If `A=[{:(2,0,1),(2,1,3),(1,-1,0):}]`. find `A^(2)-5A+6I` and hence , find a matrix `X` such that `A^(2)-5A+6I+X=O`.

Answer

To solve the problem, we need to find \( A^2 - 5A + 6I \) for the given matrix \( A \) and then determine the matrix \( X \) such that \( A^2 - 5A + 6I + X = O \).

### Step 1: Define the Matrix \( A \)

The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)

To find \( A^2 \), we perform the matrix multiplication \( A \times A \).

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- First row:
- \( (2*2 + 0*2 + 1*1) = 4 + 0 + 1 = 5 \)
- \( (2*0 + 0*1 + 1*(-1)) = 0 + 0 - 1 = -1 \)
- \( (2*1 + 0*3 + 1*0) = 2 + 0 + 0 = 2 \)

- Second row:
- \( (2*2 + 1*2 + 3*1) = 4 + 2 + 3 = 9 \)
- \( (2*0 + 1*1 + 3*(-1)) = 0 + 1 - 3 = -2 \)
- \( (2*1 + 1*3 + 3*0) = 2 + 3 + 0 = 5 \)

- Third row:
- \( (1*2 + (-1)*2 + 0*1) = 2 - 2 + 0 = 0 \)
- \( (1*0 + (-1)*1 + 0*(-1)) = 0 - 1 + 0 = -1 \)
- \( (1*1 + (-1)*3 + 0*0) = 1 - 3 + 0 = -2 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} \]

### Step 3: Calculate \( -5A \)

Now, we calculate \( -5A \):
\[ -5A = -5 \cdot \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix} \]

### Step 4: Calculate \( 6I \)

The identity matrix \( I \) for a \( 3 \times 3 \) matrix is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ 6I = 6 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

### Step 5: Combine \( A^2 - 5A + 6I \)

Now we combine the results:
\[ A^2 - 5A + 6I = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix} + \begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix} + \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

Calculating each element:
- First row:
- \( 5 - 10 + 6 = 1 \)
- \( -1 + 0 + 0 = -1 \)
- \( 2 - 5 + 0 = -3 \)

- Second row:
- \( 9 - 10 + 0 = -1 \)
- \( -2 - 5 + 6 = -1 \)
- \( 5 - 15 + 0 = -10 \)

- Third row:
- \( 0 - 5 + 0 = -5 \)
- \( -1 + 5 + 0 = 4 \)
- \( -2 + 0 + 6 = 4 \)

Thus, we have:
\[ A^2 - 5A + 6I = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix} \]

### Step 6: Find Matrix \( X \)

We need to find \( X \) such that:
\[ A^2 - 5A + 6I + X = O \]
This implies:
\[ X = - (A^2 - 5A + 6I) \]
Thus,
\[ X = - \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix} = \begin{pmatrix} -1 & 1 & 3 \\ 1 & 1 & 10 \\ 5 & -4 & -4 \end{pmatrix} \]

### Final Result

The final answer is:
\[ A^2 - 5A + 6I = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix} \]
and
\[ X = \begin{pmatrix} -1 & 1 & 3 \\ 1 & 1 & 10 \\ 5 & -4 & -4 \end{pmatrix} \]