chapter matrices 12 the grade
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In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If \displaystyle AA is an \displaystyle \text{ }m\text{ }\times \text{ }r\text{ } m × r matrix and \displaystyle BB is an \displaystyle \text{ }r\text{ }\times \text{ }n\text{ } r × n matrix, then the product matrix \displaystyle ABAB is an \displaystyle \text{ }m\text{ }\times \text{ }n\text{ } m × n matrix. For example, the product \displaystyle ABAB is possible because the number of columns in \displaystyle AA is the same as the number of rows in \displaystyle BB. If the inner dimensions do not match, the product is not defined.
A has two rows and three columns and B has three rows and three columns. Because the number of columns in A matches the number of rows in B, the product of A and B is defined.
Figure 1
We multiply entries of \displaystyle AA with entries of \displaystyle BB according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.
To obtain the entries in row \displaystyle ii of \displaystyle AB,\text{}AB, we multiply the entries in row \displaystyle ii of \displaystyle AA by column \displaystyle jj in \displaystyle BB and add. For example, given matrices \displaystyle AA and \displaystyle B,\text{}B, where the dimensions of \displaystyle AA are \displaystyle 2\text{ }\times \text{ }32 × 3 and the dimensions of \displaystyle BB are \displaystyle 3\text{ }\times \text{ }3,\text{}3 × 3, the product of \displaystyle ABAB will be a \displaystyle 2\text{ }\times \text{ }32 × 3 matrix.
Multiply and add as follows to obtain the first entry of the product matrix \displaystyle ABAB.
To obtain the entry in row 1, column 1 of \displaystyle AB,\text{}AB, multiply the first row in \displaystyle AA by the first column in \displaystyle BB, and add.
\displaystyle \left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}[
a
To obtain the entry in row 1, column 2 of \displaystyle AB,\text{}AB, multiply the first row of \displaystyle AA by the second column in \displaystyle BB, and add.
\displaystyle \left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}[
We proceed the same way to obtain the second row of \displaystyle ABAB. In other words, row 2 of \displaystyle AA times column 1 of \displaystyle BB; row 2 of \displaystyle AA times column 2 of \displaystyle BB; row 2 of \displaystyle AA times column 3 of \displaystyle BB. When complete, the product matrix will be
\displaystyle AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdotot commutative.
IS IT POSSIBLE FOR AB TO BE DEFINED BUT NOT BA?
Yes, consider a matrix A with dimension \displaystyle 3\times 43×4 and matrix B with dimension \displaystyle 4\times 24×2. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined.