Easy
Write the minors and cofactors of the elements of the following determinant: $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
(N/A) The given determinant is $A = \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$.
By the definition of minors $(M_{ij})$ and cofactors $(A_{ij})$,we calculate them for each element $a_{ij}$:
$M_{11} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{11} = (-1)^{1+1} M_{11} = 1$
$M_{12} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{12} = (-1)^{1+2} M_{12} = 0$
$M_{13} = \left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right| = 0, A_{13} = (-1)^{1+3} M_{13} = 0$
$M_{21} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{21} = (-1)^{2+1} M_{21} = 0$
$M_{22} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{22} = (-1)^{2+2} M_{22} = 1$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{23} = (-1)^{2+3} M_{23} = 0$
$M_{31} = \left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right| = 0, A_{31} = (-1)^{3+1} M_{31} = 0$
$M_{32} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{32} = (-1)^{3+2} M_{32} = 0$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{33} = (-1)^{3+3} M_{33} = 1$
By the definition of minors $(M_{ij})$ and cofactors $(A_{ij})$,we calculate them for each element $a_{ij}$:
$M_{11} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{11} = (-1)^{1+1} M_{11} = 1$
$M_{12} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{12} = (-1)^{1+2} M_{12} = 0$
$M_{13} = \left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right| = 0, A_{13} = (-1)^{1+3} M_{13} = 0$
$M_{21} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{21} = (-1)^{2+1} M_{21} = 0$
$M_{22} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{22} = (-1)^{2+2} M_{22} = 1$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{23} = (-1)^{2+3} M_{23} = 0$
$M_{31} = \left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right| = 0, A_{31} = (-1)^{3+1} M_{31} = 0$
$M_{32} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{32} = (-1)^{3+2} M_{32} = 0$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{33} = (-1)^{3+3} M_{33} = 1$
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