- 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.
THE FIRST THING WE SHOULD NOTICE IS WE HAVE A SYSTEM
OF TWO LINEAR EQUATIONS WITH TWO UNKNOWNS,
AND SINCE THE EQUATIONS ARE IN STANDARD FORM
MEANING THE X TERMS AND Y TERMS ARE ON THE LEFT
AND THE CONSTANTS ARE ON THE RIGHT
WE CAN GO AHEAD AND WRITE THIS AS AN AUGMENTED MATRIX,
WHERE THE FIRST ROW WOULD BE 5, -2, -4.
AND THE SECOND ROW WOULD BE 1, 3, -11.
SO NOTICE HOW THE FIRST COLUMN
CONTAINS THE COEFFICIENTS OF X,
THE SECOND COLUMN CONTAINS THE COEFFICIENTS OF Y
AND THE THIRD COLUMN CONTAINS THE CONSTANTS.
AND WITH MOST CASES WHEN SOLVING A SYSTEM LIKE
THIS WE COULD USE THIS FORM OF AN AUGMENTED MATRIX
AS A MODEL FOR REDUCED ROW ECHELON FORM
BUT LET'S TAKE A MOMENT AND REVIEW THE SPECIFIC CONDITIONS
IN ORDER FOR A MATRIX TO BE IN REDUCED ROW ECHELON FORM.
LOOKING AT THESE FOUR CONDITIONS,
IF THE FIRST THREE ARE MET WE HAVE ROW ECHELON FORM.
THIS FOURTH CONDITION IS WHAT MAKES REDUCED ROW ECHELON FORM.
NUMBER ONE, THE FIRST NONZERO ELEMENT IN EACH ROW
CALLED THE LEADING ENTRY IS ONE.
NUMBER TWO, EACH LEADING ENTRY OR ONE IS IN A COLUMN
TO THE RIGHT OF THE LEADING ENTRY
OR ONE IN THE PREVIOUS ROW.
NUMBER THREE, ROWS WITH ALL ZERO ELEMENTS IF ANY,
ARE BELOW ROWS HAVING A NONZERO ELEMENT AND THEN NUMBER FOUR,
IN EACH COLUMN THAT CONTAINS A LEADING ONE OF A ROW
ALL OTHER ELEMENTS IN THIS COLUMN ARE ZEROS.
SO HERE ARE SEVERAL EXAMPLES OF AUGMENTED MATRICES
IN REDUCED ROW ECHELON FORM.
NOTICE IF WE FOCUS ON THE MAIN DIAGONALS HIGHLIGHTED HERE
IN YELLOW ALL OF THE MAIN DIAGONALS CONTAIN ONES OR ZEROS.
IF THEY CONTAIN A ZERO WE HAVE EITHER NO SOLUTION
OR INFINITE SOLUTIONS BUT NOTICE
HOW WHEREVER THERE IS A ONE IN A ROW
ALL THE ELEMENTS ABOVE AND BELOW THAT ONE ARE ZERO.
SO GOING BACK TO OUR EXAMPLE,
WE'RE FIRST GOING TO FOCUS ON OBTAINING ZEROS
IN THE CORRECT POSITIONS.
SO IN THIS CASE WE WANT A ZERO IN ROW 2 COLUMN 1,
AND ROW 1 COLUMN 2.
AND TO DO THIS WE'LL PERFORM A PROCESS
CALLED GAUSSIAN ELIMINATION
WHICH MEANS TO TRANSFORM THE MATRIX,
ANY TWO ROWS CAN BE INTERCHANGED,
THE ELEMENTS OF ANY ROW
CAN BE MULTIPLIED BY A NONZERO REAL NUMBER AND NUMBER THREE,
ANY ROW CAN BE CHANGED BY ADDING
OR SUBTRACTING THE CORRESPONDING ELEMENTS WITH ANOTHER ROW.
SO LET'S FOCUS ON OBTAINING
A ZERO IN THIS POSITION HERE FIRST.
