WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In previous lessons, we talked about determinants; today, we are going to apply determinants
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in order to solve systems of equations and solve for variables.
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And we are going to do that using what is called Cramer's Rule.
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**Cramer's Rule** shows us a way to solve for systems of equations; and we can do that in two variables or three variables.
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Today, we are going to start out with two variables; and then we will go on in a minute and talk about systems of equations in three variables.
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Now, one thing to notice, before we even delve into this, is that this is broken down into this formula for x and another one for y.
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And what makes this rule helpful is that you may be in a situation where you just need to find one variable.
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You have two or three variables in your system, but you only need one variable.
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So, instead of solving the entire system, you can just hone in on the variable that you are looking for.
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OK, so looking at this: if we have a system of equations ax + by = e, and the second equation cx + dy = f; we can find x by using determinants.
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Let's first look at the determinant in the denominator; let's look at x, and I am going to call this determinant d.
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Looking at y, it is actually the same determinant; and this is comprised of the coefficients of x in this row and the coefficients of y in this row.
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So, whether you are looking for x or y, it doesn't matter; the denominator is going to be the same,
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and it is going to be a determinant that is comprised of, in this first column, coefficients of x, and then here coefficients of y in the second column.
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OK, now let's look at the numerator; and these are different; and I am going to call the numerator for x...
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d is the determinant for x, and here I am going to have d and the subscript y to indicate that it is looking for the y; OK.
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Now, if you look here, you will see that the second columns, where I had the y coefficients, are the same.
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But if I am looking for x, I am actually going to replace the column where I had the x coefficients,
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the coefficients of x, with the constants that are on the right side of the equation.
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And look over here at y: this determinant for y--I kept the x coefficients the same; they are in their place.
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But where I had the y coefficients, in that column, I replaced those with the constants.
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So, I just think of this first column as the x column, and the second column as the y column.
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And if I am looking for x, I replace the x column with the constants; if I am looking for y, I replace the y column with the constants.
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And then, I just need to solve: OK, so using an example...let's say I have a system of equations, 3x - 5y = 2 and -3x + 2y = -8.
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And from previous lessons, you know other ways to solve this.
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You know substitution; you could use elimination; but this is introducing another way that is useful in certain situations.
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So, looking at this: if I would like to find x, first I am going to find x.
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Now, this determinant in the denominator is going to consist of the coefficients of x in this first column,
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3 and -3, and the coefficients of y in the second column, which are -5 and 2.
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OK, so I have the denominator, and now I need to find the numerator.
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Recall that, in the numerator, if I am looking for x, I am going to replace the x column with constants; here, that is 2 and -8.
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The y column--I am just going to keep it as it is down there.
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OK, once I get this far, I just use my usual methods for finding a determinant.
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And remember that that is...if a determinant is a, b, c, d--a second-order determinant--I just need to find ad - bc; OK.
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So, I am going to do that: so x = 2(2) - -8(-5), over 3(2) - -3(-5).
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OK, therefore, x equals...that is 4 minus...-8 times -5 is 40; in the denominator, 3 times 2 is 6; minus -3 times -5...that gives me 15.
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This gives me x = (4 - 40), which is -36, over (6 - 15), which is -9; so my negatives cancel out, and -36/-9 is simply 4, so x equals 4.
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If I wanted to solve for y, I would already have the determinant in the denominator.
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So, if I wanted to solve for y, it would actually be even simpler, because I already know that the denominator is the same.
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So, I would just put 4 in here; then I would find the determinant that I need in the numerator;
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and I would look over here and say, "OK, my x column is the same as in the denominator; it is 3 and -3."
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y I would replace with the constants, 2 and -8.
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And then, I would just go ahead and solve, using ad - bc for the determinant, and divide by 4.
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Cramer's Rule can also be used to solve systems of equations in three variables.
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And although this looks pretty complicated, it is actually the same idea as we just covered.
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Looking in the denominator: this first column is the coefficients of x, just like we talked about with systems of two equations.
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This second column is coefficients of y, and the third is the coefficients of z.
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And that is exactly the same, whether you are looking for the variable x, y, or z.
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In the numerator, here you see that this is the same as below--coefficients of x.
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Here, this is what I think of as my x column; I think of this as my y column; and then here, my z column--
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I am going to replace that with the constants, since I am looking for z.
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If I am looking for y, I am going to replace this second column with the constants, j, k, and l.
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If I am looking for x, in the numerator, I replace that x column with the constants.
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So, the denominator is always the same; in the numerator, these two columns are the same,
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but for x this one is different; for y this one is different; and for z this one is different.
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OK, for example, solving this system of equations using Cramer's Rule: this is in the form ax + by = some constant (I am going to call it e).
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Here, I am going to call this cx + dy = f; and then, just to remind you that, if I am looking for x or y, what I am going to do...
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In the denominator, I am going to have a determinant that consists of ac (coefficients of x)
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in the first columns, and bd (coefficients of y) in the second column.
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And that is going to be the same for both.
