Tip:
Highlight text to annotate it
X
- WELCOME BACK FOR ANOTHER EXAMPLE
OF MATRIX MULTIPLICATION.
WE'RE GIVEN MATRIX A AND B,
AND ASKED TO FIND MATRIX B x MATRIX A.
IF YOU WATCHED THE PREVIOUS EXAMPLE,
WE FOUND MATRIX A x MATRIX B.
THE ORDER OF MATRIX MULTIPLICATION DOES MATTER,
SO NOTICE IN THIS CASE MATRIX B IS FIRST,
WHICH IS A 3 X 2 MATRIX.
AND MATRIX A IS SECOND, WHICH IS A 2 X 3 MATRIX.
NOW THAT WE HAVE THIS SET UP CORRECTLY,
THE NEXT STEP IS TO MAKE SURE THIS PRODUCT IS DEFINED.
WE CAN ONLY PERFORM MATRIX MULTIPLICATION
IF THE NUMBER OF COLUMNS IN THE FIRST MATRIX
IS EQUAL TO THE NUMBER OF ROWS IN THE SECOND MATRIX.
SO, AGAIN, OUR FIRST MATRIX IS A 3 X 2 MATRIX
AND THE SECOND MATRIX IS A 2 X 3 MATRIX.
SO TO SEE IF THIS IS DEFINED,
THESE TWO VALUES HERE MUST BE THE SAME, WHICH THEY ARE,
SO THIS PRODUCT IS DEFINED.
IF THESE WEREN'T EQUAL THE PRODUCT WOULD BE UNDEFINED.
WHEN WE HAVE IT SET UP LIKE THIS,
THE OUTER NUMBERS GIVE US THE DIMENSIONS OF OUR PRODUCT.
MEANING THE DIMENSION OF THE PRODUCT
WILL BE THE NUMBER OF ROWS IN THE FIRST MATRIX
BY THE NUMBER OF COLUMNS IN THE SECOND MATRIX.
SO THIS PRODUCT WILL BE A 3 X 3 MATRIX.
JUST TO COMPARE, IF WE GO BACK AND TAKE A LOOK
AT MATRIX A x MATRIX B, AS WE SEE HERE,
NOTICE HOW THE PRODUCT WAS A 2 X 2 MATRIX
WHICH EMPHASIZES THE IMPORTANCE
OF MATRIX MULTIPLICATION,
AS WELL AS THE FACT THAT MATRIX MULTIPLICATION
IS NOT A COMMUTATIVE.
NOW, TO FIND THE NINE ELEMENTS IN THIS 3 X 3 MATRIX
WE'LL BE USING ROWS FROM THE FIRST MATRIX
AND COLUMNS FROM THE SECOND MATRIX.
TO KNOW WHICH ROW AND COLUMN TO USE,
WE HAVE TO KNOW THE LOCATION OF EACH ELEMENT.
SO FOR EXAMPLE, IF WE WANT TO FIND THE ELEMENT IN ROW ONE,
COLUMN ONE, WHICH WOULD BE A SUB 1, 1,
THIS IS TELLING US THAT WE'LL USE ROW ONE
FROM THE FIRST MATRIX,
AND COLUMN ONE FROM THE SECOND MATRIX.
SO TO FIND THIS ELEMENT
WE WANT TO MULTIPLY ROW ONE BY COLUMN ONE,
WHICH MEANS WE'LL FIND THE SUM OF THE PRODUCT
OF THE FIRST ELEMENTS AND THE SECOND ELEMENTS.
SO A SUB 1, 1 IS = TO -2 x 4 + 3 x 3.
THIS WILL BE -8 + 6 WHICH IS = TO -2.
SO A SUB 1, 1 OR THIS ELEMENT HERE IS -2.
LET'S GO AHEAD AND WORK OUR WAY ACROSS THE FIRST ROW.
SO THE NEXT ELEMENT WOULD BE A SUB 1, 2,
WHICH MEANS WE'LL USE ROW ONE FROM THE FIRST MATRIX,
BUT NOW WE'LL USE ROW TWO FROM THE SECOND MATRIX.
SO NOW WE WOULD HAVE -2 x 1 + 3 x -5,
THIS WOULD BE -2 + -15 OR -17.
NEXT ELEMENT WOULD BE ROW ONE, COLUMN THREE OR A SUB 1, 3.
SO NOW WE'LL USE ROW ONE FROM THE FIRST MATRIX
BUT WE'LL USE ROW THREE FROM THE SECOND MATRIX.
SO WE'LL HAVE -2 x 3 + 3 x 7,
THIS WOULD BE -6 + 21 WHICH WILL BE 15.
NOW WE'LL GO TO ROW TWO, SO WE'LL FIND A SUB 2, 1.
SO NOW WE'LL USE ROW TWO FROM THE FIRST MATRIX
AND COLUMN ONE FROM THE SECOND MATRIX.
SO WE'LL HAVE -4 x 4 + -2 x 2.
SO WE WOULD HAVE -16 + -4 THAT'S -20.
NEXT WE HAVE ROW TWO, COLUMN TWO.
SO WE'LL USE ROW TWO FROM THE FIRST MATRIX
AND COLUMN TWO FROM THE SECOND MATRIX.
SO WE HAVE -4 x 1 + -2 x -5, WHICH WOULD BE -4 + 10 OR 6.
AND NEXT WE HAVE ROW TWO, COLUMN THREE.
SO WE'LL USE ROW TWO FROM THE FIRST MATRIX
AND NOW COLUMN THREE FROM THE SECOND MATRIX.
SO WE'LL HAVE -4 x 3 + -2 x 7.
SO WE'LL HAVE -12 + -14 WHICH IS -26.
OKAY. WE HAVE THREE MORE ELEMENTS TO FIND.
NEXT WE'LL GO TO ROW THREE, SO A SUB 3, 1.
WE'LL BE USING ROW THREE IN THE FIRST MATRIX
AND COLUMN ONE FROM THE SECOND MATRIX.
SO ROW THREE, COLUMN ONE.
SO WE'LL HAVE 5 x 4 + 1 x 2 THAT WOULD BE 20 + 2 OR 22.
NEXT WE HAVE ROW THREE, COLUMN TWO,
SO ROW THREE FROM THE FIRST MATRIX
AND COLUMN TWO FROM THE SECOND.
5 x 1 + 1 x -5 AND THAT WOULD BE 5 + -5 OR 0.
AND FINALLY, FOR THE LAST ELEMENT,
WHICH IS IN ROW THREE, COLUMN THREE,
WE'LL USE ROW THREE FROM THE FIRST MATRIX
AND COLUMN THREE FROM THE SECOND MATRIX.
SO WE HAVE 5 x 3 + 1 x 7.
SO WE HAVE 15 + 7 WHICH IS = TO 22.
SO MATRIX B x MATRIX A IS = TO THE 3 X 3 MATRIX.
AND, AGAIN, TO COMPARE B x A TO A x B,
NOTICE THAT A x B WAS A 2 X 2 AND B x A IS A 3 X 3.
I HOPE YOU FOUND THIS HELPFUL.
I'LL FINISH BY LEAVING SOME FORMAL NOTES
ON MATRIX MULTIPLICATION ON THE SCREEN.