TO OBTAIN A ZERO HERE
WE'LL ACTUALLY GOING TO ADD A MULTIPLE OF ROW 2 TO ROW 1.
IF WE WANT A ZERO HERE
THAT MEANS THE ELEMENTS ON THIS COLUMN
WOULD HAVE TO BE OPPOSITES.
WE'RE GOING TO REPLACE ROW 2 WITH -5 x ROW 2 + ROW 1.
SO WE'LL KEEP THE FIRST ROW THE SAME.
SO WE HAVE -5 x 1 + 5 THAT'S 0.
-5 x 3 THAT'S -15 + -2
THAT'S -17 AND THEN -5 x -11 THAT'S 55 + -4 THAT'S 51.
NOW LOOKING AT ROW 2,
SINCE -17 AND 51 ARE MULTIPLES OF 17
AND WE KNOW THE FIRST NONZERO ELEMENT IN A ROW MUST BE ONE
LET'S GO AHEAD AND REPLACE ROW 2 WITH -1/17 x ROW 2.
SO AGAIN, THE FIRST ROW WILL STAY THE SAME
FOR RIGHT NOW.
SO -1/17 x 0 IS 0.
-1/17 x -17 IS 1 AND -1/17 x 51 = -3.
NOW LET'S OBTAIN A ZERO IN THIS POSITION HERE.
NOTICE WE WANT TO ADD THIS ROW TO ROW 2.
WE WOULD WANT THIS ELEMENT HERE TO BE A 2,
MEANS NOW WE'RE GOING TO REPLACE ROW 1 WITH 2 x ROW 2 + ROW 1.
SO THE SECOND ROW WILL STAY THE SAME
AND NOW WE'LL HAVE 2 x 0 + 5 THAT'S 5.
AND THEN 2 x 1 + -2, THAT'S 0,
AND THEN 2 x -3 THAT'S -6 + -4 THAT'S -10
AND NOW FOR THE LAST STEP WE WANT THIS ELEMENT HERE TO BE 1
SO WE'LL REPLACE ROW 1 WITH 1/5 x ROW 1.
SECOND ROW STAYS THE SAME
AND NOW FOR THE FIRST ROW WE HAVE 1/5 x 5 THAT'S 1,
1/5 x 0 OF COURSE IS STILL 0, AND 1/5 x -10 = -2.
NOW OUR AUGMENTED MATRIX IS IN REDUCED ROW ECHELON FORM,
WHICH MEANS IF WE WRITE THIS BACK
AS THE SYSTEM OF EQUATION WE'LL HAVE OUR SOLUTION.
FOR EXAMPLE THIS FIRST ROW IS TELLING US THAT 1X + 0Y = -2
OR JUST X = -2 AND THE SECOND ROW IS TELLING US 0X + 1Y = -3
OR Y = -3.
SO THIS IS THE SOLUTION TO OUR SYSTEM OF EQUATIONS.
WE HAVE ONE SOLUTION THEREFORE;
THE SYSTEM IS CONSISTENT AND INDEPENDENT.
IF WE WANT TO WE CAN GIVE THE SOLUTION
AS AN ORDERED PAIR WHERE THE X COORDINATE WOULD BE (-2)
AND THE Y COORDINATE WOULD BE (-3),
WHICH MEANS IF WE WANT TO CHECK THIS SYSTEM GRAPHICALLY
WE CAN GRAPH THESE TWO LINEAR EQUATIONS
AND THESE TWO LINES SHOULD INTERSECT AT THE POINT (-2, -3).
LET'S GO AHEAD AND CHECK THAT.
HERE'S THE GRAPH OF THE TWO LINEAR EQUATIONS.
HERE'S THE POINT OF INTERSECTION
AND NOTICE HOW THE X COORDINATE IS (-2) THE Y COORDINATE IS (-3)
VERIFYING OUR SOLUTION IS CORRECT.
I HOPE YOU FOUND THIS EXAMPLE HELPFUL.