- WE WANT TO WRITE THE SYSTEM OF EQUATIONS
AS AN AUGMENTED MATRIX,
AND THEN SOLVE THE SYSTEM BY WRITING THE MATRIX
IN REDUCED ROW ECHELON FORM.
SO BECAUSE WE HAVE A SYSTEM OF THREE LINEAR EQUATIONS
WITH THREE UNKNOWNS,
WE'LL HAVE A 3 X 4 AUGMENTED MATRIX.
AND SINCE THE EQUATIONS ARE ALREADY IN STANDARD FORM,
LOOKING AT THE FIRST EQUATION,
THE FIRST ROW OF OUR MATRIX WILL BE 1, 2, -1, -10.
THE SECOND ROW WILL BE 2, -3, 2, 2.
THE THIRD ROW WILL BE 1, 1, 3, AND 0.
NOTICE THE FIRST THREE COLUMNS REPRESENT
THE X, Y, AND Z COEFFICIENTS,
AND THE FOURTH COLUMN REPRESENTS THE CONSTANTS.
NOW LET'S REVIEW REDUCED ROW ECHELON FORM
OF AN AUGMENTED MATRIX.
NUMBER ONE, THE FIRST NON-ZERO ELEMENT IN EACH ROW,
CALLED THE LEADING ENTRY, MUST BE 1.
NUMBER TWO, EACH LEADING ENTRY, OR 1,
IN A COLUMN IS TO THE RIGHT OF A LEADING ENTRY,
OR 1, IN THE PREVIOUS ROW.
NUMBER THREE, ROWS WITH ALL ZERO ELEMENTS,
IF ANY, ARE BELOW ROWS HAVING A NON-ZERO ELEMENT.
IF THESE FIRST THREE CONDITIONS ARE MET,
WE HAVE ROW ECHELON FORM.
THIS FOURTH CONDITION IS WHAT MAKES REDUCED ROW ECHELON FORM.
IN EACH COLUMN THAT CONTAINS A LEADING ENTRY, OR 1,
ALL OTHER ELEMENTS OF THE COLUMN ARE 0.
SO BECAUSE WE HAVE A 3 X 4 AUGMENTED MATRIX,
REDUCED ROW ECHELON FORM WILL FIT ONE OF THESE FORMS BELOW,
WHERE NOTICE IF THE MAIN DIAGONAL ONLY CONTAINS 1,
WE HAVE ONE SOLUTION.
IF THE MAIN DIAGONAL CONTAINS A 0, WE HAVE EITHER NO SOLUTION,
IN THIS CASE HERE,
OR INFINITE SOLUTIONS, IN THIS CASE HERE,
IF WE HAVE A ROW OF ZEROES.
WE'LL TRANSFORM THE MATRIX INTO THIS FORM
USING GAUSSIAN ELIMINATION,
WHICH ALLOWS ANY TWO ROWS TO BE INTERCHANGED,
THE ELEMENTS OF ANY ROW CAN BE MULTIPLIED
BY A NON-ZERO REAL NUMBER,
AND ANY ROW CAN BE CHANGED BY ADDING OR SUBTRACTING
THE CORRESPONDING ELEMENTS WITH ANOTHER ROW.
SO THIS CAN BE A FAIRLY INVOLVED PROCESS,
SO LET'S START BY FOCUSING ON OBTAINING A 0 IN ROW 2, COLUMN 1
AND ROW 3, COLUMN 1.
SO IF WE FOCUS ON JUST ROW 1 AND ROW 2,
IF WE REPLACE ROW 2 WITH -2 x ROW 1 + ROW 2,
THAT WOULD GIVE US A 0 IN THIS POSITION HERE.
NOW IF WE LOOK AT ROW 3 AND FOCUS ON ROW 2,
NOTICE IF WE REPLACE ROW 3 WITH -2 x ROW 3,
THAT WOULD GIVE US A -2 HERE,
AND IF WE ADD THIS TO ROW 2 WE WOULD GET A 0.
LET'S START BY PERFORMING THESE TWO ROW OPERATIONS,
SO WE'LL LEAVE THE FIRST ROW THE SAME FOR RIGHT NOW.
AND NOW FOR ROW 2 WE'LL HAVE -2 x 1 + 2 = 0, -2 x 2 + -3 = -7,
-2 x -1 + 2 = 4, -2 x -10 + 2 = 22.
AND NOW FOR ROW 3 WE HAVE -2 x ROW 3 + ROW 2.
SO -2 x 1 + 2 = 0, -2 x 1 + -3 = -5, -2 x 3 + 2 = -4,
AND -2 x 0 + 2 = 2.
NOW LET'S FOCUS ON OBTAINING A 0 IN THIS POSITION
AND THIS POSITION.
WE'RE LOOKING AT ROW 2 AND ROW 3.
NOTICE HOW THE ELEMENTS IN COLUMN 3 ARE ALREADY OPPOSITES,
WHICH MEANS IF WE REPLACE ROW 2 WITH ROW 2 + ROW 3
WE WOULD OBTAIN A 0 IN COLUMN 3.
