To solve the problem, we need to compute \( f(A) + \begin{pmatrix} 3 & 6 \\ -12 & -9 \end{pmatrix} \) where \( A = \begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix} \) and \( f(t) = t^2 - 3t + 7 \).
### Step 1: Calculate \( A^2 \)
First, we need to compute \( A^2 \):
\[
A^2 = A \cdot A = \begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix} \cdot \begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + (-2) \cdot 4 = 1 - 8 = -7 \)
- First row, second column: \( 1 \cdot (-2) + (-2) \cdot 5 = -2 - 10 = -12 \)
- Second row, first column: \( 4 \cdot 1 + 5 \cdot 4 = 4 + 20 = 24 \)
- Second row, second column: \( 4 \cdot (-2) + 5 \cdot 5 = -8 + 25 = 17 \)
Thus,
\[
A^2 = \begin{pmatrix} -7 & -12 \\ 24 & 17 \end{pmatrix}
\]
### Step 2: Calculate \( -3A \)
Next, we calculate \( -3A \):
\[
-3A = -3 \cdot \begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} -3 & 6 \\ -12 & -15 \end{pmatrix}
\]
### Step 3: Calculate \( f(A) \)
Now we can compute \( f(A) \):
\[
f(A) = A^2 - 3A + 7I
\]
Where \( I \) is the identity matrix:
\[
I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
Calculating \( 7I \):
\[
7I = 7 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}
\]
Now, we sum \( A^2 \), \( -3A \), and \( 7I \):
\[
f(A) = \begin{pmatrix} -7 & -12 \\ 24 & 17 \end{pmatrix} + \begin{pmatrix} -3 & 6 \\ -12 & -15 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( -7 - 3 + 7 = -3 \)
- First row, second column: \( -12 + 6 + 0 = -6 \)
- Second row, first column: \( 24 - 12 + 0 = 12 \)
- Second row, second column: \( 17 - 15 + 7 = 9 \)
Thus,
\[
f(A) = \begin{pmatrix} -3 & -6 \\ 12 & 9 \end{pmatrix}
\]
### Step 4: Add \( f(A) \) and \( \begin{pmatrix} 3 & 6 \\ -12 & -9 \end{pmatrix} \)
Now we add \( f(A) \) and the given matrix:
\[
f(A) + \begin{pmatrix} 3 & 6 \\ -12 & -9 \end{pmatrix} = \begin{pmatrix} -3 & -6 \\ 12 & 9 \end{pmatrix} + \begin{pmatrix} 3 & 6 \\ -12 & -9 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( -3 + 3 = 0 \)
- First row, second column: \( -6 + 6 = 0 \)
- Second row, first column: \( 12 - 12 = 0 \)
- Second row, second column: \( 9 - 9 = 0 \)
Thus,
\[
f(A) + \begin{pmatrix} 3 & 6 \\ -12 & -9 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\]
### Final Answer
The final result is:
\[
\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\]
