If `P=[(2, -2, -4),(-1, 3, 4),(1, -2, -3)],` then `P^(5)` equals

Options

A. P
B. 2P
C. `-P`
D. `-2P`

Answer

To find \( P^5 \) for the matrix

\[ P = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix} \]

we will first calculate \( P^2 \), and then use the properties of matrix multiplication to find \( P^5 \).

### Step 1: Calculate \( P^2 \)

To find \( P^2 \), we multiply \( P \) by itself:

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

Calculating each element of \( P^2 \):

- First row, first column:
\[ 2 \cdot 2 + (-2) \cdot (-1) + (-4) \cdot 1 = 4 + 2 - 4 = 2 \]

- First row, second column:
\[ 2 \cdot (-2) + (-2) \cdot 3 + (-4) \cdot (-2) = -4 - 6 + 8 = -2 \]

- First row, third column:
\[ 2 \cdot (-4) + (-2) \cdot 4 + (-4) \cdot (-3) = -8 - 8 + 12 = -4 \]

- Second row, first column:
\[ -1 \cdot 2 + 3 \cdot (-1) + 4 \cdot 1 = -2 - 3 + 4 = -1 \]

- Second row, second column:
\[ -1 \cdot (-2) + 3 \cdot 3 + 4 \cdot (-2) = 2 + 9 - 8 = 3 \]

- Second row, third column:
\[ -1 \cdot (-4) + 3 \cdot 4 + 4 \cdot (-3) = 4 + 12 - 12 = 4 \]

- Third row, first column:
\[ 1 \cdot 2 + (-2) \cdot (-1) + (-3) \cdot 1 = 2 + 2 - 3 = 1 \]

- Third row, second column:
\[ 1 \cdot (-2) + (-2) \cdot 3 + (-3) \cdot (-2) = -2 - 6 + 6 = -2 \]

- Third row, third column:
\[ 1 \cdot (-4) + (-2) \cdot 4 + (-3) \cdot (-3) = -4 - 8 + 9 = -3 \]

Thus, we have:

\[ P^2 = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix} = P \]

### Step 2: Calculate \( P^3 \)

Since \( P^2 = P \), we can find \( P^3 \):

\[ P^3 = P^2 \times P = P \times P = P^2 = P \]

### Step 3: Calculate \( P^4 \)

Similarly, we find \( P^4 \):

\[ P^4 = P^3 \times P = P \times P = P^2 = P \]

### Step 4: Calculate \( P^5 \)

Finally, we find \( P^5 \):

\[ P^5 = P^4 \times P = P \times P = P^2 = P \]

### Conclusion

Thus, we conclude that:

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