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Answer 1

Given :-

$\bf 3\left[\begin{array}{ccc}x&y\\z&w\end{array}\right] =\left[\begin{array}{ccc}x&6\\-1&2w\end{array}\right] +\left[\begin{array}{ccc}4&x+y&\\z+w&3\end{array}\right]$

To Find :-

Value of x, y, z and w

Solution :-

$\bf 3\left[\begin{array}{ccc}x&y\\z&w\end{array}\right] =\left[\begin{array}{ccc}x&6\\-1&2w\end{array}\right] +\left[\begin{array}{ccc}4&x+y&\\z+w&3\end{array}\right]$

$\bf 3\left[\begin{array}{ccc}x&y\\z&w\end{array}\right] =\left[\begin{array}{ccc}x+4&6+x+y\\-1+z+w&2w+3\end{array}\right]$

Using the corresponding element method

$\sf 3x=x+4 (..i)$

$\sf 3x-x=4$

$\sf 2x=4$

$\sf x=\dfrac{4}{2}$

$\sf x=2$

$\sf 3y=x+y+6(..ii)$

Substiute the value of x from 1

$\sf 3y=2+y+6$

$\sf 3y=y+8$

$\sf 3y-y=8$

$\sf 2y=8$

$\sf y=\dfrac{8}{2}$

$\sf y=4$

$\sf 3w=2w+3(..iii)$

$\sf 3w-2w=3$

$\sf w=3$

$\sf 3z=-1+z+w$

$\sf 3z=-1+z+3$

$\sf 3z=z+2$

$\sf 3z-z=2$

$\sf 2z=2$

$\sf z=\dfrac{2}{2}$

$\sf z=1$

Hence

x = 2

y = 4

w = 3

z = 1