class 12 maths determinants

Using Cofactors of elements of second row, evaluate
$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&8\\2&0&1\\1&2&3\end{array}} \right|$ .

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📘 Determinants NCERT,Ex.4.3,Q.3,Page.126 SA

Using Cofactors of elements of second row, evaluate
$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&8\\2&0&1\\1&2&3\end{array}} \right|$ .

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$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3&8\\2&0&1\\1&2&3\end{array}} \right|$

Cofactors of elements of second row are

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&8\\2&3\end{array}} \right| = - (9 - 16) = 7$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&8\\1&3\end{array}} \right| = 15 - 8 = 7$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&3\\1&2\end{array}} \right| = - (10 - 3) = - 7$

Now, $\Delta = {a_{21}}{A_{21}} + {a_{22}}{A_{22}} + {a_{33}}{A_{33}} + 2 \times 7 + 0 \times 7 + 1 \times ( - 7)$

$= 14 + 0 - 7 = 7$

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