If $A=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right], B=\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2\end{array}\right]$ and $C=\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right] ;$ verify that $A(B-C)=(A B-A C)$

Solution:

We have $A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right), B=\left(\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2\end{array}\right)$ .and
$\mathrm{C}=\left(\begin{array}{ccc} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right)$
Now sustraction of two matrices is possible if order of both the matrices are same and multiplication of two matrices is possible if
number of columns in left matrix is equals to the number of rows in right matrix
Next let us discuss the order of the matrices which are given.
The order of matrix $\boldsymbol{B}$ is $\mathbf{3} \times \mathbf{3}$ , matrix $\boldsymbol{A}$ is $\mathbf{3} \times \mathbf{3}$ and matrix $\boldsymbol{C}$ is $\mathbf{3} \times \mathbf{3}$ .
Thus the multiplications we can proceed as
(i) The multiplication $\boldsymbol{A B}$ is possible.
$A B=\left(\begin{array}{ccc} 1 & 0 & 2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right)\left(\begin{array}{ccc} 0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2 \end{array}\right)=\left(\begin{array}{ccc} -2 & 5 & -8 \\ 2 & 14 & -15 \\ -1 & -9 & 13 \end{array}\right)$
(ii) The multiplication $\boldsymbol{A} \boldsymbol{C}$ is possible.
$A C=\left(\begin{array}{ccc} 1 & 0 & 2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right)\left(\begin{array}{ccc} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right)=\left(\begin{array}{ccc} 1 & 7 & 0 \\ 4 & 16 & 6 \\ -3 & -10 & -3 \end{array}\right)$
(iii) The subtraction $\boldsymbol{B}-\boldsymbol{C}$ is possible.
$B-C=\left(\begin{array}{ccc} 0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2 \end{array}\right)-\left(\begin{array}{ccc} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right)=\left(\begin{array}{ccc} -1 & 0 & -6 \\ -1 & 0 & 3 \\ 1 & 1 & 1 \end{array}\right)$
(iv) The multiplication $\boldsymbol{A}(\boldsymbol{B}-\boldsymbol{C})$ is possible.
$\begin{aligned} A(B-C) &=\left(\begin{array}{ccc} \mathbf{1} & \mathbf{0} & \mathbf{2} \\ \mathbf{3} & -\mathbf{1} & \mathbf{0} \\ -\mathbf{2} & \mathbf{1} & \mathbf{1} \end{array}\right)\left(\begin{array}{ccc} -\mathbf{1} & \mathbf{0} & -\mathbf{6} \\ -\mathbf{1} & \mathbf{0} & \mathbf{3} \\ \mathbf{1} & \mathbf{1} & \mathbf{1} \end{array}\right) \\ &=\left(\begin{array}{ccc} -\mathbf{3} & \mathbf{- 2} & -\mathbf{8} \\ -\mathbf{2} & \mathbf{0} & -\mathbf{2 1} \\ \mathbf{2} & \mathbf{1} & \mathbf{1 6} \end{array}\right) \end{aligned}$
(v) The subtraction $\boldsymbol{A B}-\boldsymbol{A} \boldsymbol{C}$ is possible.
$\begin{aligned} A B &-A C=\left(\begin{array}{ccc} -2 & 5 & -8 \\ 2 & 14 & -15 \\ -1 & -9 & 13 \end{array}\right)-\left(\begin{array}{ccc} 1 & 7 & 0 \\ 4 & 16 & 6 \\ -3 & -10 & -3 \end{array}\right) \\ &=\left(\begin{array}{ccc} -3 & -2 & -8 \\ -2 & 0 & -21 \\ 2 & 1 & 16 \end{array}\right) \end{aligned}$
From (iv) and (v) it can be observed that both the matrices $\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})$ and $A B+A C$ are of same order $3 \times 3$ and the corresponding elements are same.
Therefore by equality of two matrices we can say here that our matrices are equal.
Hence $A(B+C)=A B+A C$ .