1.2 Matritsa va Chiziqli Algebra — Masalalar

Jami: 30 ta | Yechim bilan: ✅

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📊 Qiyinlik Darajalari

• ⭐⭐ Asosiy (1-10)
• ⭐⭐⭐ O'rtacha (11-22)
• ⭐⭐⭐⭐ Murakkab (23-30)

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Asosiy Masalalar (1-10)

Masala 1 ⭐⭐

Quyidagi matritsalarni qo'shing va ayiring:

$$A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$$

Yechim

$A + B = \begin{bmatrix} 2+5 & 3+(-1) \\ 1+2 & 4+3 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 3 & 7 \end{bmatrix}$

$A - B = \begin{bmatrix} 2-5 & 3-(-1) \\ 1-2 & 4-3 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ -1 & 1 \end{bmatrix}$

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Masala 2 ⭐⭐

$AB$ va $BA$ ni hisoblang:

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

Yechim

$AB = \begin{bmatrix} 1(0)+2(1) & 1(1)+2(0) \\ 3(0)+4(1) & 3(1)+4(0) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$

$BA = \begin{bmatrix} 0(1)+1(3) & 0(2)+1(4) \\ 1(1)+0(3) & 1(2)+0(4) \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$

$AB \neq BA$ — matritsa ko'paytirish kommutativ emas!

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Masala 3 ⭐⭐

Quyidagi matritsaning determinantini toping:

$$A = \begin{bmatrix} 3 & 8 \\ 4 & 6 \end{bmatrix}$$

Yechim

$\det(A) = 3(6) - 8(4) = 18 - 32 = -14$

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Masala 4 ⭐⭐

3×3 matritsa determinantini hisoblang:

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$

Yechim

Sarrus qoidasi: $\det(A) = 1(5)(9) + 2(6)(7) + 3(4)(8) - 3(5)(7) - 2(4)(9) - 1(6)(8)$ $= 45 + 84 + 96 - 105 - 72 - 48 = 0$

Determinant 0, ya'ni matritsa singular (teskari matritsasi yo'q).

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Masala 5 ⭐⭐

Quyidagi matritsaning teskari matritsasini toping:

$$A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$$

Yechim

$\det(A) = 2(3) - 1(5) = 6 - 5 = 1$

$A^{-1} = \frac{1}{1}\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$

Tekshirish: $AA^{-1} = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \checkmark$

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Masala 6 ⭐⭐

Transpozitsiyani toping:

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$

Yechim

$A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

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Masala 7 ⭐⭐

Tenglamalar sistemasini matritsa usulida yeching:

$$\begin{cases} 2x + 3y = 7 \\ x + 2y = 4 \end{cases}$$

Yechim

$A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 7 \\ 4 \end{bmatrix}$

$\det(A) = 4 - 3 = 1$

$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

$\vec{x} = A^{-1}\vec{b} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 7 \\ 4 \end{bmatrix} = \begin{bmatrix} 14-12 \\ -7+8 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

Javob: $x = 2$, $y = 1$

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Masala 8 ⭐⭐

$A^2$ ni hisoblang:

$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

Yechim

$A^2 = A \cdot A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$

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Masala 9 ⭐⭐

Quyidagi matritsa ortogonalmi tekshiring:

$$A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$

Yechim

Ortogonal bo'lishi uchun: $A^TA = I$

$A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$

$A^TA = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^2\theta + \sin^2\theta & -\cos\theta\sin\theta + \sin\theta\cos\theta \\ -\sin\theta\cos\theta + \cos\theta\sin\theta & \sin^2\theta + \cos^2\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Ha, ortogonal matritsa (aylanish matritsasi).

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Masala 10 ⭐⭐

Quyidagi matritsaning rangi (rank) nima?

