Matrices & Determinants
Determinants
nta_pyq_2025_jan
Grade 12
Question:
Let $\alpha,\beta\,(\alpha\ne\beta)$ be the values of $m$ for which the equations $x+y+z=1,\ x+2y+4z=m,\ x+4y+10z=m^{2}$ have infinitely many solutions. Then $\displaystyle\sum_{n=1}^{10}(n^{\alpha}+n^{\beta})$ is equal to:
Step-by-Step Solution
Key Concept: $D=0$ automatically (rows linearly dependent because $1,2,4$ and $1,4,10$ obey $z=$ linear function). Need $D_{x}=0$ (a quadratic in $m$) to find $\alpha,\beta.$ Then use $\sum n=n(n+1)/2$ and $\sum n^{2}=n(n+1)(2n+1)/6.$
$D=\begin{vmatrix}1&1&1\\1&2&4\\1&4&10\end{vmatrix}=1(20-16)-(10-4)+(4-2)=0.$ True for all $m.$
$D_{x}=\begin{vmatrix}1&1&1\\m&2&4\\m^{2}&4&10\end{vmatrix}=4-10m+4m^{2}+4m-2m^{2}=2m^{2}-6m+4=2(m-1)(m-2).$
$D_{x}=0\Rightarrow m=1$ or $m=2.$ So $\alpha=1,\beta=2.$
$\sum_{n=1}^{10}(n+n^{2})=\dfrac{10\cdot 11}{2}+\dfrac{10\cdot 11\cdot 21}{6}=55+385=440.$
Correct Answer: 4