^n$ is true for n = 1
Step 2: Let, $A^n$​​​​​​​ be true for n = k, then
$\text{A}^\text{k}=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\ \dots(\text{i})$
Step 3: We have to show that $ \text{A}^\text{k+1}=\begin{bmatrix}1&\text{k}+1\\0&1\end{bmatrix}$
So,
$\text{A}^\text{k+1}=\text{A}^\text{k}\times\text{A}$
$=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$ {using equation (i) and given}
$ =\begin{bmatrix}1+0&1+\text{k}\\0+0&0+1\end{bmatrix}$
$\text{A}^{\text{k}+1}=\begin{bmatrix}1&1+\text{k}\\0&1\end{bmatrix}$
This shows that $A^n$​​​​​​​ is true for n = k + 1 whenever it is true for n = k
Hence, by the principle of mathematical induction $A^n$​​​​​​​ is true for all positive integer.

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