- WELCOME BACK FOR A SECOND EXAMPLE
ON HOW TO FIND THE DETERMINANT OF A 3x3 MATRIX
USING THE COFACTOR METHOD.
USING THE COFACTOR METHOD TO CALCULATE THE DETERMINANT
OF A 3x3 MATRIX,
WE MULTIPLY EACH ELEMENT IN ANY ROW OR COLUMN OF THE MATRIX
BY ITS COFACTOR.
THE SUM OF THESE PRODUCTS IS EQUAL TO THE DETERMINANT
WHERE THE COFACTOR OF EACH ELEMENT IS EQUAL TO -1
RAISED TO THE POWER OF I + J WHERE I IS THE ROW,
AND J IS THE COLUMN OF THE GIVEN ELEMENT
AND M THE MINOR IS EQUAL TO THE 2 x 2 DETERMINANT
FORMED BY THE ELIMINATING THE ROW AND COLUMN
OF THE GIVEN ELEMENT.
SO GOING BACK TO OUR EXAMPLE THE FIRST STEP IS TO DECIDE
WHICH ROW OR COLUMN WE WANT TO USE TO CALCULATE
OUR DETERMINANT,
AND SINCE WE HAVE A 0 IN ROW 3, COLUMN 1,
IT'LL BE EASIEST IF WE USE EITHER ROW 3 OR COLUMN 1.
LET'S GO AHEAD AND USE ROW 3 TO CALCULATE OUR DETERMINANT.
SO WE'LL HAVE THE DETERMINANT OF MATRIX A.
IT'S GOING TO BE EQUAL TO THE FIRST ELEMENT IN OUR ROW
WHICH IS 0 x -1 RAISED TO THE POWER OF THE SUM OF THE ROW
AND COLUMN OF THIS ELEMENT.
THIS ELEMENT IS IN ROW 3, COLUMN 1, AND 3 + 1 IS 4,
SO WE'RE GOING TO RAISE THIS TO THE FOURTH POWER.
AND OF COURSE WE COULD STOP HERE
BECAUSE WE'RE MULTIPLYING BY 0,
BUT LET'S GO AHEAD AND CONTINUE JUST TO REVIEW.
SO NOW WE'RE GOING TO MULTIPLY THIS BY THE 2x2 DETERMINANT
FORMED BY ELIMINATING THE ROW AND COLUMN OF THIS ELEMENT.
SO WE'LL ELIMINATE ROW 3 AND COLUMN 1,
LEAVING US WITH THESE 4 ELEMENTS TO FORM OUR 2x2 DETERMINANT.
SO WE'LL HAVE -3, 4, 6, 8.
NOW WE'LL MOVE TO THE NEXT ELEMENT IN THIS ROW WHICH IS -2.
SO WE'LL HAVE + -2 x -1 RAISED TO THE POWER OF-
WELL, THIS IS ELEMENT IS IN ROW 3, COLUMN 2. 3 + 2 IS 5.
SO NOW WE'LL RAISE -1 TO THE FIFTH POWER,
AND WE'LL MULTIPLY THIS BY A 2 x 2 DETERMINANT
FORMED BY ELIMINATING THE ROW AND COLUMN OF THIS ELEMENT HERE.
SO WE'LL ELIMINATE ROW 3, COLUMN 2 THIS TIME,
LEAVING US WITH THE ELEMENTS 2, 4, -1, 8 +
THE LAST ELEMENT IN OUR ROW WHICH IS +1 x -1
RAISED TO THE POWER OF THE ROW PLUS THE COLUMN OF THE ELEMENT
THAT WE'RE USING.
THIS IS IN ROW 3, COLUMN 3.
3 + 3 IS EQUAL TO 6,
AND WE'LL MULTIPLY THIS BY THE 2 x 2 DETERMINANT
FORMED BY ELIMINATING THE ROW AND COLUMN OF OUR ELEMENT,
SO WE'LL ELIMINATE ROW 3 AND COLUMN 3.
SO THE ELEMENTS THAT WE'LL USE WILL BE 2, -3, -1, 6.
AND NOW WE'RE LEFT TO SIMPLIFY THIS.
THE FIRST PRODUCT WILL BE 0.
THAT'S WHY WE DECIDED TO USE ROW 3.
NOW LOOKING AT THE NEXT PRODUCT WE HAVE -1 TO THE FIFTH,
THAT'S -1 x -2.
SO WE HAVE + +2 x THE VALUE OF OUR 2 x 2 DETERMINANT
WHICH WOULD BE 2 x 8 - 4 x -1.
2 x 8 IS 16.
4 x -1 IS -4.
16 - -4 IS +20 + HERE WE HAVE -1 TO THE SIXTH,
THAT'S +1 x 1,
SO WE HAVE 1 x THE VALUE OF OUR DETERMINANT
WHICH AGAIN WOULD BE 2 x 6 - -3 x -1.
SO WE WOULD HAVE 12 - 3 OR +9.
SO WE HAVE 0 + 40 + 9 WHICH IS EQUAL TO 49.
SO THE DETERMINANT OF MATRIX A IS EQUAL TO +49.
REMEMBER, WE CAN ALSO INDICATE A DETERMINANT USING VERTICAL BARS.
SO WE COULD ALSO INDICATE THIS DETERMINANT
USING THIS NOTATION HERE.
IF YOU WANT TO SEE THE SAME DETERMINANT
USING THE DIAGONAL METHOD,
I'LL PUT A LINK ON THE SCREEN.
I HOPE YOU FOUND THIS HELPFUL.