- WE WANT TO WRITE THE SYSTEM OF EQUATIONS AS AN AUGMENTED MATRIX
AND THEN SOLVE THE SYSTEM BY WRITING THE MATRIX
IN ROW ECHELON FORM.
SO THE FIRST THING WE SHOULD NOTICE
IS WE HAVE A SYSTEM OF 3 LINEAR EQUATIONS WITH 3 UNKNOWNS,
WHICH MEANS WE'LL HAVE A 3 x 4 AUGMENTED MATRIX.
AND SINCE THE EQUATIONS ARE IN STANDARD FORM,
MEANING WE HAVE THE X TERMS, Y TERMS, AND Z TERMS ON THE LEFT,
AND THE CONSTANTS ON THE RIGHT,
IT'S READY TO WRITE AS AN AUGMENTED MATRIX.
SO LOOKING AT THE FIRST EQUATION,
THE FIRST ROW OF OUR MATRIX WILL BE 1, -2, 3, 7.
LOOKING AT THE SECOND EQUATION,
THE SECOND ROW WILL BE 3, -2, 2, 10.
AND THE THIRD ROW WILL BE 2, 1, 1, 4.
NOTICE HOW THE FIRST COLUMN
CONTAINS THE COEFFICIENTS OF X,
THE SECOND COLUMN CONTAINS THE COEFFICIENTS OF Y,
THE THIRD COLUMN CONTAINS THE COEFFICIENTS OF Z,
AND THE FOURTH ROW CONTAINS THE CONSTANTS.
NOW WE WANT TO WRITE THIS IN ROW ECHELON FORM.
WE CAN USE THIS FORM HERE AS A GUIDE,
BUT LET'S GO AHEAD AND REVIEW THE SPECIFICS
ABOUT ROW ECHELON FORM.
TO BE IN ROW ECHELON FORM,
THE FIRST NON-ZERO ELEMENT IN EACH ROW,
CALLED THE LEADING ENTRY MUST BE ONE, EACH LEADING ENTRY OR ONE,
IS IN A COLUMN TO THE RIGHT OF THE LEADING ENTRY OR ONE,
IN THE PREVIOUS ROW, AND THEN FINALLY,
ROWS WITH ALL ZERO ELEMENTS, IF ANY,
ARE BELOW ROWS HAVING A NON-ZERO ELEMENT.
SO WE HAVE SEVERAL EXAMPLES HERE,
WHICH ARE IN ROW ECHELON FORM.
BUT I THINK IT'S IMPORTANT TO ALSO NOTICE
THAT IF WE FOCUS ON THE MAIN DIAGONAL,
HIGHLIGHTED HERE IN YELLOW,
NOTICE MOST OF THE ELEMENTS IN THE MAIN DIAGONAL ARE ONE.
HOWEVER, THEY CAN ALSO BE ZERO IF THE SYSTEM HAS EITHER
NO SOLUTION OR INFINITE SOLUTIONS,
BUT ALSO NOTICE THAT ALL OF THE ELEMENTS
BELOW THE MAIN DIAGONAL ARE ZERO.
THIS IS IMPORTANT TO NOTICE
BECAUSE THIS GIVES US A STRATEGY FOR TRANSFORMING THE MATRIX
INTO ROW ECHELON FORM.
THE PROCESS FOR TRANSFORMING THE AUGMENTED MATRIX
IS CALLED GAUSSIAN ELIMINATION,
WHICH MEANS ANY TWO ROWS CAN BE INTERCHANGED,
THE ELEMENTS OF ANY ROW CAN BE MULTIPLIED
BY A NON-ZERO REAL NUMBER,
AND FINALLY, ANY ROW CAN BE CHANGED
BY ADDING OR SUBTRACTING THE CORRESPONDING ELEMENTS
WITH ANOTHER ROW.
SO GOING BACK TO OUR EXAMPLE,
WE WANT TO START BY FOCUSING
ON OBTAINING ZERO'S IN THESE 3 POSITIONS.
SO LOOKING AT OUR AUGMENTED MATRIX,
LET'S FIRST OBTAIN A ZERO IN THIS POSITION HERE.
WELL, IF WE LOOK AT ROW 1, IF THIS ELEMENT HERE WAS A -2
WE COULD THEN ADD ROW 1 AND ROW 3 TOGETHER TO REPLACE ROW 3
TO OBTAIN A 0 IN THIS POSITION.
SO OUR FIRST ROW OPERATION
IS GOING TO BE TO MULTIPLY ROW 1 BY -2 AND ADD IT TO ROW 3.
SO THE FIRST 2 ROWS WILL STAY THE SAME FOR RIGHT NOW
AND NOW FOR ROW 3 WE'LL HAVE -2 x 1 + 2, THAT'S 0.
FOR THE SECOND COLUMN WE'LL HAVE -2 x -2,
WHICH IS 4 + 1, THAT'S 5.
AND THEN WE'LL HAVE -2 x 3, THAT'S -6 + 1, THAT'S -5.
AND -2 x 7, THAT'S -14 + 4 IS -10.
NOW, LETS OBTAIN A ZERO IN THIS POSITION.
AGAIN, IF WE FOCUS ON ROW 1 AND ROW 2,
SINCE THERE'S A 1 IN THIS POSITION,
IF WE WERE TO REPLACE ROW 2 WITH -3 x ROW 1 + ROW 2,
WE WOULD HAVE A 0 IN THIS POSITION.
