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(A) According to the property of determinants,the sum of the products of elements of any row (or column) with their corresponding cofactors is equal t

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(A) According to the property of determinants,the sum of the products of elements of any row (or column) with their corresponding cofactors is equal to the value of the determinant of the matrix,denoted as $|A|$.Given the matrix $A = \begin{bmatrix} \cos 	heta & \sin 	heta & 0 \ -\sin 	heta & \cos 	heta & 0 \ 0 & 0 & 1 \end{bmatrix}$.The expression $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$ represents the expansion of the determinant $|A|$ along the second row.Calculating the determinant $|A|$ by expanding along the third column (since it has two zeros):$|A| = 1 	imes \begin{vmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{vmatrix} - 0 + 0$$|A| = (\cos 	heta)(\cos 	heta) - (\sin 	heta)(-\sin 	heta)$$|A| = \cos^2 	heta + \sin^2 	heta = 1$.Therefore,$a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = |A| = 1$.

If $A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$,where $A_{21}, A_{22}, A_{23}$ are cofactors of $a_{21}, a_{22}, a_{23}$ respectively,then the value of $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = $

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