Matrices & Determinants
Determinant with Limit
nta_pyq_2025_apr
Grade 12
Question:
For some $a$, $b$, let $f(x) = \begin{vmatrix}a + \frac{\sin z}{z} & 1 & b \\ a & 1 + \frac{\sin z}{z} & b \\ a & 1 & b + \frac{\sin z}{z}\end{vmatrix}$, $x \neq 0$, $\lim_{z \to 0} f(x) = \lambda + \mu a + \nu b$. Then $(\lambda + \mu + \nu)^2$ is equal to:
Step-by-Step Solution
Key Concept: Use $\lim_{z \to 0} \frac{\sin z}{z} = 1$ to replace the determinant, then apply row/column operations to evaluate it.
At $z \to 0$: $\frac{\sin z}{z} \to 1$. After row ops $R_1 \to R_1 - R_2$, $R_2 \to R_2 - R_3$: determinant evaluates to $a + b + 2$. So $\lambda = 2, \mu = 1, \nu = 1$. $(\lambda + \mu + \nu)^2 = (2+1+1)^2 = 16$.
Correct Answer: 16