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Matrix A =
3 2
-1 -1

Using this relation, A^2+2A+3I = 0 , find 1/A(inverse of A).

2007-10-01 04:56:03 · 3 answers · asked by aMused 2 in Science & Mathematics Mathematics

Okay here is the question. If A =
3 2
-1 -1

find the values of m and n such that A^2+mA+nI=0 . Hence, find inverse of A using the relation above.

2007-10-01 05:22:11 · update #1

3 answers

I don't want to waste your time, so I will tell you up front that I am not going to give you the complete solution. I will give you some helpful hints at the solution.
The quadratic equation is A^2+mA+nI=0 is called the characteristic polynimial or the eigenvalue equation. You calculated like this take your matrix.
3 2
-1 -1

(3-x) 2
-1 (-1-x)

and solve it by the usual methods of the determinant.
(3-x)(-1-x)-2(-1)=(-3-3x+x+x^2)+2=x^2-2*x-1=0 is quadratic.
Solve it using the quadratic formula and x= 1+√2 and
x=1-√2 likewise
(A-xI)=0 either solution another interesting property.
A^2-2A-1=0 for the matrix.
A^2=
7 4
-2 -1
-2*A=
-6 -4
+2 +2
-1*I =
-1 0
0 -1
Add all of these matrix by matrix addition and the results is
0 0
0 0
So this is a good check to see if you have the correct characteristic equation.

Now I promise you a broad hint for your solution here it goes.
(x+a)(x+b)=0 therefore
x^2+(a+b)x+ab=0
and
(1/ab)x^2+((a+b)/(ab))x+1=0
and
(1/a)(1/b)x^2+((1/a)+(1/b))x+1=0
or if you divded through all terms by the constant term you gets the reciprocal roots via.
y^2+((1/a)+(1/b))y+(1/a)(1b))=0 is the characteristic equation for the reciprocal of x, can the same be true for matrices. Yes or no?

Here is the last hint the reciprocal matrix for your problem is
1.000 2.000
-1.000 -3.000

If you still can't figure it out in day or so. Come back and I will try to help you a little more. But I think you are bright person and I wouldn't want to spoil all the fun for you.

2007-10-05 09:43:39 · answer #1 · answered by alints_2000 4 · 0 0

A^2 + 2A + 3I = 0 ( is given i dont check it )

the I is the identity matrix

with that you need to find 1/A

A^2 + 2A + 3I = 0
A(A+2) = -3I

(A+2) = -3*(1/A)*I ; here i 'divide' by A or multiply by 1/A

(-1/3)* (A+2) = 1/A ; deivide by -3.


so there you are :
1/A = (-1/3)* (A+2) : fill in the matrix A in the right hand side.

.

2007-10-01 12:59:03 · answer #2 · answered by gjmb1960 7 · 0 1

the ? is wrong or u have done some mistake please check it once again

2007-10-01 12:10:04 · answer #3 · answered by amal 2 · 0 1

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