3 and 4 .Determinants and Matrices
easy

If $A = \left| {\,\begin{array}{*{20}{c}}{ - 1}&2&4\\3&1&0\\{ - 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{*{20}{c}}{ - 2}&4&2\\6&2&0\\{ - 2}&4&8\end{array}\,} \right|$, then $B$ is given by

A

$B = 4A$

B

$B = - 4A$

C

$B = - A$

D

$B = 6A$

Solution

 (b) $B$ is obtained from $A$ by the operations ${R_1} \leftrightarrow {R_3},\,\,{R_3} \to 2{R_3}$ and ${R_2 → 2R_2}.$
Hence, $B = ( – 1)\,4A = – 4A$.

Standard 12
Mathematics

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