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In the numerator, I am going to replace this first column with the constants; the second column will remain the same (coefficients of y).
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Here, the x column still will have the coefficients of x, but the second column, where the y coefficients were--I am going to replace that with the constants, ef.
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Therefore, if I am solving for x, let's start out with the denominator.
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The denominator is going to be...the coefficient of x is 1; the other coefficient of x is 3.
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The coefficient of y is -2, and the other is 1.
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In the numerator, I am going to keep the y section the same; I am going to replace x with the constants 1 and 2, the x coefficients.
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Now, I am just going to recall that, in order to find the second-order determinant here, it is going to be...this is a, b, c, d; it is going to be ad - bc.
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Therefore, x equals...this is going to be 1 times 1, minus -2 times 2, over 1 times 1, minus -2 times 3.
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Therefore, x equals 1 minus...-2 times 2 is -4, over 1 minus...-2 times 3 is -6; x equals 1...a negative and a negative gives me a positive;
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the same here--a negative and a negative gives me a positive; so x equals 5/7.
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Solving for y is actually going to be easier, because recall that these determinants are the same: the denominator is the same for x and y.
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Well, I already did all this work, finding the denominator, which is 7.
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So, I am not going to repeat that; I am just going to get rid of this and write 7 here.
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In the numerator, in this first column, I am going to have my coefficients of x, 1 and 3, just like I did in the denominator.
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However, in this second column, I am going to replace it with the constants 1 and 2.
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Again, I am using my formula ad - bc to find this determinant; that is going to give me 1(2) - 3(1), all over 7.
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This is y = (2 - 3)/7, so y = -1/7.
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So, x = 5/7; y = -1/7; and I was able to figure that out using Cramer's rule, where the denominator is the determinant
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formed by the coefficients of x and the coefficients of y, and the numerator is...
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the first column consists of the constants; the second is the coefficients of y.
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Once I did that, y was easier to find, because I already had the denominator.
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I just plugged that in right here; for the numerator, I had a determinant consisting of the coefficients of x, and then (in the second column) the constants.
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OK, again, we are solving a system of equations with two variables for x.
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OK, recall that in the denominator, this first column is going to be coefficients of x; the second is going to be the coefficients of y.
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In the numerator, I am going to leave the y's alone; but I am going to replace x with 8 and 4, the constants.
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Now, I am going to use my rule, ad - bc, assuming that this is the set-up, to find these determinants.
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x equals a times d, minus b times c; therefore, x equals...8 times 3 is 24, minus -16, or x = 24...a negative and a negative...that is + 16.
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So, 24 and 16 is 40; x equals 40 in the numerator, over this denominator.
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Let's find the denominator; the denominator determinant is 3 times 3, so that is the numerator for x.
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Now, to find the denominator (this is just the numerator), I have 3 times 3, minus -4 times 2.
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3 times 3 is 9, and in the denominator, I have -4 times 2, so that is -8; and then, this is going give me 9.
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A negative and a negative is a positive; 9 plus 8 equals 17.
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This is my numerator right here, and I have my x denominator right here; and this is going to give me x = 40/17.
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I found my determinants; this consists of the x coefficients and the y coefficients; this one is the constants and the y coefficients.
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And then, I found the numerator right here; I found the denominator right here.
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And then, I went ahead put those together, and then x equals 40/17.
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Now, I am finding y: finding y is going to be easier, because here I have my denominator,
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which is 17--the same denominator, so that takes a lot of work out of it.
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In the numerator, in that first column, go the x coefficients.
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I replace the second column with the constants; and then, I go ahead and find this determinant,
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which is going to be 3 times 4, minus 8 times 2.
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This is going to give me 12 minus 16 over 17, and that is going to be...12 - 16 is -4, over 17.
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So, here I have...x is 40/17 y is -4/17--using Cramer's Rule.
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Now, looking at a system of equations with three variables, I am being asked only to solve for x.
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And this shows you how this rule can be useful, because I only have to focus in on that, and not solve the entire system.
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You use the same general approach as with three variables.
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So, I am going to take x; and in the denominator, I am going to say, "OK, the first column is the x coefficients."
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The coefficient here is 1; 2; 1; y column: -2, 1...there is no y here, so the coefficient is just 0.
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z--there is no z at this first one, so that is going to be 0, 1, 1.
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All right, now I am looking for x; the first column represents the x coefficients--I am going to hold on to that thought for a second.
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And I am just going to copy over the other two columns, because I am not messing with those.
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I am going to replace this first column with the constants, 3, 1, 3.
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Now that I have set this up, I have to find a third-order determinant in the numerator and the denominator.
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And you recall that we went over a couple of methods to do that: expansion of the minors and the diagonal method.
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And now, I am going to use the diagonal method.
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The first step of the diagonal method is to copy over the first two columns--so, up here, 3, 1, 3, -2, 1, 0.
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Here, it is 1, 2, 1, -2, 1, 0.
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Once I have done that, then I need to start at the upper left and go along the diagonals.
