જો A = frac{1}{pi} begin{bmatrix} sin^{-1}(xp | vedclass

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Question

(D) આપેલ શ્રેણિકો $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}

Vedclass · Accepted Answer

$\frac{1}{2}I$

Vedclass · Answer

(D) આપેલ શ્રેણિકો $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) \end{bmatrix}$ અને $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} 	an^{-1}(\pi x) \end{bmatrix}$ છે.$A$ માંથી $B$ બાદ કરતા:$A-B = \begin{bmatrix} \frac{1}{\pi}(\sin^{-1}(x\pi) + \cos^{-1}(x\pi)) & \frac{1}{\pi}(	an^{-1}(\frac{x}{\pi}) - 	an^{-1}(\frac{x}{\pi})) \ \frac{1}{\pi}(\sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi})) & \frac{1}{\pi}(\cot^{-1}(\pi x) + 	an^{-1}(\pi x)) \end{bmatrix}$.નિત્યસમ $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ અને $	an^{-1}(	heta) + \cot^{-1}(	heta) = \frac{\pi}{2}$ નો ઉપયોગ કરતા:$A-B = \begin{bmatrix} \frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \ 0 & \frac{1}{\pi} \cdot \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$.

Similar Questions

શ્રેણિક ગુણાકાર $\left[ {x\,y\,z} \right]\,\left[ {\begin{array}{{20}{c}} a&h&g\\ h&b&f\\ g&f&c \end{array}} \right]\,\left[ {\begin{array}{{20}{c}} x\\ y\\ z \end{array}} \right]$ નો ક્રમ શું છે?

આપેલ ગુણાકારની ગણતરી કરો: $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

જો $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$ હોય,તો $x$ અને $y$ શોધો.

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