The value of det A,where A = begin{bmatrix} 1 | vedclass

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(A) We have,$|A| = \begin{vmatrix} 1 & \cos 	heta & 0 \ -\cos 	heta & 1 & \cos 	heta \ -1 & -\cos 	heta & 1 \end{vmatrix}$ Expanding along the f

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in the closed interval $[1, 2]$

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(A) We have,$|A| = \begin{vmatrix} 1 & \cos 	heta & 0 \ -\cos 	heta & 1 & \cos 	heta \ -1 & -\cos 	heta & 1 \end{vmatrix}$ Expanding along the first row: $|A| = 1[1 - (-\cos 	heta)(\cos 	heta)] - \cos 	heta[-\cos 	heta - (-\cos 	heta)] + 0[\cos^2 	heta + 1]$ $|A| = 1[1 + \cos^2 	heta] - \cos 	heta[0] + 0$ $|A| = 1 + \cos^2 	heta$ Now,we know that $-1 \leq \cos 	heta \leq 1$. Therefore,$0 \leq \cos^2 	heta \leq 1$. Adding $1$ to all parts,we get $1 \leq 1 + \cos^2 	heta \leq 2$. Thus,$1 \leq |A| \leq 2$. Therefore,the value of $|A|$ lies in the closed interval $[1, 2]$.

The value of $\det A$,where $A = \begin{bmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{bmatrix}$,lies

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