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3 and 4 .Determinants and Matrices
medium
यदि $A = diag(2, - 1,\,3),B = diag( - 1,\,3,\,2)$, तो ${A^2}B = $
A
$diag\ (5, 4, 11)$
B
$diag\ (-4, 3, 18)$
C
$diag\ (3, 1, 8 )$
D
$B$
Solution
(b) ${A^2}B = (A.\,A)B = \left[ {\begin{array}{*{20}{c}}2&0&0\\0&{ – 1}&0\\0&0&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}2&0&0\\0&{ – 1}&0\\0&0&3\end{array}} \right].\,\left[ {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&3&0\\0&0&2\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}4&0&0\\0&1&0\\0&0&9\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&3&0\\0&0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 4}&0&0\\0&3&0\\0&0&{18}\end{array}} \right]$.
Standard 12
Mathematics