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix}$$

Yechim

Qatorlarni tahlil qilsak: 2-qator = 2 × (1-qator)

Gauss eliminatsiya: $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$

Nolmas qatorlar soni = 2

Rang = 2

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O'rtacha Masalalar (11-22)

Masala 11 ⭐⭐⭐

Xos qiymatlar va xos vektorlarni toping:

$$A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}$$

Yechim

Xarakteristik tenglama: $\det(A - \lambda I) = \det\begin{bmatrix} 3-\lambda & 1 \\ 1 & 3-\lambda \end{bmatrix} = (3-\lambda)^2 - 1 = 0$

$\lambda^2 - 6\lambda + 8 = 0$ $(\lambda - 4)(\lambda - 2) = 0$ $\lambda_1 = 4, \quad \lambda_2 = 2$

$\lambda_1 = 4$ uchun xos vektor: $(A - 4I)\vec{v} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}\vec{v} = 0$ $\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

$\lambda_2 = 2$ uchun xos vektor: $(A - 2I)\vec{v} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\vec{v} = 0$ $\vec{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

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Masala 12 ⭐⭐⭐

Kramer qoidasi bilan yeching:

$$\begin{cases} 3x + 2y - z = 1 \\ 2x - 2y + 4z = -2 \\ -x + \frac{1}{2}y - z = 0 \end{cases}$$

Yechim

$A = \begin{bmatrix} 3 & 2 & -1 \\ 2 & -2 & 4 \\ -1 & 0.5 & -1 \end{bmatrix}$

$\det(A) = 3(2-2) - 2(-2+4) + (-1)(1+2) = 0 - 4 - 3 = -7$

$\det(A_x) = \begin{vmatrix} 1 & 2 & -1 \\ -2 & -2 & 4 \\ 0 & 0.5 & -1 \end{vmatrix} = 1(2-2) - 2(2-0) + (-1)(-1-0) = 0 - 4 + 1 = -3$

$x = \frac{\det(A_x)}{\det(A)} = \frac{-3}{-7} = \frac{3}{7}$

(Qolgan o'zgaruvchilar ham shunga o'xshash hisoblanadi)

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Masala 13 ⭐⭐⭐

Gauss eliminatsiyasi bilan yeching:

$$\begin{cases} x + y + z = 6 \\ 2x + y - z = 1 \\ x - y + 2z = 5 \end{cases}$$

Yechim

$\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 1 & -1 & | & 1 \\ 1 & -1 & 2 & | & 5 \end{bmatrix}$

$R_2 - 2R_1$, $R_3 - R_1$: $\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -1 & -3 & | & -11 \\ 0 & -2 & 1 & | & -1 \end{bmatrix}$

$R_3 - 2R_2$: $\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -1 & -3 & | & -11 \\ 0 & 0 & 7 & | & 21 \end{bmatrix}$

Orqaga almashtirish:

• $7z = 21 \Rightarrow z = 3$
• $-y - 9 = -11 \Rightarrow y = 2$
• $x + 2 + 3 = 6 \Rightarrow x = 1$

Javob: $x = 1$, $y = 2$, $z = 3$

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Masala 14 ⭐⭐⭐

$30°$ burchakka 2D aylanish matritsasini tuzing va $(3, 4)$ nuqtani aylantiring.

Yechim

$R(30°) = \begin{bmatrix} \cos 30° & -\sin 30° \\ \sin 30° & \cos 30° \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0.866 & -0.5 \\ 0.5 & 0.866 \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix}$

$= \begin{bmatrix} 2.598 - 2 \\ 1.5 + 3.464 \end{bmatrix} = \begin{bmatrix} 0.598 \\ 4.964 \end{bmatrix}$

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Masala 15 ⭐⭐⭐

Homogen transformatsiya matritsasini tuzing: avval Z o'qi atrofida 90° aylanish, keyin $(2, 3, 1)$ ga siljish.

Yechim

Z atrofida 90° aylanish: $R_z(90°) = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Homogen transformatsiya: $T = \begin{bmatrix} 0 & -1 & 0 & 2 \\ 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

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Masala 16 ⭐⭐⭐

$\det(A) = 5$ bo'lsa, $\det(3A)$ ni toping ($A$ — $3 \times 3$ matritsa).

Yechim

$\det(kA) = k^n \det(A)$ ($n \times n$ matritsa uchun)

$\det(3A) = 3^3 \cdot \det(A) = 27 \cdot 5 = 135$

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Masala 17 ⭐⭐⭐

Matritsaning trace (iz)ini va determinantini xos qiymatlar orqali toping:

$$A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$$

Yechim

Xos qiymatlar: $\det(A - \lambda I) = (2-\lambda)^2 - 1 = 0$ $\lambda_1 = 3$, $\lambda_2 = 1$

Trace (iz): $\text{tr}(A) = \lambda_1 + \lambda_2 = 3 + 1 = 4$

Tekshirish: $\text{tr}(A) = a_{11} + a_{22} = 2 + 2 = 4$ ✓

Determinant: $\det(A) = \lambda_1 \cdot \lambda_2 = 3 \cdot 1 = 3$

Tekshirish: $\det(A) = 4 - 1 = 3$ ✓

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Masala 18 ⭐⭐⭐

Quyidagi matritsa diagonallashtiriluvchimi?

