The matrix $$ M=\left(\begin{array}{ccc}3 & -1 & 2 \ -1 & 2 & 0 \ 2 & 0 & 1\end{array} ight) $$ satisfies the equation $$ M^{3}+a M^{2}+b M+3=0 $$ if $$ (a, b) $$ are

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The matrix $$ M=\left(\begin{array}{ccc}3 & -1 & 2 \ -1 & 2 & 0 \ 2 & 0 & 1\end{array}ight) $$ satisfies the equation $$ M^{3}+a M^{2}+b M+3=0 $$ if $$ (a, b) $$ are

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To determine the values of $$ a $$ and $$ b $$ such that the matrix $$ M $$ satisfies the equation
$$
M^3 + aM^2 + bM + 3I = 0,
$$
where $$ I $$ is the identity matrix, we can use the **Cayley-Hamilton theorem**. This theorem states that every square matrix satisfies its own characteristic equation.

### Step 1: Compute the Characteristic Polynomial of $$ M $$

The characteristic polynomial of $$ M $$ is given by:
$$
\det(M - tI) = -t^3 + (\text{trace } M) t^2 - (\text{sum of principal minors}) t + \det(M).
$$
Let's compute each component:

1. **Trace of $$ M $$:**
   $$
   \text{trace } M = 3 + 2 + 1 = 6.
   $$

2. **Sum of Principal Minors:**
   The principal minors are the determinants of the $$ 2 \times 2 $$ principal submatrices:
   $$
   \begin{aligned}
   &\det\begin{pmatrix} 2 & 0 \ 0 & 1 \end{pmatrix} = 2, \
   &\det\begin{pmatrix} 3 & 2 \ 2 & 1 \end{pmatrix} = 3(1) - 2(2) = -1, \
   &\det\begin{pmatrix} 3 & -1 \ -1 & 2 \end{pmatrix} = 3(2) - (-1)(-1) = 5.
   \end{aligned}
   $$
   Thus,
   $$
   \text{sum of principal minors} = 2 + (-1) + 5 = 6.
   $$

3. **Determinant of $$ M $$:**
   $$
   \det(M) = 3(2 \cdot 1 - 0 \cdot 0) - (-1)(-1 \cdot 1 - 0 \cdot 2) + 2(-1 \cdot 0 - 2 \cdot 2) = 6 - 1 - 8 = -3.
   $$

So, the characteristic polynomial is:
$$
-t^3 + 6t^2 - 6t - 3 = 0.
$$

### Step 2: Apply the Cayley-Hamilton Theorem

By the Cayley-Hamilton theorem,
$$
M^3 - 6M^2 + 6M - 3I = 0.
$$
Comparing this with the given equation $$ M^3 + aM^2 + bM + 3I = 0 $$, we equate the coefficients:
$$
a = -6 \quad \text{and} \quad b = 6.
$$

### Final Answer

The ordered pair $$(a, b)$$ is $$(-6,\ 6)$$.
$$ a = -6 $$ and $$ b = 6 $$.

The matrix satisfies the equation if are

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