Let $ \alpha, \beta \ (\alpha \neq \beta) $ be the values of $ m $, for which the equations $ x+y+z=1 $, $ x+2y+4z=m $ and $ x+4y+10z=m^2 $ have infinitely many solutions. Then the value of $ \sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta}) $ is equal to :
Solution
<p>$$\begin{aligned}
&\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 4 & 10
\end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\
& =4-6+2=0
\end{aligned}\\
&\text { For infinite solutions }\\
&\begin{aligned}
& \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\
& \mathrm{~m}^2-3 \mathrm{x}+2=0 \\
& \mathrm{~m}=1,2 \\
& \alpha=1, \beta=2 \\
& \therefore \sum_{\mathrm{n}=1}^{10}\left(\mathrm{n}^\alpha+\mathrm{n}^\beta\right)=\sum_{\mathrm{n}=1}^{10} \mathrm{n}^1+\sum_{\mathrm{n}=1}^{10} \mathrm{n}^2 \\
&= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\
&= 55+385 \\
&= 440
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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