Matrices & Determinants
Infinite Solutions + Series Summation
nta_pyq_2025_apr
Grade 12
Question:
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 $\displaystyle\sum_{n=1}^{10}(n^\alpha + n^\beta)$ is equal to:
Step-by-Step Solution
Key Concept: Set $\Delta = 0$ (always holds for this system), then use the condition from augmented determinants to find $m$ values satisfying $m^2 - 3m + 2 = 0$.
$\Delta = 0$ always. Setting $\Delta_z = 0$ (or $\Delta_y = 0$): $m^2 - 3m + 2 = 0 \Rightarrow m = 1, 2$. So $\alpha = 1, \beta = 2$. $\sum_{n=1}^{10}(n^1 + n^2) = \frac{10 \times 11}{2} + \frac{10 \times 11 \times 21}{6} = 55 + 385 = 440$.
Correct Answer: 440