3. \( \frac{x_{1}-1}{2}=\frac{x_{2}+2}{3}=\frac{x_{3}}{-1} \) to'g'ri chiziq va \( 3 x_{1}-2 x_{2}+x_{3}-1=0 \) tekislikning kesishish \( \quad \) nuqtasini toping. 4. Quyidagi bichiziq formaning matritsasini toping: \( \varphi(x, y)=\mathrm{x}_{1} \mathrm{y}_{1}+\mathrm{x}_{2} \mathrm{y}_{2}+\ldots .+\mathrm{x}_{\mathrm{n}} \mathrm{y}_{\mathrm{n}} \)

3. Berilgan tenglikni yechish uchun, avval \( t \) ni kiritamiz: \[ \frac{x_{1}-1}{2} = t, \quad \frac{x_{2}+2}{3} = t, \quad \frac{x_{3}}{-1} = t \] Bu tengliklardan \( x_1, x_2, x_3 \) ni ifodalaymiz: 1. \( x_1 - 1 = 2t \) dan \( x_1 = 2t + 1 \) 2. \( x_2 + 2 = 3t \) dan \( x_2 = 3t - 2 \) 3. \( x_3 = -t \) Endi bu ifodalarni \( 3x_1 - 2x_2 + x_3 - 1 = 0 \) tenglamasiga qo'yamiz: \[ 3(2t + 1) - 2(3t - 2) + (-t) - 1 = 0 \] Bu tenglamani soddalashtiramiz: \[ 6t + 3 - 6t + 4 - t - 1 = 0 \] \[ 6t - 6t - t + 3 + 4 - 1 = 0 \] \[ -t + 6 = 0 \] \[ t = 6 \] Endi \( t \) ni \( x_1, x_2, x_3 \) ga qo'yamiz: 1. \( x_1 = 2(6) + 1 = 12 + 1 = 13 \) 2. \( x_2 = 3(6) - 2 = 18 - 2 = 16 \) 3. \( x_3 = -6 \) Shunday qilib, kesishish nuqtasi: \[ (x_1, x_2, x_3) = (13, 16, -6) \] 4. Bichiziq formaning matritsasini topish uchun, berilgan ifodani ko'rib chiqamiz: \[ \varphi(x, y) = x_1 y_1 + x_2 y_2 + \ldots + x_n y_n \] Bu ifoda vektorlar ko'paytmasi sifatida yozilishi mumkin. Agar \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \) va \( \mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} \) bo'lsa, unda: \[ \varphi(x, y) = \mathbf{x}^T \mathbf{y} \] Bu yerda \( \mathbf{x}^T \) - \( \mathbf{x} \) ning transpoziyasidir. Matritsa ko'rinishida ifodalash uchun, \( \varphi(x, y) \) ni quyidagi matritsa ko'rinishida yozish mumkin: \[ \mathbf{A} = \begin{pmatrix} 1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix} \] Bu yerda \( \mathbf{A} \) - birlik matritsa. Shunday qilib, bichiziq formaning matritsasi \( \mathbf{A} \) bo'ladi.