0.2 Matrices  (Page 3/11)

1. To find $\text{EF}$ , we multiply the first row $\left[\begin{array}{cc}1& 2\end{array}\right]$ of $E$ with the columns $\left[\begin{array}{c}2\\ 3\end{array}\right]$ and $\left[\begin{array}{c}-1\\ 2\end{array}\right]$ of the matrix $F$ , and then repeat the process by multiplying the other two rows of $E$ with these columns of $F$ . The result is as follows:

   $\begin{array}{ccc}\text{EF}& =& \left[\begin{array}{cc}1& 2\\ 4& 2\\ 3& 1\end{array}\right]\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}1\cdot 2+2\cdot 3& 1\cdot -1+2\cdot 2\\ 4\cdot 2+2\cdot 3& 4\cdot -1+2\cdot 2\\ 3\cdot 2+1\cdot 3& 3\cdot -1+1\cdot 2\end{array}\right]=\left[\begin{array}{cc}8& 3\\ \text{14}& 0\\ 9& -1\end{array}\right]\hfill \end{array}$

2. The product $\text{FE}$ is not possible because the matrix $F$ has two entries in each row, while the matrix $E$ has three entries in each column. In other words, the matrix $F$ has two columns, while the matrix $E$ has three rows.

3. $\text{FH}=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\left[\begin{array}{c}-3\\ -1\end{array}\right]=\left[\begin{array}{c}2\cdot -3+-1\cdot -1\\ 3\cdot -3+2\cdot -1\end{array}\right]=\left[\begin{array}{c}-5\\ -\text{11}\end{array}\right]$

4. $\text{GH}=\left[\begin{array}{cc}4& 1\end{array}\right]\left[\begin{array}{c}-3\\ -1\end{array}\right]=\left[4\cdot -3+1\cdot -1\right]=\left[-\text{13}\right]$

We multiply the matrices $R$ and $S$ .

If we multiply the matrices $A$ and $X$ , we get

If $\text{AX}=B$ then

If two matrices are equal, then their corresponding entries are equal. Therefore, it follows that

The above system of equations can be expressed in the form $\text{AX}=B$ as shown below.

We express the above information in matrix form. Since a system is entirely determined by its coefficient matrix and by its matrix of constant terms, the augmented matrix will include only the coefficient matrix and the constant matrix. So the augmented matrix we get is as follows: