If A=⎡⎢⎣201213110⎤⎥⎦ then find A2−2A+5I.

To solve the expression \( A^2 - 2A + 5I \), where \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} \) and \( I \) is the identity matrix, we'll follow these steps:

### Step 1: Find \( A^2 \)
First, we calculate the square of matrix \( A \):

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

Let's perform the multiplication:

\[
A^2 = \begin{pmatrix}
(2 \times 2) + (0 \times 2) + (1 \times 1) & (2 \times 0) + (0 \times 1) + (1 \times 1) & (2 \times 1) + (0 \times 3) + (1 \times 0) \\
(2 \times 2) + (1 \times 2) + (3 \times 1) & (2 \times 0) + (1 \times 1) + (3 \times 1) & (2 \times 1) + (1 \times 3) + (3 \times 0) \\
(1 \times 2) + (1 \times 2) + (0 \times 1) & (1 \times 0) + (1 \times 1) + (0 \times 1) & (1 \times 1) + (1 \times 3) + (0 \times 0)
\end{pmatrix}
\]

Simplifying each element:

\[
A^2 = \begin{pmatrix}
4 + 0 + 1 & 0 + 0 + 1 & 2 + 0 + 0 \\
4 + 2 + 3 & 0 + 1 + 3 & 2 + 3 + 0 \\
2 + 2 + 0 & 0 + 1 + 0 & 1 + 3 + 0
\end{pmatrix}
\]

\[
A^2 = \begin{pmatrix}
5 & 1 & 2 \\
9 & 4 & 5 \\
4 & 1 & 4
\end{pmatrix}
\]

### Step 2: Calculate \( 2A \)
Next, we find \( 2A \):

\[
2A = 2 \times \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 2 \\ 4 & 2 & 6 \\ 2 & 2 & 0 \end{pmatrix}
\]

### Step 3: Calculate \( 5I \)
Now, calculate \( 5I \), where \( I \) is the 3x3 identity matrix:

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

\[
5I = 5 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}
\]

### Step 4: Combine the results
Finally, substitute the values into the expression \( A^2 - 2A + 5I \):

\[
A^2 - 2A + 5I = \begin{pmatrix} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 2 \\ 4 & 2 & 6 \\ 2 & 2 & 0 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}
\]

Let's subtract and then add the matrices:

\[
A^2 - 2A = \begin{pmatrix} 5-4 & 1-0 & 2-2 \\ 9-4 & 4-2 & 5-6 \\ 4-2 & 1-2 & 4-0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 5 & 2 & -1 \\ 2 & -1 & 4 \end{pmatrix}
\]

\[
A^2 - 2A + 5I = \begin{pmatrix} 1 & 1 & 0 \\ 5 & 2 & -1 \\ 2 & -1 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 6 & 1 & 0 \\ 5 & 7 & -1 \\ 2 & -1 & 9 \end{pmatrix}
\]

Thus, the final result is:

\[
A^2 - 2A + 5I = \begin{pmatrix} 6 & 1 & 0 \\ 5 & 7 & -1 \\ 2 & -1 & 9 \end{pmatrix}
\]

Would you like more details on any of the steps? Here are five related questions:

1. How do you multiply two matrices?
2. What is the identity matrix and its properties?
3. How do you calculate the determinant of a matrix?
4. What are eigenvalues and eigenvectors?
5. How do you find the inverse of a matrix?

**Tip:** Remember, the order of multiplication matters in matrix operations; matrix multiplication is not commutative.

Learn how to calculate A^2 - 2A + 5I for a given matrix A = [2 0 1; 2 1 3; 1 1 0]. This problem involves matrix multiplication, identity matrix properties, and linear algebra concepts. Suitable for undergraduate level students.

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