Let A = beginbmatrix alpha & -1 \\ 6 & beta endbmatrix$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$, alpha > 0$\alpha > 0$, such that det(A) = 0$\det(A) = 0$ and alpha + beta = 1$\alpha + \beta = 1$. If I$I$ denotes the 2 times 2$2 \times 2$ identity matrix, then the matrix (I + A)^8$(I + A)^8$ is:

### Related Formula For a matrix satisfying A^2 = A$A^2 = A$ (Idempotent Matrix): (I+A)^n = I + (2^n - 1)A$$(I+A)^n = I + (2^n - 1)A$$ ### Core Logic Given det(A) = alphabeta + 6 = 0 implies alphabeta = -6$\det(A) = \alpha\beta + 6 = 0 \implies \alpha\beta = -6$ and alpha + beta = 1$\alpha + \beta = 1$. Solving these gives alpha = 3, beta = -2$\alpha = 3, \beta = -2$ (since alpha > 0$\alpha > 0$). ### Step 1: Check Powers of A Substitute values into A$A$: A = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix$$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$$ Compute A^2$A^2$: A^2 = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 9-6 & -3+2 \\ 18-12 & -6+4 endbmatrix = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = A$$A^2 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 9-6 & -3+2 \\ 18-12 & -6+4 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = A$$ ### Step 2: Expand Matrix Expression Since A^2 = A$A^2 = A$, it follows that A^n = A$A^n = A$ for all integers n ge 1$n \ge 1$. (I + A)^8 = I + sum_k=1^8 binom8k A^k = I + A sum_k=1^8 binom8k = I + (2^8 - 1)A = I + 255A$$(I + A)^8 = I + \sum_{k=1}^{8} \binom{8}{k} A^k = I + A \sum_{k=1}^{8} \binom{8}{k} = I + (2^8 - 1)A = I + 255A$$ ### Step 3: Construct the Final Matrix (I + A)^8 = beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix + 255 beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 1 + 765 & -255 \\ 1530 & 1 - 510 endbmatrix = beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix$$(I + A)^8 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 + 765 & -255 \\ 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$$ ### Pattern Recognition Whenever texttr(A) = 1$\text{tr}(A) = 1$ and det(A) = 0$\det(A) = 0$ for a 2 times 2$2 \times 2$ matrix, Cayley-Hamilton theorem gives A^2 - texttr(A)A + det(A)I = 0 implies A^2 = A$A^2 - \text{tr}(A)A + \det(A)I = 0 \implies A^2 = A$. Thus A$A$ is idempotent, simplifying polynomial expansions exponentially. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants