Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. \[ \left|\begin{array}{rrr} 1 & -1 & 8 \\ 8 & 7 & 1 \\ 7 & 8 & -7 \end{array}\right| \] Write the expression for the determinant using a cofactor expansion across the first row. Choose the correct answ below. A. Using this expansion, the determinant is \( (1)(-57)-(-1)(-63)+(8)(15) \). B. Using this expansion, the determinant is \( (1)(-41)-(-1)(71)+(8)(113) \). C. Using this expansion, the determinant is \( (1)(-57)+(-1)(-63)+(8)(15) \). D. Using this expansion, the determinant is \( (1)(-41)+(-1)(71)+(8)(113) \). Write the expression for the determinant using a cofactor expansion down the second column. Choose the corre answer below. A. Using this expansion, the determinant is \( (-1)(-63)+(7)(-63)+(8)(-63) \). B. Using this expansion, the determinant is \( (-1)(-49)+(7)(49)+(8)(65) \). C. Using this expansion, the determinant is \( -(-1)(-49)+(7)(49)-(8)(65) \). D. Using this expansion, the determinant is \( -(-1)(-63)+(7)(-63)-(8)(-63) \).

We are given the matrix \[ A=\begin{pmatrix} 1 & -1 & 8 \\ 8 & 7 & 1 \\ 7 & 8 & -7 \end{pmatrix}. \] 1. **Cofactor expansion across the first row** The formula for the determinant using the first row is \[ \det(A)=a_{11}\det(M_{11})-a_{12}\det(M_{12})+a_{13}\det(M_{13}), \] where \( M_{ij} \) is the submatrix obtained by deleting row \( i \) and column \( j \). - For \( a_{11}=1 \), delete row 1 and column 1: \[ M_{11}=\begin{pmatrix} 7 & 1 \\ 8 & -7 \end{pmatrix},\quad\det(M_{11})=7(-7)-1(8)=-49-8=-57. \] - For \( a_{12}=-1 \), delete row 1 and column 2: \[ M_{12}=\begin{pmatrix} 8 & 1 \\ 7 & -7 \end{pmatrix},\quad\det(M_{12})=8(-7)-1(7)=-56-7=-63. \] - For \( a_{13}=8 \), delete row 1 and column 3: \[ M_{13}=\begin{pmatrix} 8 & 7 \\ 7 & 8 \end{pmatrix},\quad\det(M_{13})=8(8)-7(7)=64-49=15. \] Inserting these into the expansion gives \[ \det(A)= (1)(-57)-(-1)(-63)+(8)(15). \] This corresponds to option A for the first row expansion. 2. **Cofactor expansion down the second column** For a fixed column (here, the second column with \(j=2\)), the determinant is computed as \[ \det(A)=\sum_{i=1}^3 (-1)^{i+2}a_{i2}\det(M_{i2}). \] Let us compute each term: - For \( i=1 \): \( a_{12}=-1 \) and the sign is \( (-1)^{1+2}= (-1)^3=-1 \). Removing row 1, column 2 gives \[ M_{12}=\begin{pmatrix} 8 & 1 \\ 7 & -7 \end{pmatrix},\quad\det(M_{12})=-63. \] So the term is \[ (-1)(-1)(-63)= (-1)\cdot (-63)= 63 \quad \text{but note the extra negative:} \quad -(-1)(-63)= -63. \] A more step-by-step treatment: The term is \[ (-1)^{3} \cdot (-1) \cdot (-63)= (-1)\cdot (-1)\cdot (-63) = -63. \] - For \( i=2 \): \( a_{22}=7 \) and the sign is \( (-1)^{2+2}= (-1)^4=1 \). Removing row 2, column 2 gives \[ M_{22}=\begin{pmatrix} 1 & 8 \\ 7 & -7 \end{pmatrix},\quad\det(M_{22})=1(-7)-8(7)=-7-56=-63. \] So the term is \[ 1 \cdot 7 \cdot (-63)= -441. \] - For \( i=3 \): \( a_{32}=8 \) and the sign is \( (-1)^{3+2}= (-1)^5=-1 \). Removing row 3, column 2 gives \[ M_{32}=\begin{pmatrix} 1 & 8 \\ 8 & 1 \end{pmatrix},\quad\det(M_{32})=1(1)-8(8)=1-64=-63. \] So the term is \[ (-1) \cdot 8 \cdot (-63)= 8\cdot 63=504 \quad \text{but with a minus sign:} \quad -8\cdot (-63) = 504. \] Adding the three terms: \[ \det(A)= -63 -441 +504 = 0. \] Thus, writing the expansion down the second column, we have \[ \det(A)=-(-1)(-63)+7(-63)-8(-63). \] This matches option D for the second column expansion. The correct answers are: First row expansion: Option A, and second column expansion: Option D.