Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade 12
Question:
If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}$, $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{x}(A^2 + cA + d)$ then the sum of values of $c$ and $d$ is
Step-by-Step Solution
Key Concept: The Cayley-Hamilton theorem states that every matrix satisfies its own characteristic equation.
The characteristic equation for $A$ is $|A - zI| = 0$, which gives $\begin{vmatrix} 1-z & 0 & 0 \\ 0 & 1-z & 1 \\ 0 & -2 & 4-z \end{vmatrix} = 0$. Expanding: $(1-z)[(1-z)(4-z) + 2] = 0$, which simplifies to $(1-z)(z^2 - 6z + 11z - 6) = 0$. By the Cayley-Hamilton theorem, $A^3 - 6A^2 + 11A - 6I = 0$.
Correct Answer: 5