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

Ask by Hilton Li. in India
Jan 24,2025

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Answer

\( a = -6 \) and \( b = 6 \).

Solution

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 \((-6,\ 6)\).

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To find the values of \( a \) and \( b \) that satisfy the matrix equation \( M^{3}+a M^{2}+b M+3=0 \), we first need to compute \( M^2 \) and \( M^3 \). Calculating \( M^2 \): \[ M^2 = M \cdot M = \left(\begin{array}{ccc}3 & -1 & 2 \\ -1 & 2 & 0 \\ 2 & 0 & 1\end{array}\right) \cdot \left(\begin{array}{ccc}3 & -1 & 2 \\ -1 & 2 & 0 \\ 2 & 0 & 1\end{array}\right) \] Calculating the individual entries of \( M^2 \): - First row: - \( (3*3) + (-1*-1) + (2*2) = 9 + 1 + 4 = 14 \) - \( (3*-1) + (-1*2) + (2*0) = -3 - 2 + 0 = -5 \) - \( (3*2) + (-1*0) + (2*1) = 6 + 0 + 2 = 8 \) - Second row: - \( (-1*3) + (2*-1) + (0*2) = -3 - 2 + 0 = -5 \) - \( (-1*-1) + (2*2) + (0*0) = 1 + 4 + 0 = 5 \) - \( (-1*2) + (2*0) + (0*1) = -2 + 0 + 0 = -2 \) - Third row: - \( (2*3) + (0*-1) + (1*2) = 6 + 0 + 2 = 8 \) - \( (2*-1) + (0*2) + (1*0) = -2 + 0 + 0 = -2 \) - \( (2*2) + (0*0) + (1*1) = 4 + 0 + 1 = 5 \) Thus, \[ M^2 = \begin{pmatrix} 14 & -5 & 8 \\ -5 & 5 & -2 \\ 8 & -2 & 5 \end{pmatrix} \] Next, we calculate \( M^3 = M \cdot M^2 \): \[ M^3 = \left(\begin{array}{ccc}3 & -1 & 2 \\ -1 & 2 & 0 \\ 2 & 0 & 1\end{array}\right) \cdot \left(\begin{array}{ccc} 14 & -5 & 8 \\ -5 & 5 & -2 \\ 8 & -2 & 5 \end{array}\right) \] Calculating the individual entries of \( M^3 \): - First row: - \( (3*14) + (-1*-5) + (2*8) = 42 + 5 + 16 = 63 \) - \( (3*-5) + (-1*5) + (2*-2) = -15 - 5 - 4 = -24 \) - \( (3*8) + (-1*-2) + (2*5) = 24 + 2 + 10 = 36 \) - Second row: - \( (-1*14) + (2*-5) + (0*8) = -14 - 10 + 0 = -24 \) - \( (-1*-5) + (2*5) + (0*-2) = 5 + 10 + 0 = 15 \) - \( (-1*8) + (2*-2) + (0*5) = -8 - 4 + 0 = -12 \) - Third row: - \( (2*14) + (0*-5) + (1*8) = 28 + 0 + 8 = 36 \) - \( (2*-5) + (0*5) + (1*-2) = -10 + 0 - 2 = -12 \) - \( (2*8) + (0*-2) + (1*5) = 16 + 0 + 5 = 21 \) Thus, \[ M^3 = \begin{pmatrix}

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