![\mathrm{A}=\left[\begin{array}{lll} 1 & 2 & 5 \\ 0 & 1 & 3 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 2 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{l} 1 \\ 4 \\ 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-28076805d7b71c6daa66bcab8ce41845_l3.png "Rendered by QuickLaTeX.com")
Solution:
We have 
Now if number of columns in left matrix is equals to the number of rows in right matrix, multiplication of two matrices is possible.
Next discuss the order of the matrices which are given.
The order of matrix 
Thus the multiplications we can proceed as
(i) The multiplication 
(ii) The multiplication 
(iii) The multiplication 
(iv) The multiplication 
From (iii) and (iv) it can be observed that both the matrices 
Therefore by equality of two matrices we can say here that our matrices are equal.
Hence 
