Solution

We have \(M_{11} = |\begin{smallmatrix} 0 & 4 \\ 5 & -7 \end{smallmatrix}| = 0 - 20 = -20; A_{11} = (-1)^{1+1}(-20) = -20\) \(M_{12} = |\begin{smallmatrix} 6 & 4 \\ 1 & -7 \end{smallmatrix}| = -42 - 4 = -46; A_{12} = (-1)^{1+2}(-46) = 46\) \(M_{13} = |\begin{smallmatrix} 6 & 0 \\ 1 & 5 \end{smallmatrix}| = 30 - 0 = 30; A_{13} = (-1)^{1+3}(30) = 30\) \(M_{21} = |\begin{smallmatrix} -3 & 5 \\ 5 & -7 \end{smallmatrix}| = 21 - 25 = -4; A_{21} = (-1)^{2+1}(-4) = 4\) \(M_{22} = |\begin{smallmatrix} 2 & 5 \\ 1 & -7 \end{smallmatrix}| = -14 - 5 = -19; A_{22} = (-1)^{2+2}(-19) = -19\) \(M_{23} = |\begin{smallmatrix} 2 & -3 \\ 1 & 5 \end{smallmatrix}| = 10 - (-3) = 13; A_{23} = (-1)^{2+3}(13) = -13\) \(M_{31} = |\begin{smallmatrix} -3 & 5 \\ 0 & 4 \end{smallmatrix}| = -12 - 0 = -12; A_{31} = (-1)^{3+1}(-12) = -12\) \(M_{32} = |\begin{smallmatrix} 2 & 5 \\ 6 & 4 \end{smallmatrix}| = 8 - 30 = -22; A_{32} = (-1)^{3+2}(-22) = 22\) \(M_{33} = |\begin{smallmatrix} 2 & -3 \\ 6 & 0 \end{smallmatrix}| = 0 - (-18) = 18; A_{33} = (-1)^{3+3}(18) = 18\) Now \(a_{11} = 2, a_{12} = -3, a_{13} = 5\). Also \(A_{31} = -12, A_{32} = 22, A_{33} = 18\) So \(a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 2(-12) + (-3)(22) + 5(18) = -24 - 66 + 90 = 0\)