## User: manasahegde234

manasahegde234 •

**20**- Reputation:
**20**- Status:
- Trusted
- Location:
- Last seen:
- 2 months, 3 weeks ago
- Joined:
- 8 months, 1 week ago
- Email:
- m*************@gmail.com

#### Academic profile

None

None

None

None

#### Posts by manasahegde234

<prev
• 156 results •
page 1 of 16 •
next >

0

votes

1

answer

83

views

1

answers

... 1. Characteristic equation of A is $\mid A-\lambda I \mid = 0$
$\lambda^3-(8+7+3) \lambda^2 + [(56+21+24)-(36+4+16)]\lambda - \mid A \mid =0$
$\lambda^3-18\lambda^2+45\lambda-0=0$
$\lambda=0,3,15$ are eigen values of A
As all eigen values of A are distinct. A is diagonalisable.
Now to find ei ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

83

views

1

answer

... 1. $A=$
$ \left[ {\begin{array}{cc}
8 & -6 & 2\\
-6 & 7 & -4 \\
2 & -4 & 3\\
\end{array} } \right] $
2. $ A=$
$ \left[ {\begin{array}{cc}
-9 & 4 & 4\\
-8 & 3 & 4 \\
-16 & 8 & 7\\
\end{array} } \right] $
**Subject:** Applied Mat ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

91

views

1

answers

... 1. Characteristic equation of A is $\mid A-\lambda I \mid=0$
$\lambda^3-(12)\lambda^2+36\lambda-32=0$
$\lambda=8,2,2.$
$(\lambda-8)(\lambda-2)(\lambda-2)=0$
Consider $(\lambda-8)(\lambda)=\lambda^2-0\lambda+16$
Now to check that $\lambda^2-10A+16I$
Therefore, consider $A^2-10A+16I$
$= \left[ ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

91

views

1

answer

... 1. $A=$
$ \left[ {\begin{array}{cc}
6 & -2 & 2\\
-2 & 3 & -1 \\
0 & 0 & -2\\
\end{array} } \right] $
2. $A=$
$ \left[ {\begin{array}{cc}
8 & -8 & -2\\
4 & -3 & -2 \\
3 & -4 & 1\\
\end{array} } \right] $
**Subject:** Applied Mathemat ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

111

views

1

answers

... Characteristic equation of A is $\mid A-\lambda I \mid=0$
$\lambda^2-(-1+1)\lambda+\mid A \mid=0$
$\lambda^2-0+(-1-8)=0$
$\lambda^2-9=0$
$\lambda^2=9$
$\lambda=\pm3$
Let $\phi(A)=3tanA$
Consider, $\phi(A)=\alpha_1 A+ \alpha_0 I$
$3tanA=\alpha_1 A+ \alpha_0 I$ (1)
$3tan\lambda=\alpha_1 A+ \a ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

111

views

1

answer

... $A=$
$\left[ {\begin{array}{cc}
0 & -1\\
1 & 0
\end{array} } \right] $
**Subject:** Applied Mathematics 4
**Topic:** Matrices
**Difficulty:** Medium ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

79

views

1

answers

... Characteristic equation of $\mid A-\lambda I \mid=0$
Therefore, $\lambda^2-(0)\lambda+\mid A \mid=0$
$\lambda^2+(0-(-1))=0$
$\lambda^2+1=0$
$\lambda=\pm i$
Let, $\phi(A)=e^At$
Consider, $\phi(A)=\alpha_1 A+\alpha_0 I$
$e^{At}=\alpha_1 A+\alpha_0 I $ (1)
$e^{\lambda t}=\alpha_1 \lambda+\alpha ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

79

views

1

answer

... $A= $ $ \left[ {\begin{array}{cc}
0 & -1 \\
1 & 1 \\
\end{array} } \right] $
$e^{At}$=$ \left[ {\begin{array}{cc}
cost & -sint \\
sint & cost \\
\end{array} } \right] $
**Subject:** Applied Mathematics 4
**Topic:** Matrices
**Difficulty:** Medium ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

78

views

1

answers

... Characteristic equation of A is $\mid A-\lambda I \mid=0$
$\lambda^3-5\lambda^2+[(2+2+4)-(0+1+0)]\lambda-\mid A \mid=0$
$\lambda^3-5\lambda^2+7\lambda-3=0$
Therefore, Cayley-Hamilton theorem,
$A^3-5A^2+7A-3I=0$
from given polynomial in A,
i.e $A^8-5\lambda^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$,
we ...

written 11 weeks ago by
manasahegde234 •

**20**0

votes

1

answer

78

views

1

answer

... $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
where $A=$ $ \left[ {\begin{array}{cc}
-1 & 2 & 3\\
0 & 3 & 5 \\
0 & 0 & -2\\
\end{array} } \right] $
**Subject:** Applied Mathematics 4
**Topic:** Matrices
**Difficulty:** Medium ...

written 11 weeks ago by
manasahegde234 •

**20**#### Latest awards to manasahegde234

Centurion
7 months ago,
created 100 posts.

Rising Star
7 months ago,
created 50 posts within first three months of joining.

Site

Use of this site constitutes acceptance of our User
Agreement
and Privacy
Policy.