NOTICE HOW ON THE THIRD ROW, EACH ELEMENT IS A MULTIPLE OF 5,
SO LET'S GO AHEAD AND REPLACE ROW 3 WITH 1/5 x ROW 3.
SO THE FIRST ROW STAYS THE SAME,
NOW FOR ROW 2 WE'LL HAVE -3 x 1 + 3, THAT'S 0.
THEN WE'LL HAVE -3 x -2, THAT'S 6 + -2, THAT'S 4.
AND WE HAVE -3 x 3, THAT'S -9 + 2, THAT'S -7
AND THEN WE HAVE -3 x 7, THAT'S -21 + 10, THAT'S -11.
AND NOW FOR THE THIRD ROW,
WE'RE GOING TO MULTIPLY EACH ELEMENT BY 1/5,
SO WE'LL HAVE 0, 1, -1, AND -2.
NOW, WE STILL NEED A ZERO IN THIS POSITION HERE.
NOTICE HOW WE DON'T WANT TO LOSE A 0 IN THIS POSITION,
SO WE CAN'T USE ROW 1.
WE'LL HAVE TO USE ROW 2.
AND SINCE ROW 2 HAS A 4 IN THIS POSITION,
IF WE WERE TO REPLACE ROW 3 WITH -4 x ROW 3 + ROW 2,
THIS WOULD GIVE US THE 0 IN THIS LAST POSITION,
SO THE FIRST 2 ROWS STAY THE SAME.
SO WE'LL HAVE -4 x 0 + 0, THAT'S 0,
-4 x 1, THAT'S -4 + 4, THAT'S 0,
AND THEN WE HAVE -4 x -1, THAT'S 4 + -7, THAT'S -3,
AND THEN FINALLY, WE HAVE -4 x - 2, THAT'S 8 + -11,
THAT'S -3.
NOW WE'RE ALMOST THERE,
WE HAVE THE ZEROES IN THE CORRECT POSITION,
BUT WE NEED THE FIRST ENTRY IN EACH ROW TO BE 1.
SO NOW WE'RE GOING TO REPLACE ROW 2 WITH 1/4 x ROW 2,
AND WE'LL REPLACE ROW 3 WITH -1/3 x ROW 3.
SO AGAIN, THE FIRST ROW STAYS THE SAME,
THE SECOND ROW WILL BE 0, 1, -7/4, AND -11/4.
AGAIN, WE MULTIPLIED EACH ELEMENT BY 1/4.
AND NOW FOR THE THIRD ROW,
WE'RE GOING TO MULTIPLY EACH ELEMENT BY -1/3.
SO WE'LL HAVE 0, 0, 1, 1.
NOW THAT OUR AUGMENTED MATRIX IS IN ROW ECHELON FORM,
WE'RE GOING TO WRITE THIS BACK AS A SYSTEM OF EQUATIONS
TO FIND THE SOLUTION.
SO THE FIRST ROW WOULD GIVE US THE EQUATION 1X - 2Y + 3Z = 7
OR X - 2Y + 3Z = 7.
LOOKING AT THE SECOND ROW,
THIS TELLS US THAT 0X + 1Y - 7/4Z = -11/4
OR JUST Y - 7/4Z = -11/4.
AND THE LAST ROW TELLS US THAT 0X + 0Y + 1Z = 1 OR JUST Z = 1.
NOTICE HOW THIS LAST ROW TELLS US THAT Z = 1
IS PART OF OUR SOLUTION.
NOW WE CAN PERFORM BACK SUBSTITUTION TO FIND Y AND X.
SO USING EQUATION 2 AND SUBSTITUTING 1 FOR Z,
WE KNOW THAT Y - 7/4 x 1 MUST = -11/4.
SO IF WE ADD 7/4 TO BOTH SIDES WE CAN SEE THAT Y = -4/4 OR -1.
AND NOW WE CAN USE THIS FIRST EQUATION TO FIND THE VALUE OF X.
X - 2 x Y, WHICH IS -1 + 3 x Z, WHICH IS 1 MUST = 7.
THIS IS GOING TO BE X + 2 + 3 = 7 OR X + 5 = 7.
SO NOW WE KNOW X MUST = 2.
SO IF WE WANT TO GIVE OUR SOLUTION AS AN ORDERED PAIR,
THE X COORDINATE WOULD BE + 2, THE Y COORDINATE WOULD BE -1,
AND THE Z COORDINATE WOULD BE 1.
TO VERIFY THIS, IF WE WERE TO GRAPH EACH OF THESE EQUATIONS
IN SPACE WE WOULD HAVE 3 PLAINS IN SPACE,
WHICH MEANS THESE 3 PLAINS WOULD INTERSECT IN THIS POINT.
LET'S GO AND LOOK AT THE GRAPH OF THIS.
HERE'S THE GRAPH OF THE 3 PLAINS IN SPACE,
NOTICE HOW THESE 3 PLAINS DO INTERSECT IN A SINGLE POINT,
RIGHT HERE AND THE COORDINATES OF THAT POINT WOULD BE 2, -1, 1.
OKAY, HOPE YOU FOUND THIS EXPLANATION HELPFUL.