Evaluate Delta=left|begin{array}{lll}3 & 2 & | vedclass

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(A) To evaluate the determinant $\Delta = \left|\begin{array}{lll}3 & 2 & 3 \ 2 & 2 & 3 \ 3 & 2 & 3\end{array}ight|$,we observe the rows of the ma

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(A) To evaluate the determinant $\Delta = \left|\begin{array}{lll}3 & 2 & 3 \ 2 & 2 & 3 \ 3 & 2 & 3\end{array}ight|$,we observe the rows of the matrix.Notice that the first row $R_1$ is $(3, 2, 3)$ and the third row $R_3$ is $(3, 2, 3)$.Since $R_1 = R_3$,the two rows are identical.According to the properties of determinants,if any two rows or two columns of a determinant are identical,then the value of the determinant is $0$.Alternatively,expanding along the first row:$\Delta = 3(2 	imes 3 - 3 	imes 2) - 2(2 	imes 3 - 3 	imes 3) + 3(2 	imes 2 - 3 	imes 2)$$\Delta = 3(6 - 6) - 2(6 - 9) + 3(4 - 6)$$\Delta = 3(0) - 2(-3) + 3(-2)$$\Delta = 0 + 6 - 6 = 0$.

Evaluate $\Delta=\left|\begin{array}{lll}3 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 2 & 3\end{array}\right|$

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