Question 10.2.8: Finding the Product of Two Matrices Find the products AB and...
Finding the Product of Two Matrices
Find the products AB and BA if possible.
A=\left[\begin{array}{rr}1 & 2 \\-1 & 3\end{array}\right] \text { and } B=\left[\begin{array}{rrr}3 & -2 & 1 \\1 & 2 & 3\end{array}\right]
Because A is of order 2×2 and the order of B is 2×3, the product AB is defined and has order 2×3.
Let A B=C=\left[c_{i j}\right]. By the definition of the product AB, each entry c_{i j} \text { of } C is found by multiplying the ith row of A by the jth column of B. For example, c_{11} is found by multiplying the first row of A by the first column of B.
\left[\begin{array}{ll}1 & 2 \\* & *\end{array}\right]\left[\begin{array}{lll}3 & * & * \\1 & * & *\end{array}\right]=\left[\begin{array}{ccc}1(3)+2(1) & * & * \\* & * & *\end{array}\right]
So
A B=\left[\begin{array}{rr}1 & 2 \\-1 & 3\end{array}\right]\left[\begin{array}{rrr}3 & -2 & 1 \\1 & 2 & 3\end{array}\right]
1st row of A times 1st column of B 1st row of A times 2nd column of B 1st row of A times 3rd column of B
↓ ↓ ↓
A B=\left[\begin{array}{ccc}1(3)+2(1) & 1(-2)+2(2) & 1(1)+2(3) \\-1(3)+3(1) & -1(-2)+3(2) & -1(1)+3(3)\end{array}\right]=\left[\begin{array}{ccc}5 & 2 & 7 \\0 & 8 & 8\end{array}\right]
↑ ↑ ↑
2nd row of A times 1st column of B 2nd row of A times 2nd column of B 2nd row of A times 3rd column of B
The product BA is not defined because B is of order 2×3 and A is of order 2×2; that is, the number of columns of B is not the same as the number of rows of A.
