Compute : `[(4),(2),(3)]["1 0 2"]`

Answer

To compute the product of the matrices \(\begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix}\) and \(\begin{bmatrix} 1 & 0 & 2 \end{bmatrix}\), we will follow the rules of matrix multiplication. The first matrix is a \(3 \times 1\) matrix, and the second matrix is a \(1 \times 3\) matrix. The resulting matrix will be of order \(3 \times 3\).

### Step-by-Step Solution:

1. **Identify the matrices:**
- First matrix \(A = \begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix}\)
- Second matrix \(B = \begin{bmatrix} 1 & 0 & 2 \end{bmatrix}\)

2. **Determine the order of the resultant matrix:**
- The first matrix \(A\) has 3 rows and 1 column.
- The second matrix \(B\) has 1 row and 3 columns.
- The resultant matrix \(C\) will have the order \(3 \times 3\).

3. **Multiply the matrices:**
- The element \(C_{ij}\) of the resultant matrix \(C\) is calculated by multiplying the \(i^{th}\) row of matrix \(A\) with the \(j^{th}\) column of matrix \(B\).

4. **Calculate each element of the resultant matrix:**
- \(C_{11} = 4 \times 1 = 4\)
- \(C_{12} = 4 \times 0 = 0\)
- \(C_{13} = 4 \times 2 = 8\)

- \(C_{21} = 2 \times 1 = 2\)
- \(C_{22} = 2 \times 0 = 0\)
- \(C_{23} = 2 \times 2 = 4\)

- \(C_{31} = 3 \times 1 = 3\)
- \(C_{32} = 3 \times 0 = 0\)
- \(C_{33} = 3 \times 2 = 6\)

5. **Construct the resultant matrix:**
- The resultant matrix \(C\) is:
\[ C = \begin{bmatrix} 4 & 0 & 8 \\ 2 & 0 & 4 \\ 3 & 0 & 6 \end{bmatrix} \]

### Final Result:
\[ \begin{bmatrix} 4 & 0 & 8 \\ 2 & 0 & 4 \\ 3 & 0 & 6 \end{bmatrix} \]