0.2 Matrices  (Page 2/11)

As we mentioned earlier, matrix addition and subtraction involves performing these operations entry by entry.

1. We add each element of $A$ to the corresponding entry of $B$ .

   $A+B=\left[\begin{array}{ccc}3& 1& 7\\ 4& 7& 3\\ 8& 6& 4\end{array}\right]$

2. Just like the problem above, we perform the subtraction entry by entry.

   $C-D=\left[\begin{array}{c}6\\ 5\\ -1\end{array}\right]$

3. The sum $A+D$ cannot be found because the two matrices have different sizes.

To find $\mathrm{2A}$ , we multiply each entry of matrix $A$ by $2$ , and to find $-\mathrm{3C}$ , we multiply each entry of $C$ by $-3$ . The results are given below.

1. We multiply each entry of $A$ by $2$ .

   $\mathrm{2A}=\left[\begin{array}{ccc}2& 4& 8\\ 4& 6& 2\\ \text{10}& 0& 6\end{array}\right]$

2. We multiply each entry of $C$ by $-3$ .

   $-\mathrm{3C}=\left[\begin{array}{c}-\text{12}\\ -6\\ -9\end{array}\right]$

The product is a $1\times 1$ matrix whose entry is obtained by multiplying the corresponding entries and then forming the sum.

Note that $\text{AB}$ is a $1\times 1$ matrix, and its only entry is $\mathrm{2a}+\mathrm{3b}+\mathrm{4c}$ .

Again, we multiply the corresponding entries and add.

We already know how to multiply a row matrix by a column matrix. To find the product $\text{AB}$ , in this example, we will be multiplying the row matrix $A$ to both the first and second columns of matrix $B$ , resulting in a $1\times 2$ matrix.

We have just multiplied a $1\times 3$ matrix by a matrix whose size is $3\times 2$ . So unlike addition and subtraction, it is possible to multiply two matrices with different dimensions as long as the number of entries in the rows of the first matrix are the same as the number of entries in columns of the second matrix.

This time we are multiplying two rows of the matrix $A$ with two columns of the matrix $B$ . Since the number of entries in each row of $A$ are the same as the number of entries in each column of $B$ , the product is possible. We do exactly what we did in [link] . The only difference is that the matrix $A$ has one more row.

We multiply the first row of the matrix $A$ with the two columns of $B$ , one at a time, and then repeat the process with the second row of $A$ . We get