Question

Let \( P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) and let \( I \) be the identity matrix. Then \( P^2 \) is equal to

• A. \( 2P - I \)
• B. \( P \)
• C. \( I \)
• D. \( P + I \)

Hint:

To verify matrix identities, compute both sides explicitly using matrix multiplication and subtraction. For 3x3 matrices, this is usually manageable by hand.

Answer 1

The Correct Option is A

We are given: \[ P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Now, calculate \( P^2 = P \cdot P \): \( P^2 = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2)+(1)(-1) & (2)(1)+(1)(0) & 0 \\ (-1)(2)+(0)(-1) & (-1)(1)+(0)(0) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 - 1 & 2 + 0 & 0 \\ -2 + 0 & -1 + 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) Now compute \( 2P - I \): \[ 2P = 2 \cdot \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] \[ 2P - I = \begin{bmatrix} 4 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] This matches \( P^2 \), so: \[ P^2 = 2P - I \]