Lineal Algebra: Inverse Matrix of Symmetric Matrix

In summary: I'm not sure where you're getting stuck. You noted that the inverse is just the adjoint matrix divided by the determinant. You have both. Just do the division.
  • #1
degs2k4
74
0

Homework Statement



Hello,

I need some help in the fist parts of two lineal algebra problems, specially with algebraic manipulation. I guess that if I rewrite the determinant nicely some terms get canceled and I can write the inverse nicely, but don't know how to do it...

Problem 1:

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Problem 2:

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The Attempt at a Solution



Problem 1:

(1) [tex] Det(A) = a(a^2-b^2) -b(ba-b^2) + b(b^2-ab) = b(b-a)(2b-\frac{a^2}{b})[/tex]

(2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

[tex]Adj(A) = \left[ \begin{matrix} a^2-b^2 & b(b-a) & b(b-a) \\ b(b-a) & a^2-b^2 & b(b-a) \\ b(b-a) & b(b-a) & a^2-b^2 \end{matrix} \right][/tex]

[tex] A^-1 = ? [/tex]
(I can write the terms outside the diagonal nicely because some parts get cancelled, but not the diagonal itself...)

Problem 2:

(1)

Sum of eigenvalues:
Trace(A) = a + b + c

Product of eigenvalues:
[tex] Det(A) = a(bc -b^2) -a(ac-ab) + 0 = abc - ab^2 - a^2c + a^2b [/tex]

(2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

[tex]Adj(A) = \left[ \begin{matrix} b(c-b) & a(b-c) & 0 \\ a(b-c) & a(c-a) & a(a-b) \\ 0 & a(a-b) & a(b-a) \end{matrix} \right][/tex]

[tex] A^-1 = ? [/tex]Thanks in advance...
 
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  • #2
degs2k4 said:
Problem 1:

(1) [tex] Det(A) = a(a^2-b^2) -b(ba-b^2) + b(b^2-ab) = b(b-a)(2b-\frac{a^2}{b})[/tex]
You made an algebra error when you simplified and factored. You should get (a-b)2(a+2b).
(2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

[tex]Adj(A) = \left[ \begin{matrix} a^2-b^2 & b(b-a) & b(b-a) \\ b(b-a) & a^2-b^2 & b(b-a) \\ b(b-a) & b(b-a) & a^2-b^2 \end{matrix} \right][/tex]

[tex] A^-1 = ? [/tex]
(I can write the terms outside the diagonal nicely because some parts get cancelled, but not the diagonal itself...)
That's correct. The diagonal elements already look pretty nice to me.
 
  • #3
degs2k4 said:
Problem 2:

(1)

Sum of eigenvalues:
Trace(A) = a + b + c

Product of eigenvalues:
[tex] Det(A) = a(bc -b^2) -a(ac-ab) + 0 = abc - ab^2 - a^2c + a^2b [/tex]

(2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

[tex]Adj(A) = \left[ \begin{matrix} b(c-b) & a(b-c) & 0 \\ a(b-c) & a(c-a) & a(a-b) \\ 0 & a(a-b) & a(b-a) \end{matrix} \right][/tex]

[tex] A^-1 = ? [/tex]

Thanks in advance...
This is all correct. You can factor det(A) and get some cancellation when you calculate A-1. You can always check your answer by multiplying A by the inverse you calculate and verify you get the identity matrix.
 
  • #4
vela said:
You made an algebra error when you simplified and factored. You should get (a-b)2(a+2b).

Thanks. However, my main problem is how to get the inverse, because I want to get some parts canceled but don't know how to do it...

vela said:
That's correct. The diagonal elements already look pretty nice to me.

My main problem is on getting the inverse matrix, not the adjoin matrix ...
 
  • #5
vela said:
This is all correct. You can factor det(A) and get some cancellation when you calculate A-1. You can always check your answer by multiplying A by the inverse you calculate and verify you get the identity matrix.

I am sorry it may sound very basic but... how could I factor that ? I have been trying but reach nowhere...
 
  • #6
degs2k4 said:
Thanks. However, my main problem is how to get the inverse, because I want to get some parts canceled but don't know how to do it...



My main problem is on getting the inverse matrix, not the adjoin matrix ...
I'm not sure where you're getting stuck. You noted that the inverse is just the adjoint matrix divided by the determinant. You have both. Just do the division.
 
  • #7
degs2k4 said:
I am sorry it may sound very basic but... how could I factor that ? I have been trying but reach nowhere...
[tex]abc-ab^2-a^2c+a^2b = a(bc-b^2-ac+ab) = a[b(c-b)-a(c-b)] = a(c-b)(b-a)[/tex]
 
  • #8
vela said:
You made an algebra error when you simplified and factored. You should get (a-b)2(a+2b).

Thanks! By the way, which method did you follow to reach that factorization?

vela said:
[tex]abc-ab^2-a^2c+a^2b = a(bc-b^2-ac+ab) = a[b(c-b)-a(c-b)] = a(c-b)(b-a)[/tex]

Thanks!

I used the factored determinants you showed to me to calculate the inverse matrix, but the calculations don't get as simplified as I thought at first. Are there any other properties of symmetric matrices that can be used for calculating the inverse faster ?

Thanks for your help...
 

What is a symmetric matrix?

A symmetric matrix is a square matrix that is equal to its transpose. This means that the elements on the main diagonal remain the same and the elements above and below the diagonal are mirrored.

What is the inverse matrix of a symmetric matrix?

The inverse matrix of a symmetric matrix is a matrix that, when multiplied by the original matrix, gives the identity matrix. In other words, it "undoes" the original matrix and brings it back to its original form.

Why is the inverse matrix of a symmetric matrix important?

The inverse matrix of a symmetric matrix is important because it allows us to solve systems of linear equations and perform other operations that require division by a matrix. It also has applications in fields such as computer graphics and cryptography.

How is the inverse matrix of a symmetric matrix calculated?

The inverse matrix of a symmetric matrix is calculated using a specific formula that involves taking the determinant and the adjugate (transpose of the cofactor matrix) of the original matrix. This can be done manually or with the help of software.

Can every symmetric matrix have an inverse matrix?

Yes, every symmetric matrix has an inverse matrix as long as its determinant is not equal to 0. This is because the determinant of a matrix is used in the formula for calculating the inverse matrix and a determinant of 0 would result in division by 0, which is undefined.

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