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दूसरी पंक्ति के अवयवों के सहखंडों का प्रयोग करके $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$ का मान ज्ञात कीजिए
$7$
$7$
$7$
$7$
Solution
The given determinant is $\left|\begin{array}{ccc}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$
We have:
$M_{21}=\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right|=9-16=-7$
$\therefore \mathrm{A}_{21}=$ cofactor of $a_{21}=(-1)^{2+1} \mathrm{M}_{21}=7$
$\mathrm{M}_{22}=\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right|=15-8=7$
$\therefore \mathrm{A}_{22}=$ cofactor of $a_{22}=(-1)^{2+2} \mathrm{M}_{22}=7$
$\mathrm{M}_{23}=\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=10-3=7$
$\therefore \mathrm{A}_{23}=$ cofactor of $a_{23}=(-1)^{2+3} \mathrm{M}_{23}=-7$
We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
$\therefore \Delta=a_{11} \mathrm{A}_{21}+a_{22} \mathrm{A}_{22}+a_{33} \mathrm{A}_{33}$
$=2(7)+0(7)+1(-7)=14-7=7$
