simplify:2a(4a-5)+5 and find value (1)a=0 (2)a=1.
Answers
2a * (4a - 5) + 5
8a ^ 2 - 10a + 5
Given A=
⎣
⎢
⎢
⎡
0
1
2
1
2
3
3
x
1
⎦
⎥
⎥
⎤
1) Minors matrix of A is,
M
A
=
⎣
⎢
⎢
⎡
2−3x
1−9
x−6
1−2x
0−6
0−3
3−4
0−2
0−1
⎦
⎥
⎥
⎤
∴M
A
=
⎣
⎢
⎢
⎡
2−3x
−8
x−6
1−2x
−6
−3
−1
−2
−1
⎦
⎥
⎥
⎤
2) Cofactors matrix of A is,
C
A
=
⎣
⎢
⎢
⎡
2−3x
8
x−6
−(1−2x)
−6
3
−1
2
−1
⎦
⎥
⎥
⎤
∴C
A
=
⎣
⎢
⎢
⎡
2−3x
8
x−6
2x−1
−6
3
−1
2
−1
⎦
⎥
⎥
⎤
3) Adjoint of matrix is,
Adj[A]=
⎣
⎢
⎢
⎡
2−3x
2x−1
−1
8
−6
2
x−6
3
−1
⎦
⎥
⎥
⎤
4) Determinant of matrix A is,
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
0
1
2
1
2
3
3
x
1
∣
∣
∣
∣
∣
∣
∣
∣
∴∣A∣=0[2−3x]−1[1−2x]+3[3−4]
∴∣A∣=−1[1−2x]+3[−1]
∴∣A∣=2x−1−3
∴∣A∣=2x−4
∴∣A∣=2(x−2)
Thus, inverse of a matrix A is,
[A]
−1
=
∣A∣
1
Adj.[A]
∴[A]
−1
=
2(x−2)
1
⎣
⎢
⎢
⎡
2−3x
2x−1
−1
8
−6
2
x−6
3
−1
⎦
⎥
⎥
⎤
(1)
Given [A]
−1
=
⎣
⎢
⎢
⎡
2
1
−
2
1
2
1
−4
3
y
2
5
−
2
3
2
1
⎦
⎥
⎥
⎤
(2)
Comparing a
11
in equations (1) and (2), we get,
2(x−2)
2−3x
=
2
1
∴
(x−2)
2−3x
=1
∴2−3x=(x−2)
∴4x=4
∴x=1
Comparing a
32
in equations (1) and (2), we get,
2(x−2)
2
=y
∴
(1−2)
1
=y
∴
−1
1
=y
∴y=−1