$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

Yechim

Xos qiymatlar: $\det\begin{bmatrix} 1-\lambda & 1 \\ 0 & 1-\lambda \end{bmatrix} = (1-\lambda)^2 = 0$

$\lambda = 1$ (takroriy)

$\lambda = 1$ uchun xos vektor: $(A - I)\vec{v} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\vec{v} = 0$

Faqat bitta chiziqli mustaqil xos vektor: $\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

$2 \times 2$ matritsa uchun 2 ta mustaqil xos vektor kerak, lekin faqat 1 ta bor.

Yo'q, diagonallashtirilmaydi.

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Masala 19 ⭐⭐⭐

$(A^{-1})^T$ ni toping:

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$

Yechim

$\det(A) = 4 - 6 = -2$

$A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$

$(A^{-1})^T = \begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix}$

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Masala 20 ⭐⭐⭐

LU dekompozitsiyasini toping:

$$A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$$

Yechim

Gauss eliminatsiya: $R_2 - 2R_1$

$U = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$

$L = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$

Tekshirish: $LU = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} = A \checkmark$

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Masala 21 ⭐⭐⭐

Robot qo'lining ikkita bo'g'ini bor. Birinchi bo'g'in $\theta_1 = 45°$, ikkinchi $\theta_2 = 30°$. Yakuniy aylanish matritsasini toping.

Yechim

$R_1 = \begin{bmatrix} \cos 45° & -\sin 45° \\ \sin 45° & \cos 45° \end{bmatrix} = \begin{bmatrix} 0.707 & -0.707 \\ 0.707 & 0.707 \end{bmatrix}$

$R_2 = \begin{bmatrix} \cos 30° & -\sin 30° \\ \sin 30° & \cos 30° \end{bmatrix} = \begin{bmatrix} 0.866 & -0.5 \\ 0.5 & 0.866 \end{bmatrix}$

$R_{total} = R_1 \cdot R_2 = \begin{bmatrix} 0.707 & -0.707 \\ 0.707 & 0.707 \end{bmatrix}\begin{bmatrix} 0.866 & -0.5 \\ 0.5 & 0.866 \end{bmatrix}$

$= \begin{bmatrix} 0.259 & -0.966 \\ 0.966 & 0.259 \end{bmatrix}$

Bu $75°$ aylanishga teng ($45° + 30° = 75°$).

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Masala 22 ⭐⭐⭐

Matritsa ijobiy aniqmi (positive definite)?

$$A = \begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix}$$

Yechim

Ijobiy aniq bo'lishi uchun barcha xos qiymatlar > 0.

$\det(A - \lambda I) = (4-\lambda)(2-\lambda) - 4 = \lambda^2 - 6\lambda + 4 = 0$

$\lambda = \frac{6 \pm \sqrt{36-16}}{2} = \frac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}$

$\lambda_1 = 3 + 2.236 = 5.236 > 0$ $\lambda_2 = 3 - 2.236 = 0.764 > 0$

Ha, ijobiy aniq matritsa.

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Murakkab Masalalar (23-30)

Masala 23 ⭐⭐⭐⭐

Dron IMU sensori kalibrlashda quyidagi transformatsiya kerak. $A$ matritsaning pseudoinverse'ini toping:

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$$

Yechim

Moore-Penrose pseudoinverse: $A^+ = (A^TA)^{-1}A^T$

$A^TA = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 35 & 44 \\ 44 & 56 \end{bmatrix}$

$\det(A^TA) = 35(56) - 44(44) = 1960 - 1936 = 24$

$(A^TA)^{-1} = \frac{1}{24}\begin{bmatrix} 56 & -44 \\ -44 & 35 \end{bmatrix}$

$A^+ = (A^TA)^{-1}A^T = \frac{1}{24}\begin{bmatrix} 56 & -44 \\ -44 & 35 \end{bmatrix}\begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}$

