The characteristic polynomialof 3 x 3 matrix A|λI - A|= λ3 + 3λ2 + 4λ - 3Let x - trace (A) and y = |A|, the determinant of A.Then,a)x/y = 3/4b)x/y = 4/3c)x = y = -3d)x = 3 and y = -3Correct answer is option 'C'. Can you explain this answer?

Question

The characteristic polynomialof 3 x 3 matrix A|&lambda;I - A|= &lambda;3 + 3&lambda;2 + 4&lambda; - 3Let x - trace (A) and y = |A|, the determinant of A.Then,a)x/y = 3/4b)x/y = 4/3c)x = y = -3d)x = 3 and y = -3Correct answer is option 'C'. Can you explain this answer?

Answer

We are given that the characteristic polynomial o f 3 x 3 matrix, A is
λ³ + 3λ² + 4λ - 3 = 0
and x = trace (A)
y = det(A)
The general characteristic polynomial of 3 x 3 matrix is given by,
λ³ - trace (A) λ² + λ + |A| = 0
Comparing this equation by the given equation, we get
trace = -3
or x = -3
and |A| = -3
or y = -3
Hence, x = y = -3

Answer

All the options are wrong becous the characteristics polynomial will be of the form x^3- ( trace A ) X^2+ ( A11 + A22 + A33) x - |A|=0 ....

Answer

The characteristic polynomial of a 3x3 matrix A is defined as the polynomial obtained by taking the determinant of the matrix A - λI, where λ is an eigenvalue and I is the identity matrix.

The characteristic polynomial of a 3x3 matrix A can be written as:
p(λ) = det(A - λI)

where A is the 3x3 matrix and λ is the eigenvalue.

For example, if A is a 3x3 matrix, the characteristic polynomial p(λ) is given by:
p(λ) = det(A - λI)
= |a - λ b c|
|d e - λ f|
|g h i - λ|

where a, b, c, d, e, f, g, h, and i are the elements of matrix A.

To compute the characteristic polynomial, you need to calculate the determinant of the 3x3 matrix A - λI.

For a 3x3 matrix, the determinant can be expanded as follows:
p(λ) = (a - λ)(ei - f h) - b(di - f g) + c(d h - e g)

Once you have expanded the determinant, you can simplify and rearrange the terms to obtain the characteristic polynomial in its final form.

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