Let \(\rm A = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\) and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right],\) then what is AB equal to?

Concept:

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.

Calculations:

Given \(A = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}\) and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]_{3\times1}\)

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.

i.e AB is posible and its order is \(\rm 3\times 1\)

Now, To obtain the entry in row 1, column 1 of $\displaystyle AB,\text{}$ multiply the first row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

To obtain the entry in row 2, column 1 of $\displaystyle AB,\text{}$ multiply the second row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

To obtain the entry in row 3, column 1 of $\displaystyle AB,\text{}$ multiply the third row in $\displaystyle A$ by the first column in $\displaystyle B$, and add.

\(AB = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}​​\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]_{3\times1}\)

\(AB = \left[ {\begin{array}{*{20}{c}} {{\rm{ax}} + {\rm{hy}} + {\rm{gz}}}\\ {{\rm{hx}} + {\rm{by}} + {\rm{fz}}}\\ {{\rm{gx}} + {\rm{fy}} + {\rm{cz}}} \end{array}} \right]_{3\times1}\)