$= \frac{1}{24}\begin{bmatrix} -32 & -8 & 16 \\ 26 & 8 & -10 \end{bmatrix} = \begin{bmatrix} -1.33 & -0.33 & 0.67 \\ 1.08 & 0.33 & -0.42 \end{bmatrix}$

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Masala 24 ⭐⭐⭐⭐

Kalman filtrida qo'llaniladigan Riccati tenglamasini diskret ko'rinishda ko'rib chiqing. $P$ kovarians matritsasini yangilash:

$P_{k+1} = APA^T + Q$

Agar $A = \begin{bmatrix} 1 & 0.1 \\ 0 & 1 \end{bmatrix}$, $P_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $Q = \begin{bmatrix} 0.01 & 0 \\ 0 & 0.01 \end{bmatrix}$

$P_1$ ni toping.

Yechim

$A^T = \begin{bmatrix} 1 & 0 \\ 0.1 & 1 \end{bmatrix}$

$AP_0 = \begin{bmatrix} 1 & 0.1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0.1 \\ 0 & 1 \end{bmatrix}$

$AP_0A^T = \begin{bmatrix} 1 & 0.1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0.1 & 1 \end{bmatrix} = \begin{bmatrix} 1.01 & 0.1 \\ 0.1 & 1 \end{bmatrix}$

$P_1 = AP_0A^T + Q = \begin{bmatrix} 1.01 & 0.1 \\ 0.1 & 1 \end{bmatrix} + \begin{bmatrix} 0.01 & 0 \\ 0 & 0.01 \end{bmatrix} = \begin{bmatrix} 1.02 & 0.1 \\ 0.1 & 1.01 \end{bmatrix}$

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Masala 25 ⭐⭐⭐⭐

Quadcopter uchun inertsia tensori matritsasi berilgan:

$$I = \begin{bmatrix} 0.02 & 0 & 0 \\ 0 & 0.02 & 0 \\ 0 & 0 & 0.04 \end{bmatrix} \text{ kg·m}^2$$

Burchak momentum $\vec{L} = I\vec{\omega}$ bo'lsa, $\vec{\omega} = (10, 5, 2)$ rad/s uchun $\vec{L}$ ni toping.

Yechim

$\vec{L} = \begin{bmatrix} 0.02 & 0 & 0 \\ 0 & 0.02 & 0 \\ 0 & 0 & 0.04 \end{bmatrix}\begin{bmatrix} 10 \\ 5 \\ 2 \end{bmatrix} = \begin{bmatrix} 0.2 \\ 0.1 \\ 0.08 \end{bmatrix} \text{ kg·m}^2/\text{s}$

Aylanish kinetik energiyasi: $E_{rot} = \frac{1}{2}\vec{\omega}^T I \vec{\omega} = \frac{1}{2}(10 \cdot 0.2 + 5 \cdot 0.1 + 2 \cdot 0.08)$ $= \frac{1}{2}(2 + 0.5 + 0.16) = 1.33 \text{ J}$

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Masala 26 ⭐⭐⭐⭐

Robot manipulyatorining Jacobian matritsasi:

$$J = \begin{bmatrix} -L_1\sin\theta_1 - L_2\sin(\theta_1+\theta_2) & -L_2\sin(\theta_1+\theta_2) \\ L_1\cos\theta_1 + L_2\cos(\theta_1+\theta_2) & L_2\cos(\theta_1+\theta_2) \end{bmatrix}$$

$L_1 = L_2 = 1$ m, $\theta_1 = 45°$, $\theta_2 = 45°$ da $\det(J)$ ni hisoblang. Singularlik bormi?

Yechim

$\theta_1 + \theta_2 = 90°$

$\sin 45° = \cos 45° = 0.707$ $\sin 90° = 1$, $\cos 90° = 0$

$J = \begin{bmatrix} -0.707 - 1 & -1 \\ 0.707 + 0 & 0 \end{bmatrix} = \begin{bmatrix} -1.707 & -1 \\ 0.707 & 0 \end{bmatrix}$

$\det(J) = (-1.707)(0) - (-1)(0.707) = 0.707$

$\det(J) \neq 0$, singularlik yo'q — robot bu holatda to'liq boshqariladi.

