WEBVTT
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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.
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Find π΄π΅ if possible.
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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.
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If this is not the case, the product does not exist.
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In this case, we say the product is undefined.
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Since π΄ and π΅ are both three-by-three matrices, their product will also be a three-by-three matrix.
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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΅.
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Negative three multiplied by negative two add negative four times six add four times three is negative six.
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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΅.
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Negative three multiplied by negative three add negative four multiplied by zero add four multiplied by five is 29.
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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΅.
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Negative three multiplied by two add negative four multiplied by two add four multiplied by negative four is negative 30.
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Letβs repeat this process with row two of π΄ and column one of π΅.
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Adding together the products of the entries in row two of π΄ and column one of π΅ gives us a total of 44.
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The products of row two of π΄ and column two of π΅ give us a sum of 32.
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And finally, the products of row two of π΄ and column three of π΅ give us a total of negative 16.
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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΅.
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This gives us a value of negative 19.
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The entry in row three, column two of π΄π΅ gives us a value of negative 20.
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And the sum of the products of row three of π΄ and column three of π΅ is 12.
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The product of matrices π΄ and π΅ is, therefore, π΄π΅ as shown.