Answer:
a=2
b=-3
c=2
d=0
Step-by-step explanation:
3a+b=3......1
3c-d=6.......2
a-b=5....... 3
a+b+d=2d-1
a+b=2d-d-1
a+b=d-1......4
taking equation 1 and 3 we get,
3a+b =3....... 1
a-b=5
a=5+b
substitute the value of a in equation 1 we get,
3*(5+b)+b=3
15+3b+b=3
4b=3-15
4b=-12
b=-12/4
b=-3
now putting the value of b in equation 3 we get,
a-b=5
a+3=5
a=5-3
a=2
substitute the value of a and b in equation 4 we get,
a+b=d-1
2-3=d-1
-1=d-1
d=-1+1
d=0
substitute the value of d in equation 2 we get,
3c-d=6
3c-0=6
3c=6
c=6/3
c=2
[tex]\\ \sf\longmapsto \left[{3a+b\atop a-b}\:\:\:{3c-d\atop a+b+d}\right]=\left[{3\atop 5}\:\:\;{6\atop 2d-1}\right][/tex]
Now constructing equations
[tex]\\ \sf\longmapsto 3a+b=3\dots(1)[/tex]
[tex]\\ \sf\longmapsto a-b=5\dots(2)[/tex]
[tex]\\ \sf\longmapsto 3c-d=6\dots(3)[/tex]
[tex]\\ \sf\longmapsto a+b+d=2d-1\dots(4)[/tex]
Adding eq(1) and (2)
[tex]\\ \sf\longmapsto 4a=8[/tex]
[tex]\\ \sf\longmapsto a=\dfrac{8}{4}[/tex]
[tex]\\ \bf\longmapsto a=2[/tex]
Put the value in eq(2)
[tex]\\ \sf\longmapsto a-b=5[/tex]
[tex]\\ \sf\longmapsto b=a-5[/tex]
[tex]\\ \sf\longmapsto b=2-5[/tex]
[tex]\\ \bf\longmapsto b=-3[/tex]
put the values in eq(4)
[tex]\\ \sf\longmapsto a+b+d=2d-1[/tex]
[tex]\\ \sf\longmapsto 2+(-3)+d=2d-1[/tex]
[tex]\\ \sf\longmapsto d-1=2d-1[/tex]
[tex]\\ \sf\longmapsto d-2d=-1+1[/tex]
[tex]\\ \sf\longmapsto -d=0[/tex]
[tex]\\ \bf\longmapsto d=0[/tex]
Put the value in eq(3)
[tex]\\ \sf\longmapsto 3c-d=6[/tex]
[tex]\\ \sf\longmapsto 3c-0=6[/tex]
[tex]\\ \sf\longmapsto 3c=6+0[/tex]
[tex]\\ \sf\longmapsto 3c=6[/tex]
[tex]\\ \sf\longmapsto c=\dfrac{6}{3}[/tex]
[tex]\\ \sf\longmapsto c=2[/tex]