WE ALSO WOULD NOT LOSE THE 0 IN COLUMN 1.
NOW, TO OBTAIN A 0 WHERE WE HAVE -5, WE HAVE TO USE ROW 2,
SO WE DON'T LOSE THE 0 IN COLUMN 1.
AND SINCE THE LEAST COMMON MULTIPLE OF 5 AND 7 IS 35,
WE COULD OBTAIN A 0 IN THIS POSITION
IF WE REPLACE ROW 3 WITH -5 x ROW 2 + 7 x ROW 3.
SO AGAIN, THE FIRST ROW IS GOING TO STAY THE SAME FOR RIGHT NOW.
AND NOW FOR ROW 2 WE'LL HAVE ROW 2 + ROW 3.
SO WE HAVE 0, -7 + -5 = -12, 4 + -4 = 0, 22 + 2 = 24.
AND NOW FOR ROW 3 WE'LL HAVE -5 x ROW 2 + 7 x ROW 3.
SO WE HAVE -5 x 0 + 7 x 0 = 0,
AND THEN WE HAVE -5 x -7 = 35 + 7 x -5 = 0,
AND THEN WE HAVE -5 x 4 = -20 + 7 x -4 = -48,
AND THEN WE HAVE -5 x 22 = -110 + 7 x 2 = -96.
NOW, IF WE TAKE A LOOK AT ROW 1,
LET'S OBTAIN A 0 IN THIS POSITION HERE.
LET'S GO AHEAD AND USE ROW 2.
THE LEAST COMMON MULTIPLE OF 2 AND 12 IS 12,
SO LET'S REPLACE ROW 1 WITH 6 x ROW 1 + ROW 2.
WE ALSO WANT THE FIRST ELEMENT IN EACH ROW TO BE 1,
SO LOOKING AT ROW 2,
LET'S MAKE THIS -12 A 1 BY MULTIPLYING ROW 2 BY -1/12.
AND TO MAKE THIS -48 A 1,
LET'S REPLACE ROW 3 WITH -1/48 x ROW 3.
SO LET'S BE CAREFUL HERE.
WE HAVE 6 x ROW 1 + ROW 2.
6 x 1 + 0 = 6, 6 x 2 + -12 = 0,
6 x -1 + 0 = -6, 6 x -10 + 24 = -36.
AND NOW, MULTIPLYING BY -1/12 IS THE SAME AS DIVIDING BY -12,
SO WE WOULD HAVE 0, 1, 0, -2.
FOR ROW 3 WE WOULD HAVE 0, 0, 1, AND 2.
NOW, WE'RE ALMOST IN REDUCED ROW ECHELON FORM.
WE NEED THIS ELEMENT TO BE 0,
SO LET'S GO AHEAD AND REPLACE THIS WITH 6 x ROW 3 + ROW 1.
SO THE SECOND AND THIRD ROWS WILL STAY THE SAME.
SO FOR ROW 1 WE'LL HAVE 6 x 0 + 6 = 6, 6 x 0 + 0 = 0,
AND 6 x 1 = 6 + -6 = 0, AND WE HAVE 6 x 2 = 12 + -36 = -24.
AND NOW FOR THE LAST STEP WE NEED THIS 6 TO BE A 1,
SO WE'RE GOING TO REPLACE ROW 1 WITH 1/6 x ROW 1.
SO WE'LL HAVE 1, 0, 0, -4, 0, 1, 0, -2, AND 0, 0, 1, 2.
NOW OUR MATRIX IS IN REDUCED ROW ECHELON FORM,
SO IF WE WRITE THIS BACK AS A SYSTEM OF EQUATIONS
WE'LL HAVE OUR SOLUTION.
REMEMBER THESE COLUMNS REPRESENT THE COEFFICIENTS OF X, Y, AND Z,
SO ROW 1 IS TELLING US THAT X = -4.
THE SECOND ROW IS TELLING US Y = -2,
AND THE THIRD IS TELLING US THAT Z = 2.
IF WE WANT, WE CAN WRITE THIS AS AN ORDERED TRIPLE, -4, -2, 2.
EITHER WAY, OUR SYSTEM HAS ONE SOLUTION,
WHICH MEANS THE SYSTEM IS CONSISTENT AND INDEPENDENT.
IF WE WERE TO GRAPH THESE IN SPACE,
EACH OF THESE EQUATIONS REPRESENTS A PLANE,
SO THESE PLANES SHOULD INTERSECT IN A SINGLE POINT
WITH THESE COORDINATES.
LET'S GO AHEAD AND VERIFY THAT.
HERE IS THE GRAPH OF THE THREE PLANES.
NOTICE HOW THESE PLANES DO INTERSECT IN ONE POINT,
THIS POINT HERE, WHICH WOULD HAVE THE COORDINATES -4, -2, 2.
OKAY. I HOPE YOU FOUND T HIS EXPLANATION HELPFUL.