If `P=[{:(,1,2),(,2,-1):}] and Q=[{:(,1,0),(,2,1):}]` then compute :
(i) `P^2-Q^2` (ii) `(P+Q) (P-Q)`
Is (P+Q) (P-Q) `=P^2-Q^2` true for matrix algebra?

Answer

Let's solve the problem step by step.

Given matrices:
\[ P = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \]
\[ Q = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

### Part (i): Compute \( P^2 - Q^2 \)

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

To find \( P^2 \), we multiply \( P \) by itself:
\[ P^2 = P \times P = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \times 1 + 2 \times 2 = 1 + 4 = 5 \)
- First row, second column: \( 1 \times 2 + 2 \times -1 = 2 - 2 = 0 \)
- Second row, first column: \( 2 \times 1 + -1 \times 2 = 2 - 2 = 0 \)
- Second row, second column: \( 2 \times 2 + -1 \times -1 = 4 + 1 = 5 \)

Thus,
\[ P^2 = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

**Step 2: Calculate \( Q^2 \)**

Now, we calculate \( Q^2 \):
\[ Q^2 = Q \times Q = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \times 1 + 0 \times 2 = 1 + 0 = 1 \)
- First row, second column: \( 1 \times 0 + 0 \times 1 = 0 + 0 = 0 \)
- Second row, first column: \( 2 \times 1 + 1 \times 2 = 2 + 2 = 4 \)
- Second row, second column: \( 2 \times 0 + 1 \times 1 = 0 + 1 = 1 \)

Thus,
\[ Q^2 = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \]

**Step 3: Compute \( P^2 - Q^2 \)**

Now we subtract \( Q^2 \) from \( P^2 \):
\[ P^2 - Q^2 = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 5 - 1 = 4 \)
- First row, second column: \( 0 - 0 = 0 \)
- Second row, first column: \( 0 - 4 = -4 \)
- Second row, second column: \( 5 - 1 = 4 \)

Thus,
\[ P^2 - Q^2 = \begin{pmatrix} 4 & 0 \\ -4 & 4 \end{pmatrix} \]

### Part (ii): Compute \( (P + Q)(P - Q) \)

**Step 1: Calculate \( P + Q \)**

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

**Step 2: Calculate \( P - Q \)**

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

**Step 3: Compute \( (P + Q)(P - Q) \)**

Now we multiply \( P + Q \) by \( P - Q \):
\[ (P + Q)(P - Q) = \begin{pmatrix} 2 & 2 \\ 4 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 2 \\ 0 & -2 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 2 \times 0 + 2 \times 0 = 0 \)
- First row, second column: \( 2 \times 2 + 2 \times -2 = 4 - 4 = 0 \)
- Second row, first column: \( 4 \times 0 + 0 \times 0 = 0 \)
- Second row, second column: \( 4 \times 2 + 0 \times -2 = 8 + 0 = 8 \)

Thus,
\[ (P + Q)(P - Q) = \begin{pmatrix} 0 & 0 \\ 0 & 8 \end{pmatrix} \]

### Conclusion: Verify if \( (P + Q)(P - Q) = P^2 - Q^2 \)

We have:
\[ P^2 - Q^2 = \begin{pmatrix} 4 & 0 \\ -4 & 4 \end{pmatrix} \]
\[ (P + Q)(P - Q) = \begin{pmatrix} 0 & 0 \\ 0 & 8 \end{pmatrix} \]

Since \( P^2 - Q^2 \neq (P + Q)(P - Q) \), the statement is **not true** for matrix algebra.