math pdf

ID=bc220207081
Assignment
Question No 1
t𝒙𝟐 −12x+9=0
solution;
t=?
if 𝑏2 – 4ac=0 there will be exactly one real solution, so,
a=t,
b=-12,
c=9
put the values,
(−12)2 -4(t)(9)=0
144-36t=0
36t=144
144
t=
36
t=4
for chek,
(−12)2 -4(4)(9)=0
144-144=0
there is only one real solution,
ID=bc220207081
Question NO 2
9 2
1
𝐴 = [5 −1 6 ]
4 0 −2
Find the cofactor from the matrix a which is given;
solution;
9 2
1
𝐴 = [5 −1 6 ]
4 0 −2
−1
M11=(-1)1+1|
0
6
|
2
M11= (
−1)2(−1 × −2) − (0 × 6)
M11=(1)(2 − 0)
M11=1(2)
M11=2
M12=(−1)1+2|
5
4
6
|
−2
M12=(−1)3(5 × −2) − (6 × 4)
M12=-1(−10 × −24)
ID=bc220207081
M12=-1(−34)
M12=34
M13=(−1)1+3|
5
4
−1
|
0
M13=(−1)4(5 × 0) − (4 × −1)
M13=1(4)
M13=4
2
M21=(−1)2+1|
0
1
|
−2
M21=(−1)3(−4 − 0)
M21=(−1)(−4)
M21=4
M22=(−1)2+2|
9
4
1
|
−2
M22=(−1)4(9 × −2)-(4 × 1)
M22=1(−18 − 4)
M22=1(−22)
ID=bc220207081
M22= -22
M23=(−1)2+3|
9
4
2
|
0
M23=(−1)5(9 × 0) − (4 × 2)
M23=(−1)(−8)
M23=8
M31=(−1)3+1|
2 1
|
−1 6
M31=(−1)4(2 × 6) − (−1 × 1)
M31=(1)(12 + 1)
M31=13
9
M32=(−1)3+2|
5
1
|
6
M32=(−1)5(9 × 6) − (5 × 1)
M32=(−1)(54 − 5)
M32=(−1)(49)
M32= −49
ID=bc220207081
M33=(−1)3+3|
9
5
2
|
−1
M33=(−1)6(9 × −1) − (5 × 2)
M33=1(−9 − 10)
M33=1(−19)
M33= −19
cofactor of matrix A=
𝑀11 𝑀12 𝑀13
[𝑀21 𝑀22 𝑀23]
𝑀31 𝑀32 𝑀33
2
[4
13
34
−22
−49
4
8 ]
−19
…