class 12 maths determinants

Evaluate the determinants

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&{ - 2}\\0&0&{ - 1}\\3&{ - 5}&0\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}&5\\1&1&{ - 2}\\2&3&1\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&2\\{ - 1}&0&{ - 3}\\{ - 2}&3&0\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right|$

VAVidaara Admin Asked 8d ago 0 views 0 answers
📘 Determinants NCERT,Ex.4.1,Q.5,Page.108 SA

Evaluate the determinants

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&{ - 2}\\0&0&{ - 1}\\3&{ - 5}&0\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}&5\\1&1&{ - 2}\\2&3&1\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&2\\{ - 1}&0&{ - 3}\\{ - 2}&3&0\end{array}} \right|$

• $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right|$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

(i) `
$= 3\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2 \times 0 = - 15 + 3 = - 12.$

(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}&5\\1&1&{ - 2}\\2&3&1\end{array}} \right|$

$= 3\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\3&1\end{array}} \right| + 4\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\2&1\end{array}} \right| + 5\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\2&3\end{array}} \right|$

$= 3\left( {1 + 6} \right) + 4\left( {1 + 4} \right) + 5\left( {3 - 2} \right) = 21 + 20 + 5 = 46.$

(iii)$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&2\\{ - 1}&0&{ - 3}\\{ - 2}&3&0\end{array}} \right|$

$= 0\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 3}\\3&0\end{array}} \right| - 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 3}\\{ - 2}&0\end{array}} \right| + 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&0\\{ - 2}&3\end{array}} \right|$

$= 0 - 1\left( {0 - 6} \right) + 2\left( { - 3} \right) = 6 - 6 = 0$

(iv) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right|$ $= 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}\\{ - 5}&0\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}\\3&0\end{array}} \right| - 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&2\\3&{ - 5}\end{array}} \right|$

$= 2\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2\left( {0 - 6} \right) = - 10 + 3 + 12 = 5.$

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