Matrices & Determinants
Determinants
nta_pyq_2025_jan
Grade 12
Question:
For some $a,b$, let $f(x)=\begin{vmatrix}a+\dfrac{\sin x}{x} & 1 & b\\ a & 1+\dfrac{\sin x}{x} & b\\ a & 1 & b+\dfrac{\sin x}{x}\end{vmatrix},\,x\ne 0.$ If $\displaystyle\lim_{x\to 0}f(x)=\lambda+\mu a+\nu b$, then $(\lambda+\mu+\nu)^{2}$ equals:
Step-by-Step Solution
Key Concept: $\lim_{x\to 0}\sin x/x=1$, so the determinant becomes $\begin{vmatrix}a+1&1&b\\a&2&b\\a&1&b+1\end{vmatrix}$. Reduce by row/column operations.
$\sin x/x\to 1$, so the determinant becomes $\begin{vmatrix}a+1&1&b\\a&2&b\\a&1&b+1\end{vmatrix}.$
$R_{1}\to R_{1}-R_{2},\,R_{2}\to R_{2}-R_{3}$:
$\begin{vmatrix}1&-1&0\\0&1&-1\\a&1&b+1\end{vmatrix}.$
Expand along $R_{1}$: $1\bigl[(b+1)\cdot 1-(-1)(a+1)\bigr]-(-1)\bigl[0-(-1)\cdot a\bigr]+0$
$= (b+1)+(a+1)+a=2a+b+2.$
Hmm — let's redo via $C_{2}\to C_{1}+C_{2}$ from the row-reduced form $\begin{vmatrix}1&-1&0\\0&1&-1\\a&1&b+1\end{vmatrix}$:
$\begin{vmatrix}1&0&0\\0&1&-1\\a&a+1&b+1\end{vmatrix}=1\cdot[(b+1)-(-1)(a+1)]=a+b+2.$
So $\lambda=2,\,\mu=1,\,\nu=1\Rightarrow(\lambda+\mu+\nu)^{2}=16.$
Correct Answer: 1