If `{:A=[(n,0,0),(0,n,0),(0,0,n)]and B=[(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)]:}` , then AB is equal to

Options

A. BA=I
B. nB
C. `B^n`
D. A + B

Explanation

To solve the problem, we need to find the product of the matrices A and B, where:

\[ A = \begin{pmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{pmatrix} \]
and
\[ B = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \]

### Step 1: Write down the matrices A and B

We have:
\[ A = \begin{pmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{pmatrix} \]
\[ B = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \]

### Step 2: Multiply the matrices A and B

To multiply the two matrices, we will use the formula for matrix multiplication. The element in the ith row and jth column of the resulting matrix AB is calculated as the dot product of the ith row of A and the jth column of B.

The resulting matrix AB will be:

\[ AB = \begin{pmatrix} n \cdot a_1 + 0 \cdot b_1 + 0 \cdot c_1 & n \cdot a_2 + 0 \cdot b_2 + 0 \cdot c_2 & n \cdot a_3 + 0 \cdot b_3 + 0 \cdot c_3 \\ 0 \cdot a_1 + n \cdot b_1 + 0 \cdot c_1 & 0 \cdot a_2 + n \cdot b_2 + 0 \cdot c_2 & 0 \cdot a_3 + n \cdot b_3 + 0 \cdot c_3 \\ 0 \cdot a_1 + 0 \cdot b_1 + n \cdot c_1 & 0 \cdot a_2 + 0 \cdot b_2 + n \cdot c_2 & 0 \cdot a_3 + 0 \cdot b_3 + n \cdot c_3 \end{pmatrix} \]

This simplifies to:

\[ AB = \begin{pmatrix} n a_1 & n a_2 & n a_3 \\ n b_1 & n b_2 & n b_3 \\ n c_1 & n c_2 & n c_3 \end{pmatrix} \]

### Step 3: Factor out n from the resulting matrix

We can factor out n from the resulting matrix:

\[ AB = n \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \]

This shows that:

\[ AB = nB \]

### Conclusion

Thus, the final result is:

\[ AB = nB \]