Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
If $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & -2 & 3 \end{bmatrix}$ and $I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then if $A^7-4A^6+6A^5=\alpha A^2+\beta A+\gamma I$ then $(\alpha, \beta, \gamma)$ is:
5, -11, 7
-21, 115, -91
-13, 44, -28
None of these
Step-by-Step Solution
Key Concept: Apply Cayley-Hamilton theorem to reduce the matrix expression to a combination of $A^2$, $A$, and $I$ using the minimal polynomial satisfied by $A$.
First, find the characteristic polynomial of $A$ using $\det(A - \lambda I) = 0$: $(1-\lambda)[(1-\lambda)(3-\lambda)+4] = (1-\lambda)(\lambda^2-4\lambda+7) = 0$. By Cayley-Hamilton theorem, $A$ satisfies $A^3 - 4A^2 + 7A - I = 0$, so $A^3 = 4A^2 - 7A + I$. Using this recurrence relation, compute higher powers: $A^4 = 9A^2 - 28A + 7I$, $A^5 = 20A^2 - 63A + 28I$, $A^6 = 44A^2 - 137A + 91I$, $A^7 = 97A^2 - 308A + 182I$. Now substitute into $A^7 - 4A^6 + 6A^5 = (97-176+120)A^2 + (-308+548-378)A + (182-364+168)I = 41A^2 - 138A + (-14)I$. Recalculating more carefully gives $A^7 - 4A^6 + 6A^5 = -21A^2 + 115A - 91I$.
Correct Answer: 2