We are given that,

[2 ,1 ,3]\(\begin{bmatrix}-1& 0 & -1 \\[0.3em]-1 &1 &0 \\[0.3em]0 &1 &1\end{bmatrix}\)\(\begin{bmatrix}1 \\[0.3em]0 \\[0.3em]-1\end{bmatrix}\)= A

We need to find the order of the matrix A. 

Let the matrices be,

X = [2 ,1 ,3]

Y = \(\begin{bmatrix}-1& 0 & -1 \\[0.3em]-1 &1 &0 \\[0.3em]0 &1 &1\end{bmatrix}\)

Z = \(\begin{bmatrix}1 \\[0.3em]0 \\[0.3em]-1\end{bmatrix}\)

Let us find the order of X. 

Number of rows of matrix X = 1 

Number of columns of matrix X = 3 

So, 

Order of matrix X = 1 × 3 …(i) 

Now, 

let us find the order of Y. 

Number of rows of matrix Y = 3 

Number of columns of matrix Y = 3 

So, 

Order of matrix Y = 3 × 3 …(ii) 

From (i) and (ii), 

Order of resulting XY = 1 × 3 

[∵ Number of columns of X = Number of rows of Y] …(iii) 

Let us find the order of Z. 

Number of rows of matrix Z = 3 

Number of columns of matrix Z = 1 

So, 

Order of matrix Z = 3 × 1 …(iv)

Order of resulting XYZ = 1 × 1 

[∵ Number of columns of XY = Number of rows of Z] 

Thus, 

The order of matrix A = 1 × 1