- WELCOME TO A VIDEO ON TRANSFORMING AN AUGMENTED MATRIX
INTO ROW ECHELON FORM TO SOLVE A SYSTEM OF EQUATIONS.
LET'S FIRST REVIEW WHAT ROW ECHELON FORM IS.
AN AUGMENTED MATRIX IS IN ROW ECHELON FORM
IF THE MAIN DIAGONAL CONSISTS OF ONES OR ZEROS
AND ALL OF THE ELEMENTS UNDER THE MAIN DIAGONALS ARE ZEROS.
SO HERE ARE THE MAIN DIAGONALS, AND THOSE MUST BE ONES OR ZEROS.
AND THEN THE ELEMENTS UNDER THE MAIN DIAGONAL MUST BE ZEROS
AS WE SEE HERE AND ALSO AS WE SEE HERE.
LET'S TAKE A LOOK AT HOW WE'RE GOING TO TRANSFORM
THE AUGMENTED MATRIX INTO ROW ECHELON FORM.
WHAT WE'RE ALLOWED TO DO IS INTERCHANGE TWO ROWS.
WE'RE ALSO ALLOWED TO MULTIPLY THE ELEMENTS OF ANY ROW
BY A NON-ZERO REAL NUMBER,
AND WE CAN ALSO ADD AND SUBTRACT THE CORRESPONDING ELEMENTS
OF ANY TWO ROWS.
AND THIS PROCESS IS CALLED GAUSSIAN ELIMINATION.
LET'S GO AHEAD AND GIVE IT A TRY.
LET'S START BY RUNNING THIS SYSTEM AS AN AUGMENTED MATRIX.
THE FIRST ROW WOULD BE 2, -3, 18
AND THE SECOND ROW WOULD BE 5, 2, 7.
LET'S GO AHEAD AND DRAW A VERTICAL LINE HERE
TO SEPARATE THE COEFFICIENTS FROM THE CONSTANTS.
NOW OUR FIRST GOAL IS TO GET A ZERO IN THIS POSITION.
SINCE THE CORRESPONDING ELEMENT IS A 2,
THE LEAST COMMON MULTIPLE OF 5 AND 2 WOULD BE 10.
SO IF THIS WAS A +10, THIS WAS A -10
WE COULD ADD THEM TOGETHER TO MAKE THIS 0.
SO WE'LL MULTIPLY ROW 1 BY -5, MULTIPLY ROW 2 BY 2.
SO THE FIRST ROW WOULD BECOME -10, +15, AND -90.
AND ROW 2 WOULD BE 10, 4, AND 14.
NOW WE'RE GOING TO GO AHEAD AND REPLACE ROW 2
WITH ROW 2 + ROW 1.
THIS SHOULD GIVE US THE ZERO THAT WE NEED IN THIS POSITION.
THE FIRST ROW STAYS THE SAME.
FOR ROW TWO WE'RE GOING TO HAVE 10 + -10 = 0, 5 +-- 15 = 19,
AND 14 + -90 = -76.
NOW WE WANT A MAIN DIAGONAL OF ONES,
SO WE'RE GOING TO MULTIPLY ROW 1 BY -1/10,
WHICH IS THE SAME AS DIVIDING BY -10.
AND WE'LL MULTIPLY ROW 2 BY 1/19 OR DIVIDE BY 19.
SO THIS WOULD BE A 1, -1/10 TIMES 15
OR 15 DIVIDED BY -10 = -3/2
AND -1/10 x -90 = 9.
AND THEN ROW 2 WE HAVE 0, THIS WOULD BE 1 AND 1/19 x -76 = -4.
NOW THIS IS IN ROW ECHELON FORM.
LET'S GO AHEAD AND REWRITE THIS AS A SYSTEM
AND IT SHOULD BE MUCH EASIER TO SOLVE.
THIS FIRST EQUATION WOULD BE X - 3/2Y = 9
AND THE SECOND EQUATION WOULD BE Y = -4.
SO NOW WE KNOW THAT Y = -4 WE CAN PERFORM BACK SUBSTITUTION
TO DETERMINE X.
SO WE'LL HAVE X - 12/2 OR + 6, SO WE HAVE X = 3.
SO WE FOUND THE SOLUTION TO THE SYSTEM USING A AUGMENTED MATRIX
IN ROW ECHELON FROM.
X = 3 AND Y = -4.
LET'S GO AHEAD AND TRY IT ON A SYSTEM OF 3 EQUATIONS
AND 3 VARIABLES,
AND THIS IS GOING TO TAKE A LITTLE BIT MORE WORK.
LET'S GO AHEAD AND WRITE OUR AUGMENTED MATRIX.
THE FIRST ROW WOULD BE 1, -1, 1, 2.
