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

We wish to compute the determinant of \[ \begin{pmatrix} 1 & -1 & 8 \\ 8 & 7 & 1 \\ 7 & 8 & -7 \end{pmatrix} \] using a cofactor expansion (Laplace expansion) across the first row and then down the second column. --- **1. Expansion Along the First Row** Recall that for a 3×3 matrix, the determinant by expanding across the first row is \[ \det(A)= a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}, \] where each cofactor \( C_{1j} \) equals \((-1)^{1+j}\) times the determinant of the \(2 \times 2\) submatrix obtained by deleting row 1 and column \(j\). For our matrix, label the entries as: \[ a_{11}=1,\quad a_{12}=-1,\quad a_{13}=8. \] - **First Term (\(j=1\))** Remove row 1 and column 1 to obtain \[ \begin{pmatrix} 7 & 1 \\ 8 & -7 \end{pmatrix}. \] Its determinant is \[ 7(-7)- (1)(8)= -49-8= -57. \] The sign factor is \((-1)^{1+1}=1\). Hence, the contribution is \[ (1)(-57). \] - **Second Term (\(j=2\))** Remove row 1 and column 2 to obtain \[ \begin{pmatrix} 8 & 1 \\ 7 & -7 \end{pmatrix}. \] Its determinant is \[ 8(-7)- (1)(7)= -56-7= -63. \] The sign factor is \((-1)^{1+2}=-1\). With \(a_{12}=-1\), the contribution is \[ -1 \times (-1) \times (-63)= -(-1\cdot -63)= -63. \] (In other words, the term is \(-(-1)(-63)\).) - **Third Term (\(j=3\))** Remove row 1 and column 3 to obtain \[ \begin{pmatrix} 8 & 7 \\ 7 & 8 \end{pmatrix}. \] Its determinant is \[ 8\cdot 8- 7\cdot 7= 64-49= 15. \] The sign factor is \((-1)^{1+3}=1\). Hence, the contribution is \[ (8)(15). \] Thus, the expanded expression is \[ (1)(-57)-(-1)(-63)+(8)(15). \] Among the provided choices, the correct answer for the first row expansion is: A. Using this expansion, the determinant is \(\,(1)(-57)-(-1)(-63)+(8)(15)\). Now, adding the individual contributions: \[ -57 -63 + 120 = 0. \] --- **2. Expansion Down the Second Column** To expand down column 2, recall that \[ \det(A)= \sum_{i=1}^{3} (-1)^{i+2} a_{i2}\det(M_{i2}), \] where \( M_{i2} \) is the \(2 \times 2\) submatrix formed by deleting row \( i \) and column 2. For our matrix, the entries in column 2 are: \[ a_{12}=-1,\quad a_{22}=7,\quad a_{32}=8. \] - **First Term (\(i=1\))** For \( a_{12}=-1 \) with sign \((-1)^{1+2} = -1\), remove row 1 and column 2: \[ \begin{pmatrix} 8 & 1 \\ 7 & -7 \end{pmatrix} \] whose determinant is \(-63\). The contribution is: \[ -1\cdot(-1)\cdot(-63)= -63. \] - **Second Term (\(i=2\))** For \( a_{22}=7 \) with sign \((-1)^{2+2} = 1\), remove row 2 and column 2: \[ \begin{pmatrix} 1 & 8 \\ 7 & -7 \end{pmatrix}. \] Its determinant is: \[ 1\cdot(-7)- 8\cdot 7= -7-56= -63. \] The contribution is: \[ 7\cdot1\cdot(-63)= -441. \] - **Third Term (\(i=3\))** For \( a_{32}=8 \) with sign \((-1)^{3+2} = -1\), remove row 3 and column 2: \[ \begin{pmatrix} 1 & 8 \\ 8 & 1 \end{pmatrix}. \] Its determinant is: \[ 1\cdot1- 8\cdot8= 1-64= -63. \] The contribution is: \[ -1\cdot8\cdot(-63)= 504. \] Thus, the expression for the determinant via the expansion down the second column is \[ -(-1)(-63)+(7)(-63)-(8)(-63). \] This matches option D: D. Using this expansion, the determinant is \(\,-(-1)(-63)+(7)(-63)-(8)(-63)\). Checking the sum: \[ -63-441+504= 0. \] --- **Final Answer:** - The cofactor expansion across the first row is given by \(\,(1)(-57)-(-1)(-63)+(8)(15)\). - The cofactor expansion down the second column is given by \(\,-(-1)(-63)+(7)(-63)-(8)(-63)\). - In both cases, the determinant is \(\,0\).