If `A=[{:(8,-6,2),(-6,7,-4),(2,-4,3):}]` and `X` is a non zero column matrix such that `AX=lambdaX`, where `lambda` is a scalar, then values of `lambda` can be

Options

A. `3`
B. `6`
C. `12`
D. `15`

Explanation

To solve the problem, we need to find the values of \( \lambda \) such that \( AX = \lambda X \) for the given matrix \( A \) and a non-zero column matrix \( X \). This leads us to find the eigenvalues of the matrix \( A \).

Given:
\[ A = \begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix} \]
We need to find the values of \( \lambda \) such that the equation \( AX = \lambda X \) holds true.

### Step 1: Set up the equation
We can rewrite the equation \( AX = \lambda X \) as:
\[ AX - \lambda X = 0 \]
This can be expressed as:
\[ (A - \lambda I)X = 0 \]
where \( I \) is the identity matrix of the same size as \( A \).

### Step 2: Formulate the matrix \( A - \lambda I \)
The identity matrix \( I \) for a 3x3 matrix is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus, \( \lambda I \) is:
\[ \lambda I = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} \]
Now, we can compute \( A - \lambda I \):
\[ A - \lambda I = \begin{pmatrix} 8 - \lambda & -6 & 2 \\ -6 & 7 - \lambda & -4 \\ 2 & -4 & 3 - \lambda \end{pmatrix} \]

### Step 3: Set the determinant to zero
To find the eigenvalues, we need to set the determinant of \( A - \lambda I \) to zero:
\[ \text{det}(A - \lambda I) = 0 \]

### Step 4: Calculate the determinant
We will compute the determinant:
\[ \text{det}(A - \lambda I) = \begin{vmatrix} 8 - \lambda & -6 & 2 \\ -6 & 7 - \lambda & -4 \\ 2 & -4 & 3 - \lambda \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c, d, e, f, g, h, i \) are the elements of the matrix.

Applying this to our matrix:
\[ = (8 - \lambda)((7 - \lambda)(3 - \lambda) - (-4)(-4)) - (-6)(-6(3 - \lambda) - (-4)(2)) + 2(-6(-4) - (7 - \lambda)(2)) \]

### Step 5: Simplify the determinant expression
After calculating and simplifying, we will get a polynomial in \( \lambda \). The final polynomial will be:
\[ -\lambda^3 + 18\lambda^2 - 45\lambda = 0 \]

### Step 6: Factor the polynomial
Factoring out \( \lambda \):
\[ \lambda(\lambda^2 - 18\lambda + 45) = 0 \]
This gives us \( \lambda = 0 \) or solving the quadratic equation \( \lambda^2 - 18\lambda + 45 = 0 \).

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -18, c = 45 \):
\[ \lambda = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
\[ = \frac{18 \pm \sqrt{324 - 180}}{2} \]
\[ = \frac{18 \pm \sqrt{144}}{2} \]
\[ = \frac{18 \pm 12}{2} \]
This gives us:
\[ \lambda = \frac{30}{2} = 15 \quad \text{and} \quad \lambda = \frac{6}{2} = 3 \]

### Final Result
The values of \( \lambda \) are \( 0, 3, \) and \( 15 \).