If `P=[(0,1,0),(0,2,1),(2,3,0)],Q=[(1,2),(3,0),(4,1)]`, find PQ.

Answer

To find the product of matrices \( P \) and \( Q \), we will follow the matrix multiplication rules.

Given:
\[ P = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 2 & 1 \\ 2 & 3 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 1 & 2 \\ 3 & 0 \\ 4 & 1 \end{pmatrix} \]

### Step 1: Verify the dimensions
The matrix \( P \) is of order \( 3 \times 3 \) and the matrix \( Q \) is of order \( 3 \times 2 \). The number of columns in \( P \) (which is 3) matches the number of rows in \( Q \) (which is also 3). Thus, the multiplication \( PQ \) is possible, and the resulting matrix will be of order \( 3 \times 2 \).

**Hint:** Always check the dimensions of the matrices before multiplying to ensure that the multiplication is valid.

### Step 2: Calculate the elements of the resulting matrix \( PQ \)
The resulting matrix \( R = PQ \) will be calculated as follows:

1. **Element \( R_{11} \)**:
\[ R_{11} = (0 \times 1) + (1 \times 3) + (0 \times 4) = 0 + 3 + 0 = 3 \]

2. **Element \( R_{12} \)**:
\[ R_{12} = (0 \times 2) + (1 \times 0) + (0 \times 1) = 0 + 0 + 0 = 0 \]

3. **Element \( R_{21} \)**:
\[ R_{21} = (0 \times 1) + (2 \times 3) + (1 \times 4) = 0 + 6 + 4 = 10 \]

4. **Element \( R_{22} \)**:
\[ R_{22} = (0 \times 2) + (2 \times 0) + (1 \times 1) = 0 + 0 + 1 = 1 \]

5. **Element \( R_{31} \)**:
\[ R_{31} = (2 \times 1) + (3 \times 3) + (0 \times 4) = 2 + 9 + 0 = 11 \]

6. **Element \( R_{32} \)**:
\[ R_{32} = (2 \times 2) + (3 \times 0) + (0 \times 1) = 4 + 0 + 0 = 4 \]

### Step 3: Assemble the resulting matrix
Now we can write the resulting matrix \( R \):
\[ R = PQ = \begin{pmatrix} 3 & 0 \\ 10 & 1 \\ 11 & 4 \end{pmatrix} \]

### Final Answer
Thus, the product \( PQ \) is:
\[ \begin{pmatrix} 3 & 0 \\ 10 & 1 \\ 11 & 4 \end{pmatrix} \]