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 \( (0)(-2)-(4)(-2)+(4)(0)=\square \).

We are given the matrix \[ A=\begin{pmatrix}2 & 0 & 3 \\ 2 & 4 & 3 \\ 0 & 4 & -1\end{pmatrix} \] and we will compute its determinant by two different cofactor expansions. --- **1. Cofactor Expansion Across the First Row** The formula for the determinant by expanding along the first row is \[ \det(A)= a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}, \] where \[ C_{ij}= (-1)^{i+j}\det(M_{ij}) \] and \(M_{ij}\) is the submatrix obtained by deleting row \(i\) and column \(j\). - **For \(a_{11}=2\):** Remove row 1 and column 1: \[ M_{11} = \begin{pmatrix} 4 & 3 \\ 4 & -1 \end{pmatrix}. \] Its determinant is \[ \det(M_{11}) = 4(-1) - 3(4) = -4 - 12 = -16. \] The cofactor is \[ C_{11}= (-1)^{1+1}(-16)= 1\cdot (-16)= -16. \] - **For \(a_{12}=0\):** Remove row 1 and column 2: \[ M_{12} = \begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}. \] Its determinant is \[ \det(M_{12}) = 2(-1) - 3(0) = -2. \] The cofactor is \[ C_{12}= (-1)^{1+2}(-2)= -1\cdot(-2)= 2. \] - **For \(a_{13}=3\):** Remove row 1 and column 3: \[ M_{13} = \begin{pmatrix} 2 & 4 \\ 0 & 4 \end{pmatrix}. \] Its determinant is \[ \det(M_{13}) = 2(4) - 4(0) = 8. \] The cofactor is \[ C_{13}= (-1)^{1+3}(8)= (-1)^4\cdot8= 8. \] Now, we combine the terms \[ \det(A) = 2(-16)+ 0(2)+ 3(8)= -32 + 0 + 24 = -8. \] This matches choice D for the first row expansion: \[ (2)(-16) - (0)(-2) + (3)(8) = -8. \] --- **2. Cofactor Expansion Down the Second Column** The formula for expanding along the second column is \[ \det(A) = \sum_{i=1}^{3} (-1)^{i+2}\,a_{i2}\,\det(M_{i2}). \] - **For \(a_{12}=0\) (row 1, column 2):** \[ (-1)^{1+2}\,0\,\det(M_{12}) = -0 = 0. \] - **For \(a_{22}=4\) (row 2, column 2):** Remove row 2 and column 2: \[ M_{22} = \begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}. \] Its determinant is \[ \det(M_{22}) = 2(-1)- 3(0)= -2. \] The sign factor is \[ (-1)^{2+2}= (-1)^4 = 1, \] so the contribution is \[ 4(-2)= -8. \] - **For \(a_{32}=4\) (row 3, column 2):** Remove row 3 and column 2: \[ M_{32} = \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix}. \] Its determinant is \[ \det(M_{32}) = 2(3)-3(2)= 6 - 6 = 0. \] The sign factor is \[ (-1)^{3+2}= (-1)^5 = -1, \] so the contribution is \[ -1\cdot 4\cdot 0= 0. \] Adding these contributions \[ \det(A) = 0 - 8 + 0 = -8. \] This expression corresponds to choice A for the second column expansion: \[ (2)(-16)-(0)(-2)+(3)(8) = -8. \] --- **Final Answers:** - Using the cofactor expansion across the first row (choice D): \(\det(A)=-8\). - Using the cofactor expansion down the second column (choice A): \(\det(A)=-8\).