Math, asked by deep264, 1 year ago

ch 9 matrix of class tenth of ICSE bord solution



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Answered by abhi569
14
3A =      \left[\begin{array}{ccc}-2&-2\\1&-3\end{array}\right] -  \left[\begin{array}{ccc}4&-2\\4&0\end{array}\right]



3A =   \left[\begin{array}{ccc}-2-4&-2-(-2)\\1-4&-3-0\end{array}\right]

3A =   \left[\begin{array}{ccc}-6&-2+2\\-3&-3\end{array}\right]


3A =   \left[\begin{array}{ccc}-6&0\\-3&-3\end{array}\right]


A =  \frac{1}{3} \left[\begin{array}{ccc}-6&0\\-3&-3\end{array}\right]

A =   \left[\begin{array}{ccc}-2&0\\-1&-1\end{array}\right]




i hope this will help you


-by ABHAY

abhi569: thanks to bol do
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