નીચે આપેલા નિશ્ચાયકના ઘટકોના ઉપનિશ્ચાયકો (Minors) અને સહઅવયવો (Cofactors) લખો: $\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

આપેલ નિશ્ચાયક $D = \left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$ છે.
ઉપનિશ્ચાયકો $(M_{ij})$:
$M_{11} = \left|\begin{array}{cc}5 & -1 \\ 1 & 2\end{array}\right| = (5)(2) - (-1)(1) = 10 + 1 = 11$
$M_{12} = \left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right| = (3)(2) - (-1)(0) = 6 - 0 = 6$
$M_{13} = \left|\begin{array}{cc}3 & 5 \\ 0 & 1\end{array}\right| = (3)(1) - (5)(0) = 3 - 0 = 3$
$M_{21} = \left|\begin{array}{cc}0 & 4 \\ 1 & 2\end{array}\right| = (0)(2) - (4)(1) = 0 - 4 = -4$
$M_{22} = \left|\begin{array}{cc}1 & 4 \\ 0 & 2\end{array}\right| = (1)(2) - (4)(0) = 2 - 0 = 2$
$M_{23} = \left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right| = (1)(1) - (0)(0) = 1 - 0 = 1$
$M_{31} = \left|\begin{array}{cc}0 & 4 \\ 5 & -1\end{array}\right| = (0)(-1) - (4)(5) = 0 - 20 = -20$
$M_{32} = \left|\begin{array}{cc}1 & 4 \\ 3 & -1\end{array}\right| = (1)(-1) - (4)(3) = -1 - 12 = -13$
$M_{33} = \left|\begin{array}{cc}1 & 0 \\ 3 & 5\end{array}\right| = (1)(5) - (0)(3) = 5 - 0 = 5$
સહઅવયવો $(A_{ij} = (-1)^{i+j} M_{ij})$:
$A_{11} = (-1)^{1+1} M_{11} = 11$
$A_{12} = (-1)^{1+2} M_{12} = -6$
$A_{13} = (-1)^{1+3} M_{13} = 3$
$A_{21} = (-1)^{2+1} M_{21} = -(-4) = 4$
$A_{22} = (-1)^{2+2} M_{22} = 2$
$A_{23} = (-1)^{2+3} M_{23} = -(1) = -1$
$A_{31} = (-1)^{3+1} M_{31} = -20$
$A_{32} = (-1)^{3+2} M_{32} = -(-13) = 13$
$A_{33} = (-1)^{3+3} M_{33} = 5$