Question

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Answer 1

Given :     $A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]$     $B = \left[\begin{array}{cc}p&q\\r&s\end{array}\right]$    $C = \left[\begin{array}{cc}k&l\\m&n\end{array}\right]$

To Find : Prove  A(B + C) = AB + AC

Solution:

A(B + C) = AB + AC

LHS  = A(B + C)

=  $\left[\begin{array}{cc}a&b\\c&d\end{array}\right] ( \left[\begin{array}{cc}p&q\\r&s\end{array}\right] + \left[\begin{array}{cc}k&l\\m&n\end{array}\right] )$

=  $\left[\begin{array}{cc}a&b\\c&d\end{array}\right] ( \left[\begin{array}{cc}p+k&q+l\\r+m&s+n\end{array}\right] )$

= $\left[\begin{array}{cc}ap+ak + br + bm &aq+al+bs + bn \\cp+ck + dr + dm &cq+cl+ds + dn\end{array}\right]$

RHS = AB + AC

=  $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$     $\left[\begin{array}{cc}p&q\\r&s\end{array}\right]$   +   $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$ $\left[\begin{array}{cc}k&l\\m&n\end{array}\right]$

=   $\left[\begin{array}{cc}ap + br &aq + bs\\cp + dr &cq + ds\end{array}\right]$   +       $\left[\begin{array}{cc}ak + bm &al + bn\\ck + dm &cl + dn\end{array}\right]$

= $\left[\begin{array}{cc}ap+ak + br + bm &aq+al+bs + bn \\cp+ck + dr + dm &cq+cl+ds + dn\end{array}\right]$

=> LHS = RHS

QED

Hence Proved

A(B + C) = AB + AC

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