THE SECOND ROW WOULD BE 3, -2, 1, -1
AND ROW 3, THERE'S NO Z TERM,
SO WE'RE GOING TO HAVE 1, 1, 0 FOR THE Z TERM, AND THEN -3.
WHAT WE WANT TO DO NOW IS START OFF
BY TRYING TO GET A ZERO IN THIS POSITION
AND ALSO A ZERO IN THIS POSITION.
SO IF YOU TAKE A LOOK AT ROW 1 AND ROW 3,
IF WE REPLACE ROW 3 WITH ROW 3 - ROW 1,
THAT WOULD GIVE US A 0 IN THIS POSITION.
AND THEN IF YOU TAKE A LOOK AT THE FIRST 2 ROWS,
IF WE TAKE THIS ROW AND SUBTRACT 3 x THIS ROW
THAT WOULD ALSO GIVE US A 0 HERE.
SO LET'S GO AHEAD AND DO THAT AS WELL.
LET'S GO AHEAD AND REWRITE THIS NOW,
FIRST ROW STAYS THE SAME.
SO FOR THE SECOND ROW WE'RE GOING TO HAVE 3 - 3 x 1 = 0,
-2 - 3 x -1 = -2 + 3 = 1,
AND THEN 1 - 3 x 1 = -2 AND -1 - 3 x 2 = -1 - 6 = -7.
AND THE NEW ROW 3 WILL BE 1 - 1 = 0, 1 - -1 = 2, 0 - 1 = -1,
AND THEN LASTLY WE HAVE -3 - 2 = -5.
NOW THE NEXT THING WE WANT TO DO IS HAVE A 0 IN THIS POSITION.
WE WANT TO WORK WITH ROW 2 AND ROW 3
SO WE'RE GOING TO TAKE ROW 3 AND REPLACE IT
WITH ROW 3 - 2 x ROW 2.
FIRST TWO ROWS STAY THE SAME.
SO FOR ROW 3, NOW WE'RE GOING TO HAVE 0 - 2 x 0 = 0,
2 - 2 x 1 = 0, -1 - 2 x -2 = -1 + 4 = 3,
AND THEN WE HAVE -5 - 2 x -7 = -5 + 14 = 9.
LET'S GO AHEAD AND TAKE THIS OVER TO THE NEXT SCREEN.
OUR LAST STEP HERE IS TO MAKE SURE THE MAIN DIAGONAL
CONSISTS OF ONES OR ZEROS.
THESE TWO WERE ALREADY ONE, BUT THIS ONE IS A THREE.
SO WE NEED TO MULTIPLY ROW 3 BY 1/3,
WHICH IS THE SAME AS DIVIDING BY 3.
SO AGAIN, THE FIRST 2 ROWS STAY THE SAME
AND NOW WE'LL MULTIPLY THIS BY 1/3.
SO WE HAVE 0, 0, 1, AND THEN 9 x 1/3 = 3.
NOW THAT IT'S IN ROW ECHELON FORM
WE'LL REWRITE THIS AS A SYSTEM OF EQUATIONS.
SO THE FIRST ROW WOULD BE X - Y + Z = 2.
THE SECOND ROW WOULD BE Y - 2Z = -7,
AND THE LAST ROW TELLS US THAT Z = +3.
SO WE KNOW Z = 3,
NOW WE CAN PERFORM BACK SUBSTITUTION TO FIND X AND Y.
LOOKING AT THIS SECOND EQUATION, WE HAVE Y - 2 x Z,
WHICH IS 3 MUST = -7.
SO WE HAVE Y - 6 = -7 ADD 6 TO BOTH SIDES WE HAVE Y = -1,
AND NOW WE'LL USE THE FIRST EQUATION TO DETERMINE X.
SO WE HAVE X - Y + Z = 3 = 2.
SO WE HAVE X, THIS BECOMES +1, +3 SO WE HAVE X + 4 = 2.
SO X = -2.
AND NOW WE HAVE SOLVED THE SYSTEM
BY USING AN AUGMENTED MATRIX IN ROW ECHELON FORM.
OKAY, IN THE NEXT VIDEO WE'LL TALK ABOUT HOW TO REWRITE
AN AUGMENTED MATRIX INTO REDUCED ROW ECHELON FORM.
IT'S A VERY SIMILAR PROCESS.
AS YOU BEGIN TO GET A FEEL FOR WORKING WITH AUGMENTED MATRICES,
YOU CAN PROBABLY TELL THAT IT'S VERY SIMILAR TO USING
THE ELIMINATION PROCESS.
IT'S JUST THAT NOW INSTEAD OF USING THE EQUATIONS
WE'RE USING A MATRIX.
I HOPE YOU FOUND THIS VIDEO HELPFUL.