Let the matrix mathrm A = left[ beginarrayl l l 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 endarray right]$\mathrm {A} = \left[ \begin{array}{l l l} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]$ satisfy mathrm A ^ mathrm n = mathrm A ^ mathrm n - 2 + mathrm A ^ 2 - mathrm I$\mathrm {A} ^ {\mathrm {n}} = \mathrm {A} ^ {\mathrm {n} - 2} + \mathrm {A} ^ {2} - \mathrm {I}$ for mathrm n geq 3$\mathrm {n} \geq 3$. Then the sum of all the elements of mathrmA^50$\mathrm{A}^{50}$ is :-

### Core Logic We are given the recurrence relation for the matrix power: A^n = A^n-2 + (A^2 - I)$$A^n = A^{n-2} + (A^2 - I)$$ Let's apply this equation successively down to base levels: - For n = 50$n = 50$: A^50 = A^48 + (A^2 - I)$A^{50} = A^{48} + (A^2 - I)$ - For n = 48$n = 48$: A^48 = A^46 + (A^2 - I) implies A^50 = A^46 + 2(A^2 - I)$A^{48} = A^{46} + (A^2 - I) \implies A^{50} = A^{46} + 2(A^2 - I)$ - For n = 46$n = 46$: A^50 = A^44 + 3(A^2 - I)$A^{50} = A^{44} + 3(A^2 - I)$ Following this telescoping reduction pattern down to A^2$A^2$: A^50 = A^2 + 24(A^2 - I) = 25A^2 - 24I$$A^{50} = A^2 + 24(A^2 - I) = 25A^2 - 24I$$ ### Step 1: Computing A^2 Let's perform matrix multiplication to find A^2$A^2$: A^2 = beginbmatrix 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 endbmatrix beginbmatrix 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 endbmatrix = beginbmatrix 1(1) & 0 & 0 \\ 1(1)+1(0) & 1(0)+1(1) & 0 \\ 1(0)+1(1) & 0 & 1(1) endbmatrix = beginbmatrix 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 endbmatrix$$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1(1) & 0 & 0 \\ 1(1)+1(0) & 1(0)+1(1) & 0 \\ 1(0)+1(1) & 0 & 1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$ ### Step 2: Calculating A^50 and Element Sum Substitute A^2$A^2$ back into our reduction formula: A^50 = 25beginbmatrix 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 endbmatrix - 24beginbmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 endbmatrix = beginbmatrix 25-24 & 0 & 0 \\ 25 & 25-24 & 0 \\ 25 & 0 & 25-24 endbmatrix = beginbmatrix 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 endbmatrix$$A^{50} = 25\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} - 24\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 25-24 & 0 & 0 \\ 25 & 25-24 & 0 \\ 25 & 0 & 25-24 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$ Now, sum all the individual element matrix fields: textSum = 1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$$\text{Sum} = 1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$$ ### Pattern Recognition When a matrix power formula contains a constant difference block like (A^2 - I)$(A^2 - I)$, treat it as an arithmetic progression step multiplier over successive matrix indices to bypass calculating high powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants