Find all values of \( \mu \) if \( \rho(A)=3 \), where (Hint change row \( R_{A A} \) ) \( A=\left[\begin{array}{rrrr}\mu & -1 & 0 & 0 \\ 0 & \mu & -1 & 0 \\ -6 & 11 & -6 & 1 \\ 0 & 0 & \mu & -1\end{array}\right] \)

To find the values of \( \mu \) such that the rank of the matrix \( A \) is 3, we will perform row operations to simplify the matrix and analyze its rank. The matrix \( A \) is given by: \[ A = \begin{bmatrix} \mu & -1 & 0 & 0 \\ 0 & \mu & -1 & 0 \\ -6 & 11 & -6 & 1 \\ 0 & 0 & \mu & -1 \end{bmatrix} \] We will perform row operations to simplify this matrix. Let's start by changing row \( R_1 \) to eliminate the first element of \( R_3 \). 1. **Row Operation**: Replace \( R_3 \) with \( R_3 + \frac{6}{\mu} R_1 \) (assuming \( \mu \neq 0 \)): \[ R_3 = -6 + \frac{6}{\mu} \cdot \mu = 0 \] \[ R_3 = 11 - \frac{6}{\mu} \cdot (-1) = 11 + \frac{6}{\mu} \] \[ R_3 = -6 + \frac{6}{\mu} \cdot 0 = -6 \] \[ R_3 = 1 + \frac{6}{\mu} \cdot 0 = 1 \] Thus, the new \( R_3 \) becomes: \[ R_3 = \begin{bmatrix} 0 & 11 + \frac{6}{\mu} & -6 & 1 \end{bmatrix} \] Now, the matrix \( A \) becomes: \[ A = \begin{bmatrix} \mu & -1 & 0 & 0 \\ 0 & \mu & -1 & 0 \\ 0 & 11 + \frac{6}{\mu} & -6 & 1 \\ 0 & 0 & \mu & -1 \end{bmatrix} \] Next, we will analyze the rank of this matrix. The rank will be 3 if we can find a non-zero row in the reduced matrix. 2. **Row Operation**: Now, we can focus on the second column. We can replace \( R_3 \) with \( R_3 - \frac{11 + \frac{6}{\mu}}{\mu} R_2 \): \[ R_3 = \begin{bmatrix} 0 & 11 + \frac{6}{\mu} - \frac{11 + \frac{6}{\mu}}{\mu} \cdot \mu & -6 + \frac{11 + \frac{6}{\mu}}{\mu} \cdot (-1) & 1 \end{bmatrix} \] This simplifies to: \[ R_3 = \begin{bmatrix} 0 & 0 & -6 + \frac{11 + \frac{6}{\mu}}{\mu} & 1 \end{bmatrix} \] Now, we need to ensure that the third row is non-zero for the rank to be 3. This means we need to solve: \[ -6 + \frac{11 + \frac{6}{\mu}}{\mu} \neq 0 \] This leads to: \[ -6 + \frac{11\mu + 6}{\mu^2} \neq 0 \] Multiplying through by \( \mu^2 \) (assuming \( \mu \neq 0 \)) gives: \[ -6\mu^2 + 11\mu + 6 \neq 0 \] Now we can find the roots of the quadratic equation: \[ -6\mu^2 + 11\mu + 6 = 0 \] Using the quadratic formula: \[ \mu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot (-6) \cdot 6}}{2 \cdot (-6)} \] Calculating the discriminant: \[ 11^2 - 4 \cdot (-6) \cdot 6 = 121 + 144 = 265 \] Thus, the roots are: \[ \mu = \frac{-11 \pm \sqrt{265}}{-12} \] This gives us two values for \( \mu \): \[ \mu_1 = \frac{11 + \sqrt{265}}{12}, \quad \mu_2 = \frac{11 - \sqrt{265}}{12} \] The rank of \( A \) will be 3 for all \( \mu \) except these two values. Therefore, the final answer is: \[ \mu \in \mathbb{R} \setminus \left\{ \frac{11 + \sqrt{265}}{12}, \frac{11 - \sqrt{265}}{12} \right\} \]