1. Stating the problem: Find the determinant of a 3\times3 matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \). 2. The determinant \( \det(A) \) of a 3\times3 matrix can be computed using the formula: $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ 3. Explanation: - For each element of the first row (\(a,b,c\)), we multiply it by the determinant of the 2\times2 submatrix that remains after removing the element's row and column. - We then apply a +, -, + sign pattern respectively (cofactor expansion along the first row). 4. Intermediate calculation example: If \( A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \), then $$\det(A) = 1(5\times9 - 6\times8) - 2(4\times9 - 6\times7) + 3(4\times8 - 5\times7)$$ $$= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$$ $$= 1(-3) - 2(-6) + 3(-3)$$ $$= -3 + 12 - 9 = 0$$ 5. Final answer: The determinant of the matrix \(A\) is \(0\) in this example. This method works for any 3\times3 matrix by substituting the entries accordingly.