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

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

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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:

$$
\mu \in \mathbb{R} \setminus \left\{ \frac{11 + \sqrt{265}}{12}, \frac{11 - \sqrt{265}}{12} \right\}
$$
The values of $$ \mu $$ for which the rank of matrix $$ A $$ is 3 are all real numbers except $$ \frac{11 + \sqrt{265}}{12} $$ and $$ \frac{11 - \sqrt{265}}{12} $$.

Find all values of if , where (Hint change row )

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