Solving a 3x3 System Using Cramer's Rule (Example)

here I'm going to solve this system with three equations and three unknowns and I'm going to do so using Cramer's rule so the first thing I'm going to do is to ensure that each of my equations is in standard form so X Y Z equals a number and in fact they are so the next thing I want to look at is to write my coefficient matrix and find its determinant so I'm gonna call that determinant D and I know that this determinant is going to keep coming back over and over again because each of my solutions for x y and z will be fractions and the denominator of each will actually be this determinant I'm about to find so my coefficient matrix is going to be a three by three because I have three equations and three unknowns so I'm gonna write all the X coefficients in the first column Y in the second column and Z in the third column so I'm just taking the straight from the equation and then one five and two for my y's and here my Z's right there so this is what this determinant is going to look like and in order to evaluate this determinant I'm going to copy down the first two columns over to the right of my determinant and I'm going to use the diagonal process here so these main diagonals will be multiplied and added so three times five times one will be fifteen and then one times four times three will be twelve I'm adding these and negative 1 times negative 2 times 2 will be a positive 4 okay so my minor diagonals here they'll be subtracted so 3 times 5 times negative 1 that's going to be a negative 15 minus 2 times 4 times 3 which is 24 minus 1 times negative 2 times 1 which is negative 2 so now it's just a matter of kind of simplifying this so 15 and 12 and 4 those will add up to make 31 and they're going to say plus 15 - 24 + 2 and when I do that I believe I'm gonna get 24 out of this from my determinant so this 24 is the determinant of my coefficient matrix and this is gonna keep coming back over and over again okay so in order to find my X's I have a nice equation here so my X's are going to be another determinant in the numerator and then this determinant in the denominator which I've already found so I'm just gonna go ahead and write that 24 so what happens here is I'm going to take this coefficient matrix but since I'm solving for x instead of using the X column I'm going to use my constant values over here on the right so I'm gonna replace my X column with my constants and then the rest of my coefficient matrix stays the same okay and it's the same thing when I do this I'm going to copy down these first two columns okay and I'm gonna work this determinant by diagonals so this is 0 times 5 times 1 just zeroes out 1 times 4 times 1 will be 4 and then negative 1 negative 1 & 2 will be a 2 so positive 2 I'm going to subtract my minor ones here so subtract a negative 5 that one will zero out and I'm gonna subtract a negative 1 right there okay so what I get for this is going to be 6 plus 5 plus 1 equals 12 for my numerator so my x value then becomes 12 over 24 which reduces to 1/2 so my x value is going to be 1/2 so I'm gonna go ahead and bring down my equation here I wrote them in a separate piece so I kind of drag it around so I'm gonna look for my Y value next and my y-value again will be a determinant in the numerator with this 24 in the denominator so the numerator is going to be my coefficient matrix but instead of using my Y column since I'm solving for y I'm going to replace my Y's with this constant column so my 3 negative 2 & 3 my Y's will be replaced with these values and then my X x YZ will just come on down right here so once again I'm gonna copy down this first column first two columns there and so now I'm going to use my diagonals to evaluate the determinant of the numerator so 3 times negative 1 times 1 will be negative 3 plus and that guy 0 is out and this guy is gonna be a positive 2 so minus looks like I'm gonna subtract a positive 3 minus 12 minus well 0 so that just zero is that right there so this will be a negative 3 plus 2 minus 3 minus 12 so that value right there is gonna end up being with a negative 1 negative 4 negative 16 so my Y value then is gonna be a negative 16 over that 24 which will be negative 2/3 okay let's continue on for my Z go ahead and drag my system down with us so for my value of Z I'm gonna have 20 4 in the denominator again my numerator is going to be my coefficient matrix copied but again this time since I'm solving for Z instead of my Z column I'm gonna put my constants here so my X's come down my Y's come down and my Z's are now going to be replaced with these constants okay to now evaluate this determinant I'm gonna go ahead and copy down the first two rows here and then work my diagonals again okay so here three and five and one will be 15 plus looks like a negative three plus and that guy zero is out so - looks like that guy's zero is out so - here - and now be negative six and - what looks to Libby and negative two so we've got twelve plus six plus two and that should be 24 the numerator of my Z value so when I move over here should be 20 over that denominator of 24 which will be 5/6 for my value of Z so now I can go ahead since I have all three values XY and Z and I'm gonna go ahead and write these as an ordered triplet all right X comes first then my Y and then my Z so here is my solution for this original system of equations and we solve this using Cramer's rule [Music]

Channel: James Elliott

Views: 178,728

Keywords: cramer's rule, matrices, matrix, determinants, ratios, linear system, three variables, ordered triplet

Id: X5rDjbp0t6s

Length: 8min 2sec (482 seconds)

Published: Mon Dec 10 2018