3 and 4 .Determinants and Matrices
easy

જો $A = \left[ {\begin{array}{*{20}{c}}0&2&0\\0&0&3\\{ - 2}&2&0\end{array}} \right]$અને $B = \left[ {\begin{array}{*{20}{c}}1&2&3\\3&4&5\\5&{ - 4}&0\end{array}} \right]$, તો $AB$ ની ત્રીજી હાર અને ત્રીજા સ્તંભનો ઘટક મેળવો.

A

$-18$

B

$4$

C

$-12$

D

એકપણ નહી.

Solution

(b) In the product $AB$, the required element

${C_{33}} = ( – 2)3 + 2.5 + 0.0 = – 6 + 10 = 4$.

Standard 12
Mathematics

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