To solve the problem, we need to show that \( A^2 - 7A + 10I_3 = 0 \) for the matrix \( A = \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix} \) and then find the inverse of \( A \).
### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[
A^2 = A \cdot A = \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix} \cdot \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}
\]
Calculating each element of the resulting matrix:
- First row, first column: \( 3 \cdot 3 + 2 \cdot 1 + 0 \cdot 0 = 9 + 2 + 0 = 11 \)
- First row, second column: \( 3 \cdot 2 + 2 \cdot 4 + 0 \cdot 0 = 6 + 8 + 0 = 14 \)
- First row, third column: \( 3 \cdot 0 + 2 \cdot 0 + 0 \cdot 5 = 0 + 0 + 0 = 0 \)
- Second row, first column: \( 1 \cdot 3 + 4 \cdot 1 + 0 \cdot 0 = 3 + 4 + 0 = 7 \)
- Second row, second column: \( 1 \cdot 2 + 4 \cdot 4 + 0 \cdot 0 = 2 + 16 + 0 = 18 \)
- Second row, third column: \( 1 \cdot 0 + 4 \cdot 0 + 0 \cdot 5 = 0 + 0 + 0 = 0 \)
- Third row, first column: \( 0 \cdot 3 + 0 \cdot 1 + 5 \cdot 0 = 0 + 0 + 0 = 0 \)
- Third row, second column: \( 0 \cdot 2 + 0 \cdot 4 + 5 \cdot 0 = 0 + 0 + 0 = 0 \)
- Third row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 5 \cdot 5 = 0 + 0 + 25 = 25 \)
Thus, we have:
\[
A^2 = \begin{pmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{pmatrix}
\]
### Step 2: Calculate \( 7A \)
Now, we calculate \( 7A \):
\[
7A = 7 \cdot \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{pmatrix}
\]
### Step 3: Calculate \( 10I_3 \)
The identity matrix \( I_3 \) is:
\[
I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\]
Thus,
\[
10I_3 = 10 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{pmatrix}
\]
### Step 4: Calculate \( A^2 - 7A + 10I_3 \)
Now we can compute \( A^2 - 7A + 10I_3 \):
\[
A^2 - 7A + 10I_3 = \begin{pmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{pmatrix} - \begin{pmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{pmatrix} + \begin{pmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{pmatrix}
\]
Calculating element-wise:
- First row: \( 11 - 21 + 10 = 0 \), \( 14 - 14 + 0 = 0 \), \( 0 - 0 + 0 = 0 \)
- Second row: \( 7 - 7 + 0 = 0 \), \( 18 - 28 + 10 = 0 \), \( 0 - 0 + 0 = 0 \)
- Third row: \( 0 - 0 + 0 = 0 \), \( 0 - 0 + 0 = 0 \), \( 25 - 35 + 10 = 0 \)
Thus, we have:
\[
A^2 - 7A + 10I_3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
\]
This shows that \( A^2 - 7A + 10I_3 = 0 \).
### Step 5: Find \( A^{-1} \)
From the equation \( A^2 - 7A + 10I_3 = 0 \), we can rearrange it to find \( A^{-1} \):
\[
10I_3 = 7A - A^2
\]
Multiplying both sides by \( \frac{1}{10} \):
\[
I_3 = \frac{7}{10}A - \frac{1}{10}A^2
\]
Now, multiplying both sides by \( A^{-1} \):
\[
A^{-1} = \frac{1}{10}(7I_3 - A)
\]
Substituting \( A \) and \( I_3 \):
\[
A^{-1} = \frac{1}{10} \left( 7 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix} \right)
\]
Calculating:
\[
= \frac{1}{10} \begin{pmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{pmatrix} - \frac{1}{10} \begin{pmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}
\]
\[
= \frac{1}{10} \begin{pmatrix} 7 - 3 & 0 - 2 & 0 \\ 0 - 1 & 7 - 4 & 0 \\ 0 & 0 & 7 - 5 \end{pmatrix}
\]
\[
= \frac{1}{10} \begin{pmatrix} 4 & -2 & 0 \\ -1 & 3 & 0 \\ 0 & 0 & 2 \end{pmatrix}
\]
Thus, the inverse of \( A \) is:
\[
A^{-1} = \begin{pmatrix} \frac{4}{10} & \frac{-2}{10} & 0 \\ \frac{-1}{10} & \frac{3}{10} & 0 \\ 0 & 0 & \frac{2}{10} \end{pmatrix} = \begin{pmatrix} 0.4 & -0.2 & 0 \\ -0.1 & 0.3 & 0 \\ 0 & 0 & 0.2 \end{pmatrix}
\]
