To solve the problem, we need to find \( A^3 - 2A^2 + A - I_3 \) for the given matrix \( A \).
Given:
\[
A = \begin{pmatrix}
1 & 3 & 2 \\
2 & 0 & 3 \\
1 & -1 & 1
\end{pmatrix}
\]
### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[
A^2 = A \cdot A = \begin{pmatrix}
1 & 3 & 2 \\
2 & 0 & 3 \\
1 & -1 & 1
\end{pmatrix} \cdot \begin{pmatrix}
1 & 3 & 2 \\
2 & 0 & 3 \\
1 & -1 & 1
\end{pmatrix}
\]
Calculating each element of \( A^2 \):
- First row, first column:
\[
1 \cdot 1 + 3 \cdot 2 + 2 \cdot 1 = 1 + 6 + 2 = 9
\]
- First row, second column:
\[
1 \cdot 3 + 3 \cdot 0 + 2 \cdot -1 = 3 + 0 - 2 = 1
\]
- First row, third column:
\[
1 \cdot 2 + 3 \cdot 3 + 2 \cdot 1 = 2 + 9 + 2 = 13
\]
- Second row, first column:
\[
2 \cdot 1 + 0 \cdot 2 + 3 \cdot 1 = 2 + 0 + 3 = 5
\]
- Second row, second column:
\[
2 \cdot 3 + 0 \cdot 0 + 3 \cdot -1 = 6 + 0 - 3 = 3
\]
- Second row, third column:
\[
2 \cdot 2 + 0 \cdot 3 + 3 \cdot 1 = 4 + 0 + 3 = 7
\]
- Third row, first column:
\[
1 \cdot 1 + -1 \cdot 2 + 1 \cdot 1 = 1 - 2 + 1 = 0
\]
- Third row, second column:
\[
1 \cdot 3 + -1 \cdot 0 + 1 \cdot -1 = 3 + 0 - 1 = 2
\]
- Third row, third column:
\[
1 \cdot 2 + -1 \cdot 3 + 1 \cdot 1 = 2 - 3 + 1 = 0
\]
Thus,
\[
A^2 = \begin{pmatrix}
9 & 1 & 13 \\
5 & 3 & 7 \\
0 & 2 & 0
\end{pmatrix}
\]
### Step 2: Calculate \( A^3 \)
Now we calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):
\[
A^3 = A^2 \cdot A = \begin{pmatrix}
9 & 1 & 13 \\
5 & 3 & 7 \\
0 & 2 & 0
\end{pmatrix} \cdot \begin{pmatrix}
1 & 3 & 2 \\
2 & 0 & 3 \\
1 & -1 & 1
\end{pmatrix}
\]
Calculating each element of \( A^3 \):
- First row, first column:
\[
9 \cdot 1 + 1 \cdot 2 + 13 \cdot 1 = 9 + 2 + 13 = 24
\]
- First row, second column:
\[
9 \cdot 3 + 1 \cdot 0 + 13 \cdot -1 = 27 + 0 - 13 = 14
\]
- First row, third column:
\[
9 \cdot 2 + 1 \cdot 3 + 13 \cdot 1 = 18 + 3 + 13 = 34
\]
- Second row, first column:
\[
5 \cdot 1 + 3 \cdot 2 + 7 \cdot 1 = 5 + 6 + 7 = 18
\]
- Second row, second column:
\[
5 \cdot 3 + 3 \cdot 0 + 7 \cdot -1 = 15 + 0 - 7 = 8
\]
- Second row, third column:
\[
5 \cdot 2 + 3 \cdot 3 + 7 \cdot 1 = 10 + 9 + 7 = 26
\]
- Third row, first column:
\[
0 \cdot 1 + 2 \cdot 2 + 0 \cdot 1 = 0 + 4 + 0 = 4
\]
- Third row, second column:
\[
0 \cdot 3 + 2 \cdot 0 + 0 \cdot -1 = 0 + 0 + 0 = 0
\]
- Third row, third column:
\[
0 \cdot 2 + 2 \cdot 3 + 0 \cdot 1 = 0 + 6 + 0 = 6
\]
Thus,
\[
A^3 = \begin{pmatrix}
24 & 14 & 34 \\
18 & 8 & 26 \\
4 & 0 & 6
\end{pmatrix}
\]
### Step 3: Calculate \( -2A^2 \)
Now we calculate \( -2A^2 \):
\[
-2A^2 = -2 \cdot \begin{pmatrix}
9 & 1 & 13 \\
5 & 3 & 7 \\
0 & 2 & 0
\end{pmatrix} = \begin{pmatrix}
-18 & -2 & -26 \\
-10 & -6 & -14 \\
0 & -4 & 0
\end{pmatrix}
\]
### Step 4: Calculate \( A - I_3 \)
The identity matrix \( I_3 \) is:
\[
I_3 = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]
Now, calculate \( A - I_3 \):
\[
A - I_3 = \begin{pmatrix}
1 & 3 & 2 \\
2 & 0 & 3 \\
1 & -1 & 1
\end{pmatrix} - \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} = \begin{pmatrix}
0 & 3 & 2 \\
2 & -1 & 3 \\
1 & -1 & 0
\end{pmatrix}
\]
### Step 5: Combine all parts to find \( A^3 - 2A^2 + A - I_3 \)
Now we combine all the results:
\[
A^3 - 2A^2 + A - I_3 = \begin{pmatrix}
24 & 14 & 34 \\
18 & 8 & 26 \\
4 & 0 & 6
\end{pmatrix} + \begin{pmatrix}
-18 & -2 & -26 \\
-10 & -6 & -14 \\
0 & -4 & 0
\end{pmatrix} + \begin{pmatrix}
0 & 3 & 2 \\
2 & -1 & 3 \\
1 & -1 & 0
\end{pmatrix}
\]
Calculating each element:
- First row:
\[
(24 - 18 + 0) = 6, \quad (14 - 2 + 3) = 15, \quad (34 - 26 + 2) = 10
\]
- Second row:
\[
(18 - 10 + 2) = 10, \quad (8 - 6 - 1) = 1, \quad (26 - 14 + 3) = 15
\]
- Third row:
\[
(4 + 0 + 1) = 5, \quad (0 - 4 - 1) = -5, \quad (6 + 0 + 0) = 6
\]
Thus, the final result is:
\[
A^3 - 2A^2 + A - I_3 = \begin{pmatrix}
6 & 15 & 10 \\
10 & 1 & 15 \\
5 & -5 & 6
\end{pmatrix}
\]
