Thread: How to find the determinant of matrix if given characteristic equation only?

hey there..

The question is..
lambda = x (i dont know how to type the symbol..)

Let A be n x n square matrix. Suppose that
det(xI-A) = c(x) = x^3- 3x^2 + x + 6..

I know the answer is -6 by substitute lambda@x with 0..
However.. i just want to know.. why should i substitute it with 0..?

Thanks in advance..

Originally Posted by nameck

hey there..

The question is..
lambda = x (i dont know how to type the symbol..)

Let A be n x n square matrix. Suppose that
det(xI-A) = c(x) = x^3- 3x^2 + x + 6..

I know the answer is -6 by substitute lambda@x with 0..
However.. i just want to know.. why should i substitute it with 0..?

Thanks in advance..

You must be careful.



If you use x=0 you get

yup2... so.. -(det A) = 6.. so.. det A = -6 right?
ok then.. why it is 0?

Got one more question..
if i've got 4 x 4 matrix,
the sign will be positive?
(-1)^4 det(A) correct?

Originally Posted by nameck

hey there..

The question is..
lambda = x (i dont know how to type the symbol..)

Let A be n x n square matrix. Suppose that
det(xI-A) = c(x) = x^3- 3x^2 + x + 6..

I know the answer is -6 by substitute lambda@x with 0..
However.. i just want to know.. why should i substitute it with 0..?

Thanks in advance..

Since the det(xI- A), with x= 0 is just det(0-A). Multiplying one row of a matrix by -1 multiplies the determinant -1. Multiplying A by -1 multiplies all three rows by -1 and so multiplies the determinant by -1.

Yes, if you multiply a 4 by 4 matrix by -1, you have multiplied all 4 rows by -1 and so multiply the determinant by

Thanks HallsofIvy.. U are always answering my question perfectly..