Finding x such that 1 is eigenvalue

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Let A=[1 2 3;3 x 4; 1 2 5], find x such that 1 is eigenvalue of the matrix
syms x
det(A=[1 2 3;3 x 4; 1 2 5]-eye(3));
Which function can I use to find the roots of the polynomial I got

Accepted Answer

John D'Errico
John D'Errico on 17 Jun 2018
You are trying to do too much in one line. Making your code super compact does nothing, except make it unreadable. And it did not work here for you, as you found out.
First, you cannot assign something to a variable like A in the middle of a line of code.
If 1 is an eigenvalue of A, then it is true that det(A-1*eye(3)) must be zero.
syms x
A = [1 2 3;3 x 4; 1 2 5];
xsol = solve(det(A - eye(3)) == 0)
xsol =
5/3
Testing that result, we see that:
eig(subs(A,x,xsol))
ans =
1
10/3 - 178^(1/2)/3
178^(1/2)/3 + 10/3
So, you were close. What you did wrong was to try to do too much in one line of code. And then you failed to try to apply solve. Since your goal was to solve something, why not consider solve?

More Answers (1)

James Tursa
James Tursa on 17 Jun 2018
Edited: James Tursa on 17 Jun 2018
>> syms x
>> det([1 2 3;3 x 4; 1 2 5]-eye(3))
ans =
5 - 3*x
>> solve(ans)
ans =
5/3
>> A = [1 2 3;3 5/3 4; 1 2 5]
A =
1.0000 2.0000 3.0000
3.0000 1.6667 4.0000
1.0000 2.0000 5.0000
>> eig(A)
ans =
7.7806
-1.1139
1.0000

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