If $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$, then AB is equals to :

We are given:

$A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}, \quad B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$

Notice that matrix $A$ has the same number $n$ on every diagonal entry and $0$ everywhere else. This means we can write it as:

$A = n \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = nI$

where $I$ is the identity matrix — think of it as the "1" of the matrix world. Multiplying any matrix by $I$ gives back the same matrix, just like $1 \times x = x$.

So:

$AB = (nI) \cdot B = n(I \cdot B) = nB$

Since $I \cdot B = B$, every element of $B$ simply gets multiplied by $n$.

$AB = nB = \begin{bmatrix} na_1 & na_2 & na_3 \\ nb_1 & nb_2 & nb_3 \\ nc_1 & nc_2 & nc_3 \end{bmatrix}$

Tip: Whenever you see a matrix of the form $A = kI$ (a constant times the identity matrix), you don't need to do full matrix multiplication. Just multiply every entry of the other matrix by that constant: $(kI) \cdot B = kB$.