Question

solution We have
[
begin{array}{l}
Rightarrowleft[begin{array}{cc}
x & y
3 y & x
end{array}right]left[begin{array}{l}
1
2
end{array}right]=left[begin{array}{l}
3
5
end{array}right]
Rightarrowleft{begin{array}{l}
x+2 y
3 y+2 x
end{array}right]=left[begin{array}{l}
3
5
end{array}right]
Rightarrowleft{begin{array}{l}
x+2 y=3
2 x+3 y=5
end{array}right.
end{array}
]
Multiplying (i) by 2 and subtracting (ii) from it, we get ( y=1 ). Putting ( y=1 ) in (i), we get ( x=1 ) Hence, ( x=1 ) and ( y=1 )

# EXAMPLE 12 Solve for x and y, given that

Solution