Video Transcript
Consider the matrices π΄ equals
negative three, negative four, four, negative four, four, four, five, negative one,
negative one and π΅ equals negative two, negative three, two, six, zero, two, and
three, five, negative four. Find π΄π΅ if possible.
We can multiply two matrices β if
and only if β the number of columns in the first matrix equals the number of rows in
the second. If this is not the case, the
product does not exist. In this case, we say the product is
undefined.
Since π΄ and π΅ are both
three-by-three matrices, their product will also be a three-by-three matrix. To find the first entry in π΄π΅, we
find the sum of the products of the entries in row one of π΄ and column one of
π΅. Negative three multiplied by
negative two add negative four times six add four times three is negative six.
To find the entry for row one,
column two of π΄π΅, we again find the sum of the products of row one of π΄ and
column two of π΅. Negative three multiplied by
negative three add negative four multiplied by zero add four multiplied by five is
29.
To find the entry for row one,
column three of π΄π΅, we now find the sum of the products of row one of π΄ and
column three of π΅. Negative three multiplied by two
add negative four multiplied by two add four multiplied by negative four is negative
30.
Letβs repeat this process with row
two of π΄ and column one of π΅. Adding together the products of the
entries in row two of π΄ and column one of π΅ gives us a total of 44. The products of row two of π΄ and
column two of π΅ give us a sum of 32. And finally, the products of row
two of π΄ and column three of π΅ give us a total of negative 16.
To find the first entry in the
final row of π΄π΅, we find the sum of the products of row three of π΄ and column one
of π΅. This gives us a value of negative
19. The entry in row three, column two
of π΄π΅ gives us a value of negative 20. And the sum of the products of row
three of π΄ and column three of π΅ is 12.
The product of matrices π΄ and π΅
is, therefore, π΄π΅ as shown.
