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2(1) + 3b = 11 \implies 3b = 9 \implies b = 3
استكمالًا، لنفترض $c = 3$:
2(3) + 3d = 15 \implies 3d = 9 \implies d = 3
\mathbf{M} = \begin{pmatrix} 1 & 3 \\ 3 & 3 \end{pmatrix}
\boxed{\begin{pmatrix} 1 & 3 \\ 3 & 3 \end{pmatrix}}
الإجابة النهائية:** $\boxed{\begin{pmatrix} 1 & 3 \\ 3 & 3 \end{pmatrix}}$
السؤال: أوجد جميع الزوايا $z \in [0^\circ, 360^\circ]$ التي تحقق المعادلة $2\sin z + \sqrt{3} = 0$.
2\sin z + \sqrt{3} = 0 \implies \sin z = -\frac{\sqrt{3}}{2}
الزوايا $z$ التي تحقق $\sin z = -\frac{\sqrt{3}}{2}$ تقع في الربعين الثالث والرابع. الزاوية المرجعية هي $60^\circ$ لأن $\sin 60^\circ = \frac{\sqrt{3}}{2}$. لذلك، الحلول في الفترة $[0^\circ, 360^\circ]$ هي:
z = 180^\circ + 60^\circ = 240^\circ \quad \text{و} \quad z = 360^\circ - 60^\circ = 300^\circ