If `P=[(lambda,0),(7,1)]` and `Q=[(4,0),(-7,1)]` such that `P^(2)=Q`, then `P^(3)` is equal to

Options

A. `[(-8,0),(21,1)]`
B. `[(10,1),(8,0)]`
C. `[(7,0),(8,1)]`
D. `[(6,0),(4,1)]`

Answer

To solve the problem, we need to find \( P^3 \) given that \( P^2 = Q \) where:

\[ P = \begin{pmatrix} \lambda & 0 \\ 7 & 1 \end{pmatrix}, \quad Q = \begin{pmatrix} 4 & 0 \\ -7 & 1 \end{pmatrix} \]

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

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

\[ P^2 = P \cdot P = \begin{pmatrix} \lambda & 0 \\ 7 & 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda & 0 \\ 7 & 1 \end{pmatrix} \]

Calculating the product:

- First row, first column: \( \lambda \cdot \lambda + 0 \cdot 7 = \lambda^2 \)
- First row, second column: \( \lambda \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( 7 \cdot \lambda + 1 \cdot 7 = 7\lambda + 7 \)
- Second row, second column: \( 7 \cdot 0 + 1 \cdot 1 = 1 \)

Thus, we have:

\[ P^2 = \begin{pmatrix} \lambda^2 & 0 \\ 7\lambda + 7 & 1 \end{pmatrix} \]

### Step 2: Set \( P^2 \) equal to \( Q \)

Since \( P^2 = Q \), we equate:

\[ \begin{pmatrix} \lambda^2 & 0 \\ 7\lambda + 7 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ -7 & 1 \end{pmatrix} \]

From this, we can derive two equations:

1. \( \lambda^2 = 4 \)
2. \( 7\lambda + 7 = -7 \)

### Step 3: Solve for \( \lambda \)

From the first equation:

\[ \lambda^2 = 4 \implies \lambda = 2 \text{ or } \lambda = -2 \]

From the second equation:

\[ 7\lambda + 7 = -7 \implies 7\lambda = -14 \implies \lambda = -2 \]

Thus, we have \( \lambda = -2 \).

### Step 4: Substitute \( \lambda \) back into \( P \)

Now substituting \( \lambda = -2 \) into matrix \( P \):

\[ P = \begin{pmatrix} -2 & 0 \\ 7 & 1 \end{pmatrix} \]

### Step 5: Calculate \( P^3 \)

Now, we need to find \( P^3 = P^2 \cdot P \).

We already calculated \( P^2 \):

\[ P^2 = \begin{pmatrix} 4 & 0 \\ -7 & 1 \end{pmatrix} \]

Now we calculate \( P^3 \):

\[ P^3 = P^2 \cdot P = \begin{pmatrix} 4 & 0 \\ -7 & 1 \end{pmatrix} \cdot \begin{pmatrix} -2 & 0 \\ 7 & 1 \end{pmatrix} \]

Calculating the product:

- First row, first column: \( 4 \cdot -2 + 0 \cdot 7 = -8 \)
- First row, second column: \( 4 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( -7 \cdot -2 + 1 \cdot 7 = 14 + 7 = 21 \)
- Second row, second column: \( -7 \cdot 0 + 1 \cdot 1 = 1 \)

Thus, we have:

\[ P^3 = \begin{pmatrix} -8 & 0 \\ 21 & 1 \end{pmatrix} \]

### Final Answer

Therefore, \( P^3 \) is:

\[ \begin{pmatrix} -8 & 0 \\ 21 & 1 \end{pmatrix} \]