Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. \( \left|\begin{array}{rrr}2 & 0 & 3 \\ 2 & 4 & 3 \\ 0 & 4 & -1\end{array}\right| \) Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is \( -(0)(-2)+(4)(-2)-(4)(0)=\square \). Using this expansion, the determinant is \( -(2)(-16)+(0)(-2)-(3)(8)=\square \). C. Using this expansion, the determinant is \( (0)(-2)-(4)(-2)+(4)(0)=\square \). D. Using this expansion, the determinant is \( (2)(-16)-(0)(-2)+(3)(8)=\square \).
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To compute the determinant using a cofactor expansion across the first row, we apply the formula: \[ \text{det} = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} \] where \(C_{ij}\) is the cofactor of element \(a_{ij}\). For the matrix given: \[ \left|\begin{array}{rrr}2 & 0 & 3 \\ 2 & 4 & 3 \\ 0 & 4 & -1\end{array}\right| \] - The first element \(a_{11} = 2\) (cofactor \(C_{11} = \text{det}\left( \begin{array}{rr}4 & 3 \\ 4 & -1\end{array} \right) = (4)(-1) - (3)(4) = -4 - 12 = -16\)) - The second element \(a_{12} = 0\) (cofactor \(C_{12} = \text{det}\left( \begin{array}{rr}2 & 3 \\ 0 & -1\end{array} \right) = (2)(-1) - (3)(0) = -2\)) - The third element \(a_{13} = 3\) (cofactor \(C_{13} = \text{det}\left( \begin{array}{rr}2 & 4 \\ 0 & 4\end{array} \right) = (2)(4) - (4)(0) = 8\)) Putting it all together, we find: \[ \text{det} = 2(-16) + 0(-2) + 3(8) = -32 + 0 + 24 = -8 \] So the correct choice, filling in the answer box, is: D. Using this expansion, the determinant is \( (2)(-16)-(0)(-2)+(3)(8)=-8 \). Now, let's compute the determinant by a cofactor expansion down the second column: - The first element \(a_{12} = 0\) (cofactor irrelevant) - The second element \(a_{22} = 4\) (cofactor \(C_{22} = (-1)^{2+2} \text{det}\left( \begin{array}{rr}2 & 3 \\ 0 & -1\end{array} \right) = \text{det}\left( \begin{array}{rr}2 & 3 \\ 0 & -1\end{array} \right) = -2\)) - The third element \(a_{32} = 4\) (cofactor \(C_{32} = (-1)^{3+2} \text{det}\left( \begin{array}{rr}2 & 3 \\ 2 & 3\end{array} \right) = -(2 \cdot 3 - 3 \cdot 2) = 0\))) Putting this expansion together: \[ \text{det} = 0 + 4(-2) + 4(0) = -8 \] So the determinant is also confirmed as \( -8 \).
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