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<p>Determinant calculation of a 3x3 matrix using the Laplace expansion.</p> <p>Given Matrix A:</p> <p>\[ A = \begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ 1 & 3 & 5 \end{bmatrix} \]</p> <p>Expanding along the first row:</p> <p>\[ \text{det}(A) = (-1)^{1+1} \cdot (-1) \cdot \begin{vmatrix} -3 & -5 \\ 3 & 5 \end{vmatrix} + (-1)^{1+2} \cdot 3 \cdot \begin{vmatrix} 1 & -5 \\ 1 & 5 \end{vmatrix} + (-1)^{1+3} \cdot 5 \cdot \begin{vmatrix} 1 & -3 \\ 1 & 3 \end{vmatrix} \]</p> <p>\[ \text{det}(A) = -1 \cdot ((-3)\cdot(5) - (-5)\cdot(3)) + 3 \cdot (1\cdot(5) - (-5)\cdot(1)) + (-5) \cdot (1\cdot(3) - (-3)\cdot(1)) \]</p> <p>\[ \text{det}(A) = -1 \cdot (-15 + 15) + 3 \cdot (5 + 5) - 5 \cdot (3 + 3) \]</p> <p>\[ \text{det}(A) = -1 \cdot 0 + 3 \cdot 10 - 5 \cdot 6 \]</p> <p>\[ \text{det}(A) = 0 + 30 - 30 \]</p> <p>\[ \text{det}(A) = 0 \]</p> <p>Hence, the determinant of matrix A is 0.</p>
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