Ex - Solve a system of three equations using cramer's rule

- WELCOME TO AN EXAMPLE ON HOW TO SOLVE A SYSTEM OF THREE LINEAR EQUATIONS WITH THREE UNKNOWNS USING CRAMER'S RULE. CRAMER'S RULE PROVIDES A METHOD FOR SOLVING A SYSTEM OF EQUATIONS USING DETERMINANTS. SO BEFORE APPLYING CRAMER'S RULE YOU WANT TO MAKE SURE THAT OUR EQUATIONS ARE IN STANDARD FORM OR THIS FORM HERE. IF THE EQUATIONS ARE IN THIS FORM THEN THE SOLUTION OR THE VALUE OF X, (Y) AND (Z) THAT SATISFY THE SYSTEM ARE GIVEN BY THESE QUOTIENTS OF 3 BY 3 DETERMINANTS. - NOTICE THAT THE "As" ARE THE COEFFICIENTS OF (X). THE "Bs" ARE THE COEFFICIENTS OF (Y). THE "Cs" ARE THE COEFFICIENTS OF (Z) AND THE "Ds"ARE THE CONSTANTS. SO IF YOU LOOK AT THE FRACTIONS FOR (X,Y,Z) NOTICE THAT THE THREE DETERMINANTS IN THE DENOMINATOR ARE ALL THE SAME. WHERE THE FIRST COLUMN CONTAINS THE COEFFICIENTS OF (X). THE SECOND COLUMN CONTAINS THE COEFFICIENTS OF (Y) AND THE THIRD COLUMN CONTAINS THE COEFFICIENTS OF (Z). AND NOW THEY HELP US REMEMBER OUR NUMERATORS WILL COMPARE THEM TO OUR DENOMINATORS. SO LOOKING AT THE VALUE OF (X) NOTICE HOW THE FIRST COLUMN WHICH IN THE DENOMINATOR CONTAINS THE COEFFICIENTS OF (X) ARE REPLACED WITH THE CONSTANTS BUT THE SECOND AND THIRD COLUMNS STAY THE SAME. FOR (Y) NOTICE HOW THE SECOND COLUMN WHICH IN THE DENOMINATOR CONTAINS THE COEFFICIENTS OF (Y) IS REPLACED WITH THE CONSTANTS BUT THE FIRST AND THIRD COLUMN STAY THE SAME. AND THEN FOR (Z), NOTICE, IN THE DENOMINATOR THE THIRD COLUMN CONTAINS THE COEFFICIENTS OF (Z) BUT IN THE NUMERATOR THE THIRD COLUMN CONTAINS THE CONSTANTS BUT THE FIRST AND SECOND COLUMNS STAY THE SAME. SO LET'S GO AHEAD AND APPLY THIS FORMULA TO SOLVE OUR SYSTEM. NOTICE, FOR OUR SYSTEM THE "As" WILL BE (2, 4, 6), THE "Bs" WILL BE (-1,1,-2), THE "Cs" WILL BE (-1,-1,+2) AND THE "Ds" WILL BE (2,-5,17). LET'S GO AHEAD AND COMPLETE THE 3 BY 3 DETERMINANTS FOR (X,Y,Z) IN THE DENOMINATOR BECAUSE THEY'RE ALL THE SAME. THE FIRST COLUMN WILL BE (2,4,6). THE SECOND COLUMN WILL BE (-1,1,-2). - AND THE THIRD COLUMN WILL BE (-1,-1,2). - AND NOW FOR THE 3 BY 3 DETERMINANT IN THE NUMERATOR OF (X) WE'RE GOING TO REPLACE THE FIRST COLUMN WITH THE CONSTANTS. SO WE'LL HAVE (2,-5,17) AND THE SECOND AND THIRD COLUMNS WILL BE THE SAME AS THE DENOMINATOR. - THE NUMERATOR OF Y, THE SECOND COLUMN IS REPLACED WITH THE CONSTANTS. SO WE'LL HAVE (2,-5,17) HERE BUT THE FIRST AND THIRD COLUMNS MATCH THE FIRST AND THIRD COLUMNS OF THE DENOMINATOR. SO WE'LL HAVE (2,4,6) AND (-1,-1,2). AND THEN FOR THE NUMERATOR OF (Z) THE THIRD COLUMN I REPLACED WITH THE CONSTANTS. AND THE FIRST AND SECOND COLUMNS ARE THE SAME AS THE DENOMINATOR. SO WE'LL HAVE (2,4,6) AND (-1,1,-2). NOW, WE'RE NOT GOING TO EVALUATE THESE 3 BY 3 DETERMINANTS BY HAND IN THIS VIDEO BUT I WILL PUT A LINK