Using properties of determinants solve this matrix: R1:(x+1),-5,-6 and R2:-1,x,2 and R3:-3,2,(x+1) all of that equals to 0

We are given the following \( 3 \times 3 \) matrix \( A \) with determinant equal to 0, and we need to solve for \( x \):

\[
A = \begin{bmatrix}
x+1 & -5 & -6 \\
-1 & x & 2 \\
-3 & 2 & x+1
\end{bmatrix}
\]

We will find the determinant of this matrix, set it equal to 0, and then solve for \( x \).

### Step 1: Write the determinant of the matrix

The determinant of a \( 3 \times 3 \) matrix can be found using the following formula:
\[
\text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
\]
where the matrix is:
\[
A = \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
\]

For the given matrix \( A \), we can substitute:

\[
\text{det}(A) = (x+1) \left( x(x+1) - (2 \cdot 2) \right) - (-5) \left( -1(x+1) - (-3 \cdot 2) \right) + (-6) \left( -1(2) - (-3x) \right)
\]

### Step 2: Simplify each part of the determinant

Let's now simplify each term.

- First term: \( (x+1) \left( x(x+1) - 4 \right) = (x+1)(x^2 + x - 4) = (x+1)(x^2 + x - 4) \)
- Second term: \( -(-5) \left( -(x+1) + 6 \right) = 5 \left( -x - 1 + 6 \right) = 5(5 - x) \)
- Third term: \( -6 \left( -2 + 3x \right) = -6(3x - 2) = -18x + 12 \)

### Step 3: Expand and simplify the determinant expression

Now let's combine all terms:
\[
(x+1)(x^2 + x - 4) + 5(5 - x) + (-18x + 12)
\]
Expanding the first term:
\[
(x+1)(x^2 + x - 4) = x^3 + x^2 - 4x + x^2 + x - 4 = x^3 + 2x^2 - 3x - 4
\]
Now, the full expression is:
\[
x^3 + 2x^2 - 3x - 4 + (25 - 5x) + (-18x + 12)
\]
Combine like terms:
\[
x^3 + 2x^2 - 3x - 5x - 18x - 4 + 25 + 12 = x^3 + 2x^2 - 26x + 33
\]

### Step 4: Set the determinant to 0

Since the determinant is given to be 0, we set the expression equal to 0:
\[
x^3 + 2x^2 - 26x + 33 = 0
\]

### Step 5: Solve the cubic equation

We need to solve the cubic equation \( x^3 + 2x^2 - 26x + 33 = 0 \). This can be done using various methods like factoring or numerical methods. First, we try factoring by checking possible rational roots using the Rational Root Theorem.

Let me compute this.The solutions to the cubic equation \( x^3 + 2x^2 - 26x + 33 = 0 \) are:

\[
x = 3, \quad x = \frac{-5 + \sqrt{69}}{2}, \quad x = \frac{-5 - \sqrt{69}}{2}
\]

Thus, the values of \( x \) are \( 3 \) and two irrational roots: \( \frac{-5 + \sqrt{69}}{2} \) and \( \frac{-5 - \sqrt{69}}{2} \).

Would you like further details on the solving process or a specific explanation for any part?

Here are five related questions for further exploration:
1. How do you calculate the determinant of a matrix in general?
2. What are some methods for solving cubic equations?
3. How can the Rational Root Theorem be applied to cubic polynomials?
4. How do irrational roots affect the behavior of polynomials?
5. What is the relationship between eigenvalues and the determinant of a matrix?

**Tip:** Always verify potential roots by substituting them back into the original equation to ensure correctness.

Discover how to solve a 3x3 matrix equation with determinant equal to zero using key algebraic methods. This problem involves determinants, cubic equations, and the Rational Root Theorem. Ideal for advanced students in Grades 11-12.

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