Algebra of Matrices — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
If $A =$ diag$(a, b, c),$ show that $A^n =$ diag$(a^n, b^n, c^n)$ for all positive integer $n.$
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Answer
Given, $A =$ diag$(a, b, c)$ Show that, $A^n =$ diag$(a^n, b^n, c^n)$ Step 1: Put $n = 1 A^1 =$ diag$(a^1, b^1, c^1) A =$ diag$(a, b, c)$ So, $A^n$ is true for $n = 1$ Step 2: Let, An be true for $n = k,$ so, $A^k =$ diag$(a^k, b^k, c^k) ...(i)$ Step 3: Now, we have to show that, $A^{k+1} =$ diag$(a^{k+1}, b^{k+1}, c^{k+1})$
Now,$ A^{k+1} = A^k \times k^3 =$ diag$(a^k, b^k, c^k) \times$ diag$(a, b, c)$
{using equation (i) and given}$\text{A}^{\text{k}+1}=\begin{bmatrix}\text{a}^\text{k}&0&0\\0&\text{b}^\text{k}&0\\0&0&\text{c}^\text{k}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{a}^\text{k}\times\text{a}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{b}^\text{k}\times\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{c}^\text{k}\times\text{c}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^\text{k+1}&0&0\\0&\text{b}^\text{k+1}&0\\0&0&\text{c}^\text{k+1}\end{bmatrix}$
$\text{A}^{\text{k}+1}=\text{diag}\big(\text{a}^{\text{k}+1},\text{b}^{\text{k}+1},\text{c}^{\text{k}+1}\big)$
So, $P(n)$ is true for $n = k + 1$ whenever P(n) is true for $n = k$
Hence, by principle of mathematical induction $A^n$ is true for all positive integer.
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