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Masala 27 ⭐⭐⭐⭐

SVD dekompozitsiya. Quyidagi matritsa uchun singular qiymatlarni toping:

$$A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$$

Yechim

Diagonal matritsa uchun singular qiymatlar = diagonal elementlarning absolyut qiymatlari.

$\sigma_1 = 3, \quad \sigma_2 = 2$

Umumiy usul: $A^TA$ ning xos qiymatlarining ildizlari.

$A^TA = \begin{bmatrix} 9 & 0 \\ 0 & 4 \end{bmatrix}$

Xos qiymatlar: $\lambda_1 = 9$, $\lambda_2 = 4$

Singular qiymatlar: $\sigma_1 = \sqrt{9} = 3$, $\sigma_2 = \sqrt{4} = 2$

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Masala 28 ⭐⭐⭐⭐

Matritsa eksponentasini hisoblang ($e^A$):

$$A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

Yechim

Bu matritsa 90° aylanishga mos keladi. Xos qiymatlar: $\lambda = \pm i$ (kompleks).

$e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots$

$A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$

$A^3 = A^2 \cdot A = -A$ $A^4 = -A^2 = I$

Pattern: $I, A, -I, -A, I, A, ...$

$e^A = I(1 - \frac{1}{2!} + \frac{1}{4!} - \cdots) + A(1 - \frac{1}{3!} + \frac{1}{5!} - \cdots)$ $= I\cos(1) + A\sin(1)$

$e^A = \begin{bmatrix} \cos 1 & \sin 1 \\ -\sin 1 & \cos 1 \end{bmatrix} \approx \begin{bmatrix} 0.540 & 0.841 \\ -0.841 & 0.540 \end{bmatrix}$

Bu 1 radianga aylanish matritsasi!

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Masala 29 ⭐⭐⭐⭐

State-space modelda barqarorlikni tekshiring:

$$A = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$$

Xos qiymatlar real qismlarini toping va barqarorlikni aniqlang.

Yechim

$\det(A - \lambda I) = \lambda^2 + 3\lambda + 2 = 0$

$\lambda = \frac{-3 \pm \sqrt{9-8}}{2} = \frac{-3 \pm 1}{2}$

$\lambda_1 = -1$, $\lambda_2 = -2$

Har ikkala xos qiymat manfiy real qismga ega.

Tizim barqaror (stable) — barcha trajectoryalar 0 ga intiladi.

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Masala 30 ⭐⭐⭐⭐

Dron uchish dinamikasi lineerlashtirish. Kichik burchaklar ($\phi, \theta, \psi \approx 0$) uchun Euler burchaklaridan aylanish matritsasiga o'tish:

$$R \approx I + \begin{bmatrix} 0 & -\psi & \theta \\ \psi & 0 & -\phi \\ -\theta & \phi & 0 \end{bmatrix}$$

$\phi = 0.1$, $\theta = 0.05$, $\psi = 0.02$ rad uchun $R$ ni hisoblang va to'liq formula bilan solishtiring.

Yechim

Lineerlashtirilgan: $R_{approx} = \begin{bmatrix} 1 & -0.02 & 0.05 \\ 0.02 & 1 & -0.1 \\ -0.05 & 0.1 & 1 \end{bmatrix}$

To'liq formula (ZYX Euler): $R_{exact} = R_z(\psi)R_y(\theta)R_x(\phi)$

$\cos 0.1 \approx 0.995$, $\sin 0.1 \approx 0.0998$ $\cos 0.05 \approx 0.999$, $\sin 0.05 \approx 0.05$ $\cos 0.02 \approx 0.9998$, $\sin 0.02 \approx 0.02$

$R_{exact} \approx \begin{bmatrix} 0.998 & -0.020 & 0.050 \\ 0.025 & 0.994 & -0.100 \\ -0.048 & 0.101 & 0.994 \end{bmatrix}$

Xatolik: < 1% — kichik burchaklarda lineerlashtirilgan model yaxshi ishlaydi!

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✅ Tekshirish Ro'yxati

• 1-10: Asosiy operatsiyalar
• 11-22: O'rtacha masalalar
• 23-30: Murakkab masalalar

Maqsad: Kamida 24 ta masalani mustaqil yechish

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Keyingi Qadam

🔬 Amaliyot — NumPy/MATLAB da kod yozish!