Let A = `[{:(3, 0, 3),(0, 3,0),(3, 0,3):}]` . Then the roots of the equation det
`(A - lambda I_(3)) = 0 ` (where `I_(3)` is the identity matrix of order 3) are

Options

A. 3, 0 , 3
B. 0, 3, 6
C. 1, 0 , - 6
D. 3, 3, 6

Answer

To find the roots of the equation \( \text{det}(A - \lambda I_3) = 0 \), where \( A = \begin{pmatrix} 3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{pmatrix} \) and \( I_3 \) is the identity matrix of order 3, we will follow these steps:

### Step 1: Write the matrix \( A - \lambda I_3 \)

The identity matrix \( I_3 \) is given by:

\[ I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus, \( \lambda I_3 \) is:

\[ \lambda I_3 = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} \]

Now, we can compute \( A - \lambda I_3 \):

\[ A - \lambda I_3 = \begin{pmatrix} 3 - \lambda & 0 & 3 \\ 0 & 3 - \lambda & 0 \\ 3 & 0 & 3 - \lambda \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A - \lambda I_3 \)

To find the determinant, we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A - \lambda I_3 \):

\[ B = \begin{pmatrix} 3 - \lambda & 0 & 3 \\ 0 & 3 - \lambda & 0 \\ 3 & 0 & 3 - \lambda \end{pmatrix} \]

Calculating the determinant:

\[ \text{det}(A - \lambda I_3) = (3 - \lambda) \left( (3 - \lambda)(3 - \lambda) - 0 \cdot 0 \right) - 0 + 3 \left( 0 - 3(3 - \lambda) \right) \]

This simplifies to:

\[ = (3 - \lambda)((3 - \lambda)^2) - 9(3 - \lambda) \]

### Step 3: Simplify the determinant expression

Expanding the expression:

\[ = (3 - \lambda)((3 - \lambda)^2 - 9) \]

Now, we can factor \( (3 - \lambda)^2 - 9 \):

\[ = (3 - \lambda)((3 - \lambda - 3)(3 - \lambda + 3)) = (3 - \lambda)(-\lambda)(6 - \lambda) \]

### Step 4: Set the determinant to zero

Now we set the determinant equal to zero:

\[ (3 - \lambda)(-\lambda)(6 - \lambda) = 0 \]

### Step 5: Solve for \( \lambda \)

The solutions to this equation are:

1. \( 3 - \lambda = 0 \) → \( \lambda = 3 \)
2. \( -\lambda = 0 \) → \( \lambda = 0 \)
3. \( 6 - \lambda = 0 \) → \( \lambda = 6 \)

Thus, the roots of the equation \( \text{det}(A - \lambda I_3) = 0 \) are:

\[ \lambda = 0, 3, 6 \]

### Final Answer:

The roots of the equation are \( \lambda = 0, 3, 6 \).

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