Related Formula
|AB| = |A||B|$|AB| = |A||B|$
|Aⁿ| = |A|ⁿ$|A^n| = |A|^n$
Core Logic
Given A = bmatrix2 & 3 3 & 5 bmatrix$A = \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix}$.
The determinant |A| = (2)(5) - (3)(3) = 10 - 9 = 1$|A| = (2)(5) - (3)(3) = 10 - 9 = 1$.
We need to find the determinant of X = A²⁰²⁵ - 3A²⁰²⁴ + A²⁰²³$X = A^{2025} - 3A^{2024} + A^{2023}$.
Factor out the lowest power of A:
X = A²⁰²³(A² - 3A + I)$$X = A^{2023}(A^2 - 3A + I)$$
Step 1: Determinant Factoring
Using properties of determinants:
|X| = |A²⁰²³| · |A² - 3A + I|$$|X| = |A^{2023}| \cdot |A^2 - 3A + I|$$
Since |A| = 1$|A| = 1$, |A|²⁰²³ = 1$|A|^{2023} = 1$.
Therefore, |X| = |A² - 3A + I|$|X| = |A^2 - 3A + I|$.
Step 2: Matrix Computation
Calculate A²$A^2$:
A² = bmatrix2 & 3 3 & 5 bmatrix bmatrix2 & 3 3 & 5 bmatrix = bmatrix4+9 & 6+15 6+15 & 9+25 bmatrix = bmatrix13 & 21 21 & 34 bmatrix$$A^2 = \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix} \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix} = \begin{bmatrix}4+9 & 6+15 \\ 6+15 & 9+25\end{bmatrix} = \begin{bmatrix}13 & 21 \\ 21 & 34\end{bmatrix}$$
Calculate -3A$-3A$:
-3A = bmatrix-6 & -9 -9 & -15 bmatrix$$-3A = \begin{bmatrix}-6 & -9 \\ -9 & -15\end{bmatrix}$$
Assemble A² - 3A + I$A^2 - 3A + I$:
A² - 3A + I = bmatrix13 & 21 21 & 34 bmatrix + bmatrix-6 & -9 -9 & -15 bmatrix + bmatrix1 & 0 0 & 1 bmatrix$$A^2 - 3A + I = \begin{bmatrix}13 & 21 \\ 21 & 34\end{bmatrix} + \begin{bmatrix}-6 & -9 \\ -9 & -15\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$
= bmatrix13 - 6 + 1 & 21 - 9 + 0 21 - 9 + 0 & 34 - 15 + 1 bmatrix = bmatrix8 & 12 12 & 20 bmatrix$$= \begin{bmatrix}13 - 6 + 1 & 21 - 9 + 0 \\ 21 - 9 + 0 & 34 - 15 + 1\end{bmatrix} = \begin{bmatrix}8 & 12 \\ 12 & 20\end{bmatrix}$$
Step 3: Final Determinant
|A² - 3A + I| = vmatrix8 & 12 12 & 20 vmatrix$$|A^2 - 3A + I| = \begin{vmatrix}8 & 12 \\ 12 & 20\end{vmatrix}$$
= (8)(20) - (12)(12) = 160 - 144 = 16$$= (8)(20) - (12)(12) = 160 - 144 = 16$$
Pattern Recognition
High-powered matrix polynomials simplify radically by factoring out the lowest power. Since |A| = 1$|A| = 1$, the leading coefficient collapses completely, leaving only a basic A²$A^2$ evaluation.
Chapter Mix
Class 12 Maths: Matrices and Determinants