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 t ow and fill in the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is \( -(0)(-2)+(4)(-2)-(4)(0)=\square \) B. Using this expansion, the determinant is \( -(2)(-16)+(0)(-2)-(3)(8)=\square \) C. Using this expansion, the determinant is \( (0)(-2)-(4)(-2)+(4)(0)= \) (D. Using this expansion, the determinant is \( (2)(-16)-(0)(-2)+(3)(8)=-8 \). Compute the determinant using a cofactor expansion down the second column. 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 \( (2)(-16)-(0)(-2)+(3)(8)=\square \). B. Using this expansion, the determinant is \( -(\mathbb{1})(-2)+(4)(-2)-(4)(0)=\square \) C. Using this expansion, the determinant is \( -(2)(-16)+(0)(-2)-(3)(8)=\square \). O D. Using this expansion, the determinant is \( (0)(-2)-(4)(-2)+(4)(0)=\square \).

To compute the determinant of the matrix \[ \begin{pmatrix} 2 & 0 & 3 \\ 2 & 4 & 3 \\ 0 & 4 & -1 \end{pmatrix}, \] let's start with the cofactor expansion across the first row. 1. The first element in the first row is 2. Its minor, obtained by removing the first row and column, gives us the matrix \[ \begin{pmatrix} 4 & 3 \\ 4 & -1 \end{pmatrix}. \] The determinant of this \(2 \times 2\) matrix is \( (4)(-1) - (4)(3) = -4 - 12 = -16 \). Thus, the first term is \(2 \cdot (-16) = -32\). 2. The second element is 0, which will contribute 0 to the determinant. 3. The third element is 3. Its minor gives us \[ \begin{pmatrix} 2 & 4 \\ 0 & 4 \end{pmatrix}. \] The determinant of this matrix is \( (2)(4) - (4)(0) = 8\). Therefore, the contribution from this term is \(3 \cdot 8 = 24\). Adding these together (accounting for signs), we have: \[ -(2)(-16) + 0 + (3)(8) = -32 + 0 + 24 = -8. \] So, in answer form, we choose: **D. Using this expansion, the determinant is \( (2)(-16)-(0)(-2)+(3)(8)=-8 \).** Next, let's compute the determinant using a cofactor expansion down the second column. 1. The first element in the second column is 0, so it contributes 0. 2. The second element is 4. Its minor matrix is \[ \begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}. \] The determinant of this matrix is \( (2)(-1) - (3)(0) = -2 \). Thus, the contribution is \(4(-2) = -8\). 3. The third element is 4, which similarly has a minor matrix \[ \begin{pmatrix} 2 & 0 \\ 2 & 4 \end{pmatrix}, \] leading to a determinant of \( (2)(4)-(0)(2) = 8\), contributing \(4 \cdot 8 = 32\). So we sum these contributions: \[ 0 + 4 \cdot (-2) + (-4) \cdot 8 = 0 - 8 + -0 = -8. \] Thus, the answer is: **A. Using this expansion, the determinant is \( (2)(-16)-(0)(-2)+(3)(8)=-8 \).**