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Problem 221

# (A) Define the characteristic polynomial of the matrix $$\mathrm{A}$$, (B) Let $\mathrm{A}=\begin{array}{ccc}\mid 1 & 2 & -1 \mid \\ \mid 1 & 0 & 1 \mid \\ \mid 4 & -4 & 5\end{array}$ Find the characteristic polynomial of $$\mathrm{A}$$.

Expert verified
The characteristic polynomial of the given matrix A is $$p(λ) = -λ^3 + 6λ^2 - 11λ + 6$$.
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## Step 1: Definition of Characteristic Polynomial

The characteristic polynomial of an n x n matrix A is the polynomial det(A - λI), where λ is the eigenvalue and I is the identity matrix of size n x n. When the polynomial is written in the form of an equation, called the characteristic equation, the determinant of (A - λI) is set to zero and its solutions are the eigenvalues of A. (B) Find the characteristic polynomial of the given matrix A = $$\begin{array}{ccc}\mid 1 & 2 & -1 \mid \\\ \mid 1 & 0 & 1 \\\ \mid 4 & -4 & 5\end{array}$$

## Step 1: Calculating (A - λI)

To find the characteristic polynomial, we first need to calculate the matrix (A - λI), where I is the 3x3 identity matrix: $$(A - λI) = \begin{bmatrix} 1-λ & 2 & -1 \\ 1 & -λ & 1 \\ 4 & -4 & 5-λ\end{bmatrix}$$

## Step 2: Determinant of (A - λI)

Now find the determinant of (A - λI): $$\operatorname{det}(A - λI) = (1-λ)\begin{vmatrix} -λ & 1 \\ -4 & 5-λ \end{vmatrix} - 2\begin{vmatrix} 1 & 1 \\ 4 & 5-λ \end{vmatrix} - (-1)\begin{vmatrix} 1 & -λ \\ 4 & -4 \end{vmatrix}.$$

## Step 3: Expanding the Determinant

Next, expand the determinant using the cofactor expansion: $$\operatorname{det}(A - λI) = (1-λ)((-λ)(5-λ) - 1(-4)) - 2((1)(5-λ) - 1(4)) + (1(1) - 4(-λ)).$$

## Step 4: Simplifying the Expression

Simplify the expression to obtain the characteristic polynomial, p(λ): $$p(λ) = (1-λ)(λ^2 - 5λ + 4) - 2(-λ + 1) + 1 + 4λ = -λ^3 + 6λ^2 - 11λ + 6.$$ So, the characteristic polynomial of matrix A is $$p(λ) = -λ^3 + 6λ^2 - 11λ + 6$$.

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