Assignment Maths Problem and Solutions

Problem 1
a. A=
(1 0 1
2 −1 1
1 3 2 ) and B=
( 1 0 −3
−6 7 −3
−3 −9 −2 )
Solution
A2=
(1 0 1
2 −1 1
1 3 2 )(1 0 1
2 −1 1
1 3 2)=
(2 3 3
1 4 3
9 3 8 ) and AB=
(1 0 1
2 −1 1
1 3 2 )( 1 0 −3
−6 7 −3
−3 −9 −2 )=
( −2 −9 −5
5 −16 −5
−23 3 −16 )
From x A2+ yAB+ zI=0 we replace A2, AB and I;
x (2 3 3
1 4 3
9 3 8 )+ y ( −2 −9 −5
5 −16 −5
−23 3 −16 )+z (1 0 0
0 1 0
0 0 1 )=
(0 0 0
0 0 0
0 0 0 )
( 2 x−2 y+ z 3 x−9 y 3 x−5 y
x+5 y 4 x −16 y+ z 3 x−5 y
9 x+23 y 3 x+3 y 8 x−16 y + z ) =
( 0 0 0
0 0 0
0 0 0 )
Which gives x=y=z=0
b. Finding adj(A) and A-1 for matrix
𝐀 = ( 1 0 a
2 b c
−1 1 1 )
( A11=b c
1 1=b−c A12=0 a
1 1=−a A13=0 a
b c=−ab
A12= 2 c
−1 1=2+c A22= 1 a
−1 1=1+ a A23=1 a
2 c =c−2 a
A13= 2 b
−1 1=2+b A23= 1 0
−1 1=1 A33=1 0
2 b=b )
Matrix of minors Aij=
(b−c 2+c 2+ b
−a 1+ a 1
−ab c−2 a b )
adj(A)= ( b−c 2+c 2+ b
−a 1+ a 1
−ab c−2 a b )

Matrix of cofactors
Bij=-1i+j=Aij= -1
(b−c 2+c 2+ b
−a 1+ a 1
−ab c−2 a b ) note if i+j =even consider – (sign),if odd take + (sign)
Bi j =
(b−c −2−c 2+b
a 1+ a −1
−ab −c+ 2 a b ), the transpose of Bi j= Bi jT=
( b−c a −ab
−2−c 1+a −c +2 a
2+b −1 b ) which is also
considered as ad joint of a matrix A.
The Det (A) = (A) = ( 1 0 a
2 b c
−1 1 1 ) = a (b+2) +b - c =ab + 2a + b - c =0
Hence the inverse of A; A-1 = adj A
( A ) = (b−c 2+c 2+ b
−a 1+ a 1
−ab c−2 a b )
ab+2 a+ b−c
=
( b−c
ab+2 a+ b−c
a
ab+ 2 a+b−c
−ab
ab+2 a+b−c
−2−c
ab+2 a+ b−c
1+ a
ab+ 2 a+b−c
−c +2 a
ab+2 a+b−c
2+ b
ab+2 a+ b−c
−1
ab+ 2 a+b−c
b
ab+2 a+b−c
)c. To find ¿
Consider Aε M 3× 3 ( R) with |A| = 2
Since A-1 = adj A
| A| hence adj A is same as | A|A-1
Therefore (adj A)-1 = | A|A-1 = 1
|A| (A-1)-1= A
2 since |A |=2
=2 ¿
=¿
Problem 2
a. A=
( 1 −1 4
−2 1 −11
−1 2 2
5
−8
2 ) and B=
( 1 −1 3
0 1 4
1 0 8
2
1
6 ) B=
(1 −1 3
0 1 4
1 0 8
2
1
6 )A=
( 1 −1 4
−2 1 −11
−1 2 2
5
−8
2 )

b. Determining values of ∝ such that
X + 2y – z = 2
−T 1+T 3
2 T1 +T 2
T 1+T 3 −T 2+T 3
−4 T3 +T 2
−1 ×T 2
(1 −1 4
0 1 3
0 0 1
5
−2
3 )
(1 −1 4
0 1 0
0 0 1
5
−11
3 )
( 1 −1 0
0 1 0
0 0 1
−7
−11
3 )
( 1 0 0
0 1 0
0 0 1
−18
−11
3 )
−3 T 3+T 1
T 3−T 2
1
3 T 3
Since both of the matrix’s echelon form are
equal. The matrix A and B are equivalent.
(1 0 0
0 1 0
0 0 1
18
−11
3 )
−3 T 3+T 2
−4 T3 +T 1
T 2+T 1