Related Formula
For a 2 × 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states:
M² - Tr(M)M + det(M)I = O$$M^2 - \operatorname{Tr}(M)M + \operatorname{det}(M)I = O$$
Also, |adj(M)| = |M|ⁿ⁻¹$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 × 2$2 \times 2$ matrix, |adj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
Core Logic
Using the characteristic equation A² - Tr(A)A + (A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$:
Comparing with A² - 4A + 2I = O$A^2 - 4A + 2I = O$:
Tr(A) = 4 ⇒ α + 2 = 4 ⇒ α = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$
Comparing with B² - 3B + I = O$B^2 - 3B + I = O$:
Tr(B) = 3 ⇒ 1 + β = 3 ⇒ β = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$
Step 1: Compute A³$A^3$ and B³$B^3$ via reduction
A = bmatrix 2 & 2 1 & 2 bmatrix$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$
A² = 4A - 2I$A^2 = 4A - 2I$
A³ = A(4A - 2I) = 4A² - 2A = 4(4A - 2I) - 2A = 14A - 8I$$A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$
A³ = 14 bmatrix 2 & 2 1 & 2 bmatrix - 8 bmatrix 1 & 0 0 & 1 bmatrix = bmatrix 28 & 28 14 & 28 bmatrix - bmatrix 8 & 0 0 & 8 bmatrix = bmatrix 20 & 28 14 & 20 bmatrix$$A^3 = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$
Similarly, B = bmatrix 1 & 1 1 & 2 bmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
B² = 3B - I$B^2 = 3B - I$
B³ = 3B² - B = 3(3B - I) - B = 8B - 3I$$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$
B³ = 8 bmatrix 1 & 1 1 & 2 bmatrix - bmatrix 3 & 0 0 & 3 bmatrix = bmatrix 8 & 8 8 & 16 bmatrix - bmatrix 3 & 0 0 & 3 bmatrix = bmatrix 5 & 8 8 & 13 bmatrix$$B^3 = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$
Step 2: Difference and Determinant
A³ - B³ = bmatrix 20 & 28 14 & 20 bmatrix - bmatrix 5 & 8 8 & 13 bmatrix = bmatrix 15 & 20 6 & 7 bmatrix$$A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$
(A³ - B³) = (15 × 7) - (20 × 6) = 105 - 120 = -15$$\det(A^3 - B^3) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$$
Step 3: Final Answer
For a 2 × 2$2 \times 2$ matrix, |adj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
Thus, (adj(A³ - B³)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$.
( (adj(A³ - B³)))² = (-15)² = 225$$(\det(\operatorname{adj}(A^3 - B^3)))^2 = (-15)^2 = 225$$
Pattern Recognition
Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M³$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic.
Chapter Mix
Class 12 Maths: Matrices