To solve the problem, we need to calculate the matrix products \( EF \) and \( FE \), and then show that \( E^2F + FE^2 = E \).
### Step 1: Define the matrices
Let:
\[
E = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}, \quad
F = \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\]
### Step 2: Calculate \( EF \)
To find \( EF \), we perform matrix multiplication:
\[
EF = E \cdot F = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 \)
- First row, second column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0 \)
- First row, third column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, first column: \( 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1 \)
- Second row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 = 0 \)
- Third row, first column: \( 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \)
- Third row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
Thus, we have:
\[
EF = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]
### Step 3: Calculate \( FE \)
Now, we calculate \( FE \):
\[
FE = F \cdot E = \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, second column: \( 0 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, third column: \( 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
- Second row, first column: \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, second column: \( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 \)
- Second row, third column: \( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
- Third row, first column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, third column: \( 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 \)
Thus, we have:
\[
FE = \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]
### Step 4: Calculate \( E^2 \)
Next, we calculate \( E^2 \):
\[
E^2 = E \cdot E = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, second column: \( 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, third column: \( 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 \)
- Second row, first column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0 \)
- Second row, third column: \( 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Third row, first column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, third column: \( 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
Thus, we have:
\[
E^2 = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]
### Step 5: Calculate \( E^2F \) and \( FE^2 \)
Now we calculate \( E^2F \):
\[
E^2F = E^2 \cdot F = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 = 0 \)
- First row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1 \)
- First row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 = 0 \)
- Second row, first column: \( 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, first column: \( 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \)
- Third row, third column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
Thus, we have:
\[
E^2F = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]
Now we calculate \( FE^2 \):
\[
FE^2 = F \cdot E^2 = \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, second column: \( 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- First row, third column: \( 0 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, first column: \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, second column: \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, third column: \( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 \)
- Third row, first column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
- Third row, third column: \( 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0 \)
Thus, we have:
\[
FE^2 = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
### Step 6: Calculate \( E^2F + FE^2 \)
Now we add \( E^2F \) and \( FE^2 \):
\[
E^2F + FE^2 = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} + \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 0 + 0 = 0 \)
- First row, second column: \( 1 + 0 = 1 \)
- First row, third column: \( 0 + 0 = 0 \)
- Second row, first column: \( 0 + 0 = 0 \)
- Second row, second column: \( 0 + 0 = 0 \)
- Second row, third column: \( 0 + 1 = 1 \)
- Third row, first column: \( 0 + 0 = 0 \)
- Third row, second column: \( 0 + 0 = 0 \)
- Third row, third column: \( 0 + 0 = 0 \)
Thus, we have:
\[
E^2F + FE^2 = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
### Step 7: Conclusion
We compare \( E^2F + FE^2 \) with \( E \):
\[
E = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\]
Since \( E^2F + FE^2 = E \), we have shown that:
\[
E^2F + FE^2 = E
\]
