Fsc Part 1 Mathematics (Complete Solution)

# Q2 Use matrices to solve the following systems

(i)
 x-2y+z = -1 3x+y-2z = 4 y-z = 1
}

SOLUTION:

The matrix form of the system

 1 -2 1 3 1 -2 0 1 -1

 x y z

=

 -1 4 1

Let   A=

 1 -2 1 3 1 -2 0 1 -1

; X=

 x y z

;    B=

 -1 4 1

then system becomes  AX=B

=> X=A-1B

lAl=

 1 -2 1 3 1 -2 0 1 -1

=1(-1+2)+2(-3+0)+1(3-0)

=1-6+3

=-5+3

=-2

Cofactors of A =

 A11 A12 A13 A21 A22 A23 A31 A32 A33

Aij=(-1)i+jMij

A11  =  (-1)1+1

 1 -2 1 -1

=(-1)2 (-1+2) =1
A12  =  (-1)1+2

 3 -2 0 -1

=(-1)3 (-3+2) =3
A13  =  (-1)1+3

 3 1 0 1

=(-1)4 (3-0) =3
A21  =  (-1)2+1

 -2 1 1 -1

=(-1)3 (2-1) =-1
A22  =  (-1)2+2

 1 1 0 -1

=(-1)4 (-1-0) =-1
A23  =  (-1)2+3

 1 -2 0 1

=(-1)5 (1+0) =-1
A31  =  (-1)3+1

 -2 1 1 -2

=(-1)4 (4-1) =3
A32  =  (-1)3+2

 1 1 3 -2

=(-1)5 (-2-3) =5
A33  =  (-1)3+3

 1 -2 3 1

=(-1)6 (1+6) =7
Cofactors of A =

 1 3 3 -1 -1 -1 3 5 7

 1 -1 3 3 -1 5 3 -1 7

SO,  A-1=
=
 1 -2

 1 -1 3 3 -1 5 3 -1 7

=

 -1/2 1/2 -3/2 -3/2 -1/2 -5/2 -3/2 1/2 -7/2

then X=A-1B

X= A-1B=

 -1/2 1/2 -3/2 -3/2 -1/2 -5/2 -3/2 1/2 -7/2

 -1 4 1

 x y z

=

 1 1 0

x=1   , y=1  , z=0

(ii)
 2x1+x2+3x3 = 3 x1+x2-2x3 = 0 -3x1-x2+2x3 = -4
}

SOLUTION:

The matrix form of the system

 2 1 3 1 1 -2 -3 -1 2

 x1 x2 x3

=

 -3 0 -4

Let   A=

 2 1 2 1 1 -2 -3 -1 2

; X=

 x1 x2 x3

;    B=

 -3 0 -4

then system becomes  AX=B

=> X=A-1B

lAl=

 2 1 3 1 1 -2 -3 -1 2

=2(2-2)-1(2-5)+3(-1+3)

=0+4+6

=4+6

=10

Cofactors of A =

 A11 A12 A13 A21 A22 A23 A31 A32 A33

Aij=(-1)i+jMij

A11  =  (-1)1+1

 1 -2 -1 2

=(-1)2 (2-2) =0
A12  =  (-1)1+2

 -1 2 3 -2

=(-1)3 (2-6) =4
A13  =  (-1)1+3

 1 1 -3 -1

=(-1)4 (-1+3) =2
A21  =  (-1)2+1

 -1 -3 1 -2

=(-1)3 (2+3) =-5
A22  =  (-1)2+2

 2 3 -3 2

=(-1)4 (4+9) =13
A23  =  (-1)2+3

 -2 -1 3 1

=(-1)5 (-2+3) =-1
A31  =  (-1)3+1

 1 3 1 -2

=(-1)4 (-2-3) =-5
A32  =  (-1)3+2

 -2 -3 -1 2

=(-1)5 (-4-3) =7
A33  =  (-1)3+3

 2 1 1 1

=(-1)6 (2-1) =1
Cofactors of A =

 0 4 2 -5 13 -1 -5 7 1

 0 -5 -5 4 13 7 2 -1 1

SO,  A-1=
=
 1 10

 0 -5 -5 4 13 7 2 -1 1

=

 0 -1/2 -1/2 2/5 13/10 7/10 1/5 -1/10 1/10

then X=A-1B

X= A-1B=

 0 -1/2 -1/2 2/5 13/10 7/10 1/5 -1/10 1/10

 -3 0 -4

 x1 x2 x3

=

 2 -4 -1

x1=2   , x2= -4  , x3= -1

(iii)
 x+y = 2 2x-z = 1 2y-3z = -1
}

SOLUTION:

The matrix form of the system

 1 1 0 2 0 -1 0 2 -3

 x y z

=

 2 1 -1

Let   A=

 1 1 0 2 0 -1 0 2 -3

; X=

 x y z

;    B=

 2 1 -1

then system becomes  AX=B

=> X=A-1B

lAl=

 1 1 0 2 0 -1 0 2 -3

=1(0+2)-1(-6+0)+0(4-0)

=2+6+0

=2+6

=8

Cofactors of A =

 A11 A12 A13 A21 A22 A23 A31 A32 A33

Aij=(-1)i+jMij

A11  =  (-1)1+1

 0 -1 2 -3

=(-1)2 (0+2) =2
A12  =  (-1)1+2

 2 -1 0 -3

=(-1)3 (-6+0) =6
A13  =  (-1)1+3

 2 0 0 2

=(-1)4 (4-0) =4
A21  =  (-1)2+1

 1 0 2 -3

=(-1)3 (-3-0) =3
A22  =  (-1)2+2

 1 0 0 -3

=(-1)4 (-3-0) = -3
A23  =  (-1)2+3

 1 1 0 2

=(-1)5 (2-0) =-2
A31  =  (-1)3+1

 1 1 0 -1

=(-1)4 (-1-0) =-1
A32  =  (-1)3+2

 1 0 2 -1

=(-1)5 (-1-0) =1
A33  =  (-1)3+3

 1 1 2 0

=(-1)6 (0-2) =-2
Cofactors of A =

 2 6 4 3 -3 -2 -1 1 -2

 2 3 -1 6 -3 1 4 -2 -2

SO,  A-1=
=
 1 8

 2 3 -1 6 -3 1 4 -2 -2

=

 2/8 3/8 -1/8 6/8 -3/8 1/8 4/8 -2/8 -2/8

then X=A-1B

X= A-1B=

 2/8 3/8 -1/8 6/8 -3/8 1/8 4/8 -2/8 -2/8

 2 1 -1

 x y z

=

 1 1 1

x=1  , y=1  , z=1       ANSWER