`if A=[{:(0),(1),(2):}]and B=[3" "2" "1],`then show that :
`(AB)'=B'A'`

Explanation

To prove that \((AB)' = B'A'\) for the given matrices \(A\) and \(B\), we will first define the matrices and then perform the necessary calculations step by step.

Given:
\[ A = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & 2 & 1 \end{pmatrix} \]

### Step 1: Calculate the product \(AB\)

To calculate \(AB\), we multiply matrix \(A\) (which is a \(3 \times 1\) matrix) with matrix \(B\) (which is a \(1 \times 3\) matrix):

\[ AB = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \begin{pmatrix} 3 & 2 & 1 \end{pmatrix} \]

Using the rules of matrix multiplication, we calculate:

\[ AB = \begin{pmatrix} 0 \cdot 3 & 0 \cdot 2 & 0 \cdot 1 \\ 1 \cdot 3 & 1 \cdot 2 & 1 \cdot 1 \\ 2 \cdot 3 & 2 \cdot 2 & 2 \cdot 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 2 & 1 \\ 6 & 4 & 2 \end{pmatrix} \]

### Step 2: Calculate the transpose \((AB)'\)

Now, we need to find the transpose of the resulting matrix \(AB\):

\[ (AB)' = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 2 & 1 \\ 6 & 4 & 2 \end{pmatrix}' = \begin{pmatrix} 0 & 3 & 6 \\ 0 & 2 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]

### Step 3: Calculate the transpose of \(B\) and \(A\)

Next, we calculate the transposes of \(B\) and \(A\):

\[ B' = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}, \quad A' = \begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \]

### Step 4: Calculate the product \(B'A'\)

Now, we multiply \(B'\) and \(A'\):

\[ B'A' = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \]

Calculating this product:

\[ B'A' = \begin{pmatrix} 3 \cdot 0 & 3 \cdot 1 & 3 \cdot 2 \\ 2 \cdot 0 & 2 \cdot 1 & 2 \cdot 2 \\ 1 \cdot 0 & 1 \cdot 1 & 1 \cdot 2 \end{pmatrix} = \begin{pmatrix} 0 & 3 & 6 \\ 0 & 2 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]

### Step 5: Conclusion

Now we compare the two results:

\[ (AB)' = \begin{pmatrix} 0 & 3 & 6 \\ 0 & 2 & 4 \\ 0 & 1 & 2 \end{pmatrix} \quad \text{and} \quad B'A' = \begin{pmatrix} 0 & 3 & 6 \\ 0 & 2 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]

Since \((AB)' = B'A'\), we have shown that:

\[ (AB)' = B'A' \]