4. Hisoblang. \( \left[\begin{array}{cccc}1 & 1 & 0 & 0 \\ -2 & 1 & 1 & 0 \\ 0 & -2 & 1 & 1 \\ 0 & 0 & -2 & 1\end{array}\right] \)

Berilgan matritsaning determinantini hisoblaymiz: \[ \begin{vmatrix} 1 & 1 & 0 & 0 \\ -2 & 1 & 1 & 0 \\ 0 & -2 & 1 & 1 \\ 0 & 0 & -2 & 1 \end{vmatrix} \] Bu tridiagonal matritsa bo‘lib, asoslash uchun rekursiv usuldan foydalanamiz. Har bir kichik determinant quyidagicha hisoblanadi: \[ D_n = a_n D_{n-1} - b_{n-1}c_{n-1} D_{n-2} \] Bu yerda: - \( D_n \) – \( n \times n \) o‘lchamdagi matritsaning determiantasi, - \( a_n \) – asosiy diagonaldagi element, - \( b_{n-1} \) – yuqori diagonaldagi element, - \( c_{n-1} \) – pastki diagonaldagi element. Bizning matritsamiz uchun: - \( a_i = 1 \), - \( b_i = 1 \), - \( c_i = -2 \). Bosqichma-bosqich hisoblaymiz: 1. **\( D_1 \):** \[ D_1 = 1 \] 2. **\( D_2 \):** \[ D_2 = 1 \cdot D_1 - 1 \cdot (-2) \cdot D_0 = 1 \cdot 1 - 1 \cdot (-2) \cdot 1 = 1 + 2 = 3 \] (Chunki \( D_0 = 1 \) deb olinadi.) 3. **\( D_3 \):** \[ D_3 = 1 \cdot D_2 - 1 \cdot (-2) \cdot D_1 = 1 \cdot 3 - 1 \cdot (-2) \cdot 1 = 3 + 2 = 5 \] 4. **\( D_4 \):** \[ D_4 = 1 \cdot D_3 - 1 \cdot (-2) \cdot D_2 = 1 \cdot 5 - 1 \cdot (-2) \cdot 3 = 5 + 6 = 11 \] Shunday qilib, berilgan matritsaning determinant qiymati **11** ga teng.