यदि A = begin{bmatrix} cos alpha & -sin alpha | vedclass

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(C) दिया गया है $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$ और $B = \begin{bmatrix} \cos \beta & -\sin

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$AB = BA$

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(C) दिया गया है $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$ और $B = \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix}$.$AB$ की गणना करने पर:$AB = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix}$$= \begin{bmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -\cos \alpha \sin \beta - \sin \alpha \cos \beta \ \sin \alpha \cos \beta + \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix}$$= \begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix}$.इसी प्रकार,$BA$ की गणना करने पर:$BA = \begin{bmatrix} \cos \beta & -\sin \beta \ \sin \beta & \cos \beta \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$$= \begin{bmatrix} \cos \beta \cos \alpha - \sin \beta \sin \alpha & -\cos \beta \sin \alpha - \sin \beta \cos \alpha \ \sin \beta \cos \alpha + \cos \beta \sin \alpha & -\sin \beta \sin \alpha + \cos \beta \cos \alpha \end{bmatrix}$$= \begin{bmatrix} \cos(\beta + \alpha) & -\sin(\beta + \alpha) \ \sin(\beta + \alpha) & \cos(\beta + \alpha) \end{bmatrix}$.चूंकि $\cos(\alpha + \beta) = \cos(\beta + \alpha)$ और $\sin(\alpha + \beta) = \sin(\beta + \alpha)$,इसलिए $AB = BA$ है।

यदि $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ और $B = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}$ है,तो सही संबंध है

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यदि $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ है,तो $A$ है

दिए गए गुणनफल की गणना करें: $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$.

यदि $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ और $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$ है,तो $(BB^TA)^5$ का मान ज्ञात कीजिए।

यदि $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ है,तो ${A^4} = $

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