If A=[(1,0,3),(-2,4,5),(2,0,1)] then find A^2−2A+5I.

Given the matrix \( A = \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix} \), we need to find \( A^2 - 2A + 5I \), where \( I \) is the identity matrix.

Let's break it down step by step:

### 1. Compute \( A^2 \)
First, we need to multiply \( A \) by itself:

\[
A^2 = A \times A = \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix}
\]

\[
= \begin{pmatrix}
(1 \times 1) + (0 \times -2) + (3 \times 2) & (1 \times 0) + (0 \times 4) + (3 \times 0) & (1 \times 3) + (0 \times 5) + (3 \times 1) \\
(-2 \times 1) + (4 \times -2) + (5 \times 2) & (-2 \times 0) + (4 \times 4) + (5 \times 0) & (-2 \times 3) + (4 \times 5) + (5 \times 1) \\
(2 \times 1) + (0 \times -2) + (1 \times 2) & (2 \times 0) + (0 \times 4) + (1 \times 0) & (2 \times 3) + (0 \times 5) + (1 \times 1)
\end{pmatrix}
\]

\[
= \begin{pmatrix}
1 + 0 + 6 & 0 + 0 + 0 & 3 + 0 + 3 \\
-2 - 8 + 10 & 0 + 16 + 0 & -6 + 20 + 5 \\
2 + 0 + 2 & 0 + 0 + 0 & 6 + 0 + 1
\end{pmatrix}
\]

\[
= \begin{pmatrix} 7 & 0 & 6 \\ 0 & 16 & 19 \\ 4 & 0 & 7 \end{pmatrix}
\]

### 2. Compute \( 2A \)
Now, multiply \( A \) by 2:

\[
2A = 2 \times \begin{pmatrix} 1 & 0 & 3 \\ -2 & 4 & 5 \\ 2 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 6 \\ -4 & 8 & 10 \\ 4 & 0 & 2 \end{pmatrix}
\]

### 3. Compute \( 5I \)
Here, \( I \) is the identity matrix:

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

Now multiply it by 5:

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

### 4. Compute \( A^2 - 2A + 5I \)
Finally, subtract \( 2A \) from \( A^2 \) and add \( 5I \):

\[
A^2 - 2A + 5I = \begin{pmatrix} 7 & 0 & 6 \\ 0 & 16 & 19 \\ 4 & 0 & 7 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 6 \\ -4 & 8 & 10 \\ 4 & 0 & 2 \end{pmatrix} + \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}
\]

\[
= \begin{pmatrix} (7 - 2 + 5) & (0 - 0 + 0) & (6 - 6 + 0) \\ (0 + 4 + 0) & (16 - 8 + 5) & (19 - 10 + 0) \\ (4 - 4 + 0) & (0 - 0 + 0) & (7 - 2 + 5) \end{pmatrix}
\]

\[
= \begin{pmatrix} 10 & 0 & 0 \\ 4 & 13 & 9 \\ 0 & 0 & 10 \end{pmatrix}
\]

Thus, \( A^2 - 2A + 5I = \begin{pmatrix} 10 & 0 & 0 \\ 4 & 13 & 9 \\ 0 & 0 & 10 \end{pmatrix} \).

Would you like further details or have any questions about this solution?

### Related Questions:
1. How can you verify that the calculation of \( A^2 \) is correct?
2. What would change if the scalar multiplication was by another number instead of 2 in \( 2A \)?
3. How is the identity matrix \( I \) important in linear algebra?
4. How would the result change if the matrix \( A \) had different dimensions?
5. What are the properties of matrix multiplication that apply to this problem?

### Tip:
Always double-check matrix multiplications by verifying each element calculation to avoid small errors that could lead to incorrect results.

Learn how to compute A^2 - 2A + 5I for the matrix A = [[1, 0, 3], [-2, 4, 5], [2, 0, 1]]. This problem involves matrix algebra, including matrix multiplication and the use of the identity matrix. Suitable for undergraduate level study.

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