Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 39 Question 39 of 472 Medium MCQ
Let α , β ( α ≠ β ) \alpha ,\beta (\alpha \ne \beta ) α , β ( α = β ) be the values of m m m , for which the equations x + y + z = 1 x+y+z=1 x + y + z = 1 , x + 2 y + 4 z = m x+2y+4z=m x + 2 y + 4 z = m and x + 4 y + 10 z = m 2 x+4y+10z={m}^{2} x + 4 y + 10 z = m 2 have infinitely many solutions. Then the value of ∑ n = 1 10 ( n α + n β ) \sum _{n=1}^{10}({n}^{\alpha }+{n}^{\beta }) ∑ n = 1 10 ( n α + n β ) is equal to :
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\[ \begin{aligned}
&\begin{aligned}
\Delta & =\left|\[\begin{array}{|ccccccc|}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10\end{array}\end{aligned} \end{aligned}\]\right|=1(20-16)-1(10-4)+1(4-2) \\
\& =4-6+2=0
\\
\&\text { For infinite solutions }\\
\& Δ _ x = Δ _ y = Δ _ z = 0 m 2 ˆ − 3 x + 2 = 0 m = 1 , 2 α = 1 , β = 2 ∴ ∑ _ n = 1 10 ˆ ( n α ˆ + n β ˆ ) = ∑ _ n = 1 10 ˆ n 1 ˆ + ∑ _ n = 1 10 ˆ n 2 ˆ = 10 ( 11 ) 2 + 10 ( 11 ) ( 21 ) 6 = 55 + 385 = 440 \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} Δ_ x = Δ_ y = Δ_ z = 0 m 2 ˆ − 3 x + 2 = 0 m = 1 , 2 α = 1 , β = 2 ∴ ∑ _ n = 1 10 ˆ ( n α ˆ + n β ˆ ) = ∑ _ n = 1 10 ˆ n 1 ˆ + ∑ _ n = 1 10 ˆ n 2 ˆ = 2 10 ( 11 ) + 6 10 ( 11 ) ( 21 ) = 55 + 385 = 440
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