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And I am going to find, actually, right along this way...the product of that and add
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the product of this diagonal to the product of the next diagonal to the product of the next diagonal.
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OK, so x equals...the first product...this is 3 times 1 times 1, plus -2 times 1 times 3, plus 0 times 1 times 0.
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OK, I have taken care of those diagonals; now I start at the lower left and go along those diagonals.
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And I am going to subtract those products: so minus 3 times 1 times 0.
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Next, 0, 1, 3--minus 0 times 1 times 3--minus the last diagonal: and that is 1, times 1, times -2.
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OK, that takes care of the numerator; in the denominator, I am going to start at the upper left.
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My first diagonal is 1 times 1 times 1; and I am going to add that to the next diagonal, -2 times 1 times 1.
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I am adding that to this diagonal, 0 times 2 times 0.
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Now, subtracting: we are starting down here: 1 times 1 times 0; we are subtracting 0 times 1 times 1,
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and then finally subtracting 1 times 2 times -2.
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I'm running out of room; OK.
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1 times 2 times -2; OK, I am working this out...
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Therefore, x equals 3, and this is -2 times 1 times 3, which gives me -6, plus 0, minus 0, minus 0.
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1 times 1 times -2...so that is minus -2.
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In the denominator: the 1's multiply and give me 1; plus -2 times 1 is -2, times 1...-2 times 1 times 1 is -2;
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plus...this is going to come out to 0; minus 1 times 1 times 0; minus 0 times 1 times 1 (that is another 0); minus 1 times 2 times -2.
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OK, let's clean this up: this is 3 minus 6; get rid of all these 0's; and this is minus -2, and that is the same as plus 2.
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In the denominator, I have 1 minus 2; get rid of all the 0's; and this is minus -2; actually, that is 2 times -2--this should be -4; minus -4 gives me plus 4.
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Now, x equals...well, 3 minus 6 plus 2 is going to give me -1;
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in the denominator, I have 1 minus 2 (that is -1), plus 4: that is going to give me 3.
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So, x is -1/3; and as you can see, once you get it set up, then it is a matter of just keeping track of all of that multiplication and these signs.
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So, I set this up by forming determinants with coefficients of x, y, and z in the denominator.
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In the numerator, there are coefficients of y and z, and then the constants in the first row.
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Then, I solve this by using the diagonal method; and I did that right here to come up with x = -1/3.
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OK, again, we are asked to solve a system of three equations.
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I am going to start out by looking for x, and looking at the denominator.
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For the denominator, I am going to use the coefficients of x: 2, -6, and 4; the coefficients of y: -3, 9, -1; and the coefficients of z: 1, -3, 1.
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OK, in the numerator, I am looking for x, so I am going to keep the y and z columns4 just as they were.
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And I am going to go over here to the x area, and I am going to substitute the constants 4, 11, and 10.
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Now, I am going to, again, use the diagonal method; and I am actually going to start with the denominator.
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And you will see why in a minute--why you should always start with the denominator, and how it can save you work.
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So, I need to find this third-order determinant; and I am going to use the diagonal method.
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I am going to rewrite these first two columns: 2, -6, 4, -3, 9, -1.
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After I have done that, I am just going ahead and rewriting this; I am not working with the numerator yet.
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I am just going to leave that like this.
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And I am only working at finding this determinant in the denominator.
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Starting at the upper left, using the diagonal method, I am going to find the products of this first diagonal: that is 2 times 9 times 1,
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plus this next number right here--next set of numbers: -3, -3, 4--that product, plus, coming down right here, 1, -6, -1.
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Then, from that, I am going to start subtracting this other set of diagonals, starting down here, and going up:
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the product of 4 times 9 times 1; continuing on: minus -1, -3, and 2, minus this last diagonal right here: 1, -6, -3.
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So, x equals the same determinant; let's just work with the denominator right now.
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2 times 9 times 1 is 18; plus -3 times -3 (is 9) times 4--that is 36; -6 and 1--that is 6.
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OK, and now subtracting: 4 times 9 times 1--that is 36; minus...-1 and -3 is 3, times 2 is 6, minus 1 times -6 is -6, times -3 is 18.
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Now, you can see what is happening here: x equals this determinant in the numerator.
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But I have 18 and 18, so those cancel; that gives me 0.
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I have 36 and 36, so those cancel: 0; 6 and 6--those cancel; so I have a 0 in the denominator, and that is not allowed.
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Since that is not defined, what I have ended up with is a situation where Cramer's Rule does not work.
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It doesn't mean that there is not necessarily a solution to this system of equations, but what it means that there is no unique solution.
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It may be that there is an infinite number of solutions (it is a dependent system).
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It may mean that there is no solution.
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But I can't go any farther, and I can't use Cramer's Rule in this situation.
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And that is why the first thing you should do, if you are using Cramer's Rule, is to check out the denominator and make sure that it is not 0.
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If the determinant in the denominator, d, is 0, you can't use this method.
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And that concludes this session of Educator.com on Cramer's Rule; I will see you next lesson.