ON THE SCREEN THAT WILL SHOW YOU HOW TO DO THIS BY HAND USING THE COFACTOR AND DIAGONAL METHODS. IN THIS VIDEO WE'LL GO AHEAD AND USE THE CALCULATOR TO FIND THESE DETERMINANTS. TO DO THIS WE'RE GOING TO USE THE ELEMENTS IN THIS DETERMINANT TO FORM MATRIX "A". WE'LL USE THESE ELEMENTS TO FORM MATRIX "B". THESE ELEMENTS TO FORM MATRIX "C". REMEMBER, THE DENOMINATORS ARE ALL THE SAME. SO WE'LL USE THESE ELEMENTS TO FORM MATRIX "D". SPEAKING OF THESE AS MATRIXES AND THEN CALCULATE THESE DETERMINANTS. AND, AGAIN, TO SAVE TIME I'VE ALREADY ENTERED MATRIXES A, B AND C. BUT TO REVIEW WE'LL GO AHEAD AND SHOW HOW TO ENTER MATRIX "D". SO FROM THE HOME SCREEN I'M GOING TO PRESS SECOND MATRIX OR SECOND (X) TO THE -1. I'M GOING TO -- TWICE TO EDIT. WE'RE GOING TO USE THESE ELEMENTS HERE TO FORM MATRIX "D". SO I'M GOING TO SELECT OPTION 4 FOR MATRIX "D", TYPE IN THE DIMENSIONS, 3 ENTER, 3 ENTER AND THEN TYPE IN THE ELEMENTS ROW BY ROW. SO WE'LL HAVE 2, ENTER, -1, ENTER, -1, ENTER AND SO ON. - IT'S IMPORTANT TO DOUBLE CHECK HERE BECAUSE IF WE ENTER ONE ELEMENT INCORRECTLY IT WILL GIVE US THE WRONG ANSWER. SO THIS LOOKS GOOD. SO LET'S GO AHEAD AND CALCULATE OUR DETERMINANTS. AND WE'LL START WITH FINDING THE DETERMINANT IN THE DENOMINATORS. SO WE WANT TO CALCULATE THE DETERMINANT OF MATRIX "D". SO WE'LL PRESS SECOND MATRIX, RIGHT ARROW ONCE TO MATH, SELECT 1 FOR DETERMINANT. NOW WE HAVE TO GO BACK AND SELECT MATRIX "D". SO PRESS SECOND MATRIX AGAIN AND THEN SELECT OPTION 4 FOR MATRIX "D". (X,Y,Z) HAVE A DENOMINATOR OF 28. - AND NOW WE NEED TO FIND THE DETERMINANT OF MATRIX "A", "B" AND "C" TO FIND OUR NUMERATORS. SO SECOND MATRIX, RIGHT ARROW ONCE FOR MATH, SELECT ENTER FOR DETERMINANT AND THEN WE WANT TO SELECT MATRIX "A". SO I'LL PRESS SECOND MATRIX. A IS THE FIRST OPTION SO I'LL PRESS ENTER. CLOSE PARENTHESIS, ENTER. SO THIS DETERMINANT HERE IS EQUAL TO 14. SO (X) IS EQUAL TO 14/28 OR ONE HALF. NOW, WE'LL FIND THIS DETERMINANT. SO THAT WILL BE THE DETERMINANT OF MATRIX "B". SO THE SECOND MATRIX, RIGHT ARROW FOR MATH, ENTER FOR DETERMINANT, SECOND MATRIX OPTION 2 FOR MATRIX "B", CLOSE PARENTHESIS, ENTER, -112. -112 DIVIDED BY 28 IS EQUAL TO -4 SO Y = -4. AND NOW WE STILL HAVE TO FIND THE VALUE OF THIS DETERMINANT HERE WHICH ON THE CALCULATOR WILL BE DETERMINANT OF MATRIX "C". SO SECOND MATRIX, RIGHT ARROW, ENTER, SECOND MATRIX OPTION 3, CLOSE PARENTHESIS ENTER. WE HAVE 84. SO THIS SIMPLIFIES NICELY AS WELL. 84 DIVIDED BY 28 IS EQUAL TO POSITIVE 3. SO HERE IS THE SOLUTION TO OUR SYSTEM OF EQUATIONS. SINCE WE HAVE ONE SOLUTION WE CAN SAY THE SYSTEM IS CONSISTENT AND INDEPENDENT. WE CAN ALSO GIVE THE SOLUTION AS AN ORDER TRIPLE WITHER THE (X) QUOTIENT IS 1/2, THE Y QUOTIENT IS -4 AND THE Z QUOTIENT IS +3. - I HOPE YOU FOUND THIS EXPLANATION HELPFUL.