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निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखंड लिखिए।:
$\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$
Solution
The given determinant is $\left|\begin{array}{lll}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$
By definition of minors and cofactors, we have:
$\mathrm{M}_{11}=$ minor of $a_{11}=\left|\begin{array}{cc}5 & -1 \\ 1 & 2\end{array}\right|=10+1=1$
$\mathrm{M}_{12}=$ minor of $a_{12}=\left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right|=6-0=6$
$\mathrm{M}_{13}=$ minor of $a_{13}=\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|=3-0=3$
$\mathrm{M}_{21}=$ minor of $a_{21}=\left|\begin{array}{ll}0 & 4 \\ 1 & 2\end{array}\right|=0-4=-4$
$\mathrm{M}_{22}=$ minor of $a_{22}=\left|\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right|=2-0=2$
$\mathrm{M}_{23}=$ minor of $a_{23}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1-0=1$
$\mathrm{M}_{31}=$ minor of $a_{31}=\left|\begin{array}{cc}0 & 4 \\ 5 & -1\end{array}\right|=0-20=-20$
$\mathrm{M}_{32}=$ minor of $a_{32}=\left|\begin{array}{cc}1 & 4 \\ 3 & -1\end{array}\right|=-1-12=-13$
$\mathrm{M}_{33}=$ minor of $a_{33}=\left|\begin{array}{ll}1 & 0 \\ 3 & 5\end{array}\right|=5-0=5$
$\mathrm{A}_{\mathrm{II}}=$ cofactor of $a_{11}=(-1)^{1+1} \mathrm{M}_{\mathrm{II}}=11$
$\mathrm{A}_{12}=$ cofactor of $a_{12}=(-1)^{1+2} \mathrm{M}_{12}=-6$
$\mathrm{A}_{13}=$ cofactor of $a_{13}=(-1)^{1+3} \mathrm{M}_{13}=3$
$\mathrm{A}_{21}=$ cofactor of $a_{21}=(-1)^{2+1} \mathrm{M}_{21}=4$
$\mathrm{A}_{22}=\mathrm{cofactor}$ of $a_{22}=(-1)^{2+2} \mathrm{M}_{22}=2$
$\mathrm{A}_{23}=$ cofactor of $a_{23}=(-1)^{2+3} \mathrm{M}_{23}=-1$
$\mathrm{A}_{31}=$ cofactor of $a_{31}=(-1)^{3+1} \mathrm{M}_{31}=-20$
$\mathrm{A}_{32}=$ cofactor of $a_{32}=(-1)^{3+2} \mathrm{M}_{32}=13$
$\mathrm{A}_{33}=$ cofactor of $a_{33}=(-1)^{3+3} \mathrm{M}_{33}=5$
