4. Hisoblang. $$ \left[\begin{array}{cccc}1 & 1 & 0 & 0 \ -2 & 1 & 1 & 0 \ 0 & -2 & 1 & 1 \ 0 & 0 & -2 & 1\end{array} ight] $$

Question

4. Hisoblang. $$ \left[\begin{array}{cccc}1 & 1 & 0 & 0 \ -2 & 1 & 1 & 0 \ 0 & -2 & 1 & 1 \ 0 & 0 & -2 & 1\end{array}ight] $$

UpStudy AI · Accepted Answer

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.
Matritsaning determinanti 11 ga teng.

Hisoblang.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions