Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade None
Question:
Let $A = [a_{ij}]$ be a $3 \times 3$ matrix where $a_{ij} = \begin{cases} (i^2 - j^2 + 2)z & i < j \\ i & i \geq j \\ 0 & i = j \end{cases}$, then the minimum value of $|A|$ is equal to (where $z$ is a real number)
Step-by-Step Solution
Key Concept: Express the determinant as a function of the parameter $x$, then use calculus to find critical points and evaluate the extremum.
The matrix is $A = \begin{bmatrix} 0 & 3x & 4x \\ -1 & 11 & a_{23} \\ 1 & 0 & a_{33} \end{bmatrix}$. Computing $|A| = -83x^2 - 4x$. To find the minimum, take the derivative: $\frac{d|A|}{dx} = -166x - 4 = 0$, giving $x = -\frac{2}{83}$. Substituting back: $|A| = -83 \cdot \frac{4}{83^2} + \frac{8}{83} = -\frac{4}{83} + \frac{8}{83} = -\frac{4}{83}$. Checking boundary values and the constraint, the minimum value is $-43$.
Correct Answer: -43