If A=`[{:(,1,0,2),(,0,2,1),(,2,0,3):}]` is a root of polynomial `x^(3)-6x^(2)+7x+k=0` then the value of k is

Options

A. 2
B. 4
C. `-2`
D. 1

Answer

To find the value of \( k \) in the polynomial \( x^3 - 6x^2 + 7x + k = 0 \) given that the matrix \( A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \) is a root, we will follow these steps:

### Step 1: Define the polynomial function
Let \( f(x) = x^3 - 6x^2 + 7x + k \).

### Step 2: Substitute the matrix \( A \) into the polynomial
We need to evaluate \( f(A) \):
\[ f(A) = A^3 - 6A^2 + 7A + kI_3 \]
where \( I_3 \) is the \( 3 \times 3 \) identity matrix.

### Step 3: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \]

Calculating the elements:
- First row:
- \( 1 \cdot 1 + 0 \cdot 0 + 2 \cdot 2 = 1 + 0 + 4 = 5 \)
- \( 1 \cdot 0 + 0 \cdot 2 + 2 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 1 \cdot 2 + 0 \cdot 1 + 2 \cdot 3 = 2 + 0 + 6 = 8 \)

- Second row:
- \( 0 \cdot 1 + 2 \cdot 0 + 1 \cdot 2 = 0 + 0 + 2 = 2 \)
- \( 0 \cdot 0 + 2 \cdot 2 + 1 \cdot 0 = 0 + 4 + 0 = 4 \)
- \( 0 \cdot 2 + 2 \cdot 1 + 1 \cdot 3 = 0 + 2 + 3 = 5 \)

- Third row:
- \( 2 \cdot 1 + 0 \cdot 0 + 3 \cdot 2 = 2 + 0 + 6 = 8 \)
- \( 2 \cdot 0 + 0 \cdot 2 + 3 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 2 \cdot 2 + 0 \cdot 1 + 3 \cdot 3 = 4 + 0 + 9 = 13 \)

Thus,
\[ A^2 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} \]

### Step 4: Calculate \( A^3 \)
Now, we calculate \( A^3 = A^2 \cdot A \):
\[ A^3 = \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \]

Calculating the elements:
- First row:
- \( 5 \cdot 1 + 0 \cdot 0 + 8 \cdot 2 = 5 + 0 + 16 = 21 \)
- \( 5 \cdot 0 + 0 \cdot 2 + 8 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 5 \cdot 2 + 0 \cdot 1 + 8 \cdot 3 = 10 + 0 + 24 = 34 \)

- Second row:
- \( 2 \cdot 1 + 4 \cdot 0 + 5 \cdot 2 = 2 + 0 + 10 = 12 \)
- \( 2 \cdot 0 + 4 \cdot 2 + 5 \cdot 0 = 0 + 8 + 0 = 8 \)
- \( 2 \cdot 2 + 4 \cdot 1 + 5 \cdot 3 = 4 + 4 + 15 = 23 \)

- Third row:
- \( 8 \cdot 1 + 0 \cdot 0 + 13 \cdot 2 = 8 + 0 + 26 = 34 \)
- \( 8 \cdot 0 + 0 \cdot 2 + 13 \cdot 0 = 0 + 0 + 0 = 0 \)
- \( 8 \cdot 2 + 0 \cdot 1 + 13 \cdot 3 = 16 + 0 + 39 = 55 \)

Thus,
\[ A^3 = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} \]

### Step 5: Substitute \( A \), \( A^2 \), and \( A^3 \) into \( f(A) \)
Now we will substitute \( A \), \( A^2 \), and \( A^3 \) into the polynomial:
\[ f(A) = A^3 - 6A^2 + 7A + kI_3 \]
Substituting the values:
\[ f(A) = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} - 6 \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} + 7 \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} + k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating \( -6A^2 \):
\[ -6A^2 = -6 \begin{pmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{pmatrix} = \begin{pmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{pmatrix} \]

Calculating \( 7A \):
\[ 7A = 7 \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix} \]

Now, adding these matrices together:
\[ f(A) = \begin{pmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{pmatrix} + \begin{pmatrix} -30 & 0 & -48 \\ -12 & -24 & -30 \\ -48 & 0 & -78 \end{pmatrix} + \begin{pmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{pmatrix} + k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the sum:
\[ = \begin{pmatrix} 21 - 30 + 7 & 0 + 0 + 0 & 34 - 48 + 14 \\ 12 - 12 + 0 & 8 - 24 + 14 & 23 - 30 + 7 \\ 34 - 48 + 14 & 0 + 0 + 0 & 55 - 78 + 21 \end{pmatrix} + k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This simplifies to:
\[ = \begin{pmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{pmatrix} + k \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 6: Set \( f(A) = 0 \)
For \( A \) to be a root, we need \( f(A) = 0 \):
\[ \begin{pmatrix} -2 + k & 0 & 0 \\ 0 & -2 + k & 0 \\ 0 & 0 & -2 + k \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

This gives us the equation:
\[ -2 + k = 0 \]
Thus, solving for \( k \):
\[ k = 2 \]

### Final Answer
The value of \( k \) is \( \boxed{2} \).