Multiplying Matrices Lesson

In this lesson, we will learn how to find the product of two matrices. First and foremost, in order to multiply two matrices together, the number of columns from the first matrix (leftmost) must be equal to the number of rows from the second matrix (rightmost). If these two numbers do not match, we stop and state that the product does not exist. If the two numbers match, we are able to find our product. Let's look at an example.
Example #1: Find AB. $$A=\left[ \begin{array}{cc}5&6&3\\ 4&2 &1\end{array}\right] $$ $$B=\left[ \begin{array}{cc}8&4\\ -1&1 \\ 8 & 10\end{array}\right] $$ A is a 2 x 3 matrix.
B is a 3 x 2 matrix.
The columns from A (3) match up with the rows from B (3). Therefore, we can find the product.
To set up the product, we take the rows from A and the columns from B. So AB will be a 2 x 2 matrix. $$AB=\left[ \begin{array}{cc}a&b\\ c&d\end{array}\right] $$ To find each entry in the answer, we find the product of the corresponding row from the first matrix and the corresponding column from the second matrix. In other words, if we are in the first row and first column of the answer, we will look at the first row from the first matrix and the first column from the second matrix. We will find the sum of the product of all corresponding entries. Lowercase a represents the entry in the first row and first column. So we will look at the entries in the first row of the leftmost matrix and the entries in the first column of the rightmost matrix. $$a=5 \cdot 8 + 6 \cdot -1 + 3 \cdot 8$$ $$a=40 - 6 + 24=58$$ Continuing this pattern: $$b=5 \cdot 4 + 6 \cdot 1 + 3 \cdot 10$$ $$b=20 + 6 + 30=56$$ $$c=4 \cdot 8 + 2 \cdot -1 + 1 \cdot 8$$ $$c=32 - 2 + 8=38$$ $$d=4 \cdot 4 + 2 \cdot 1 + 1 \cdot 10$$ $$d=16 + 2 + 10=28$$ This gives us a final answer of: $$AB=\left[ \begin{array}{cc}58&56\\ 38&28\end{array}\right] $$

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