Given P = $\begin{bmatrix}1 & 2\\2 & 3\end{bmatrix}$, find P$^2$ - 4P - I where I is the identity matrix

1. Analyze the Question

We are given a $2 \times 2$ matrix $P = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$. We need to evaluate the expression $P^2 - 4P - I$, where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

2. Explain the Concept

• Matrix Multiplication ($P^2$): To find $P^2$, we multiply matrix $P$ by itself ($P \times P$). Remember the row-by-column rule: multiply the elements of the row of the first matrix by the corresponding elements of the column of the second matrix and sum them.
• Scalar Multiplication ($4P$): Multiply every element inside matrix $P$ by the scalar 4.
• Matrix Subtraction: Subtract corresponding elements of the matrices.

3. Show the Solution

Step 1: Calculate $P^2$ $P^2 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$ $= \begin{bmatrix} (1 \times 1 + 2 \times 2) & (1 \times 2 + 2 \times 3) \\ (2 \times 1 + 3 \times 2) & (2 \times 2 + 3 \times 3) \end{bmatrix}$ $= \begin{bmatrix} (1 + 4) & (2 + 6) \\ (2 + 6) & (4 + 9) \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$

Step 2: Calculate $4P$ $4P = 4 \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix}$

Step 3: Evaluate $P^2 - 4P - I$ $= \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Perform the subtraction element by element: Top-left: $5 - 4 - 1 = 0$ Top-right: $8 - 8 - 0 = 0$ Bottom-left: $8 - 8 - 0 = 0$ Bottom-right: $13 - 12 - 1 = 0$

Result: $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

4. Identify the Answer

The result is the zero matrix. Comparing this to the given options, the correct choice is B.

5. Tip

When dealing with matrix expressions like $P^2 - 4P - I$, always double-check your arithmetic in the multiplication step. A single sign error in $P^2$ will ruin the entire calculation. Also, remember that $I$ is always $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ for a $2 \times 2$ matrix; don't mistake it for a zero matrix!