Determinants and Cramer's Rule - Video Tutorials & Practice Problems

Determinants of 2×2 Matrices

Hey, everyone, welcome back. So throughout our discussion on matrices, one of the things that you may be asked to calculate is something called the determinant of a matrix. Now the first time you learn determinants, it may be kind of scary because there's lots of letters and numbers flying around everywhere, but I'm gonna break it down for you. And I'm gonna show you that the determinant is really just a number that you calculate by doing some subtraction and multiplication. So I'm gonna show you that the determinant of this matrix over here is actually just the number two, I'm gonna break it down for you show exactly how we get that. Now, later on, we're gonna use this determinant to solve systems of equations. But for now we only just need to worry about how to calculate it. Let's go ahead and get started here. Now, we're gonna look at a two by two matrix because the formula depends on the type of matrix. And this is the simplest version, the way we calculate this is basically by subtracting the products of the diagonals. Now that sounds like a lot. But let me just break it down for you, right? So we've got this, determine this matrix here. A which is 325 and four. So we've got columns and rows, right? This is, this is like the rows and this is like the columns or whatever. But the diagonals are really just like the opposite corners of your matrix. So the diagonals here are just basically these two numbers over here and then these two numbers over here. So the way that you calculate this matrix and by the way, you're gonna see some different uh notation for this like debt a or sometimes even these little straight bars or whatever is you're gonna multiply these diagonals. You're gonna multiply these two numbers. You always do the blue ones first and then you go through these two numbers. All right. So you always go down to the right and then down to the left and you subtract those. All right. So really what happens is this just becomes a subtraction of two products. So I've got that the first two are numbers had multiplied are gonna be the three and the four. So this is gonna be three times four minus, you're gonna subtract and then two times five, you always do the blue ones first, the red one second. All right. So this is gonna be two times five. It doesn't matter which way which way you, which order you multiply these because the answer is the same. And so what, this ends up being is, this ends up being 12. So this is gonna be 12 minus and this ends up being 10. And that's why we got our answer of two. So the determinants of this matrix, which is really just a number is just the number two. All right. So that's really all there is to it. In fact, the general formula for doing this or the determinant of some two by two matrix is if you have these numbers ranged in like a grid, then you again, always just do these two numbers. So in other words, A times D minus B times C, that's the general formula for any determinant of A two by two matrix, right? So hopefully you guys remember that, that's really all there's to it. So let's go ahead and just get some more practice here. We're gonna take these uh examples and we're gonna evaluate the determinants of these matrices. So here we have matrix A which is given to us uh as 845 and zero. So the way we would write this is again, the determinants of A or you could also just use the square bracket notation where the straight lines, the determinant of this is we're gonna take a look at again, we're gonna subtract these two products over here. Remember these numbers will go first. So in other words, we're gonna do eight times zero and then we're gonna do minus four times five. All right. Now, one of the things you might notice here is that if you do eight times zero, that just ends up being zero. So it's just zero minus. And then this over here is just gonna be 20 right? In other words, the determinant of this matrix is just the number negative 20. All right. That's really all there is to it. Uh really nothing else to sort of read into in terms of what does this number mean? It's just a number and we'll use it later on. All right. So let's take a look at the second matrix we have B and here we have some negative numbers. So instead of using this, get a notation, you also may see this written like this, you may see this written as just these little straight lines would be and again, this is just gonna be the subtraction of two products. All right. So again, you're gonna go down to the right and then down to the left and you're going to do this, the multiplication in that order. All right. So this is gonna be negative three times negative two. And then this is gonna, you're gonna subtract negative seven times one. All right. So just be careful with the negative signs here. What we're gonna see here is that this first product ends up being negative three, negative two, which actually just works out to just positive six because the negative is canceled. So it's just positive six. And then we have minus and this is just gonna be negative seven, right? So this is gonna be negative seven. So here we have six minus negative seven. And so in other words, the determinant of B is just gonna be the number 13. All right. So that's your answer. That's all, really all there to it. Folks. Let me know if you have any questions. Thanks for watching.

Evaluate the determinant of the matrix.

• A. $\frac{107}{3}$
• B. $\frac{227}{6}$
• C. 33
• D. 12

Cramer's Rule - 2 Equations with 2 Unknowns

Hey, everyone, welcome back. So in a previous video, we learned how to calculate the determinant of a two by two matrix. And I mentioned that we would eventually use these to solve a system of equations. Well, that's exactly what we're gonna do in this video. In this video, we're gonna see how to solve a system of two equations with two unknowns using something called Kramer's rule. Now the first time you learn this, it might be kind of intimidating. There's lots of numbers flying around everywhere, but I'm gonna break it down for you and show you that it's really just a formula. It's a formula that directly gives you the solution to a system of equations. We're gonna pop a bunch of numbers in calculates and terminates and it'll just spit out our numbers for X and Y are numbers that make these equations true. That's really all there is. Let me go ahead and break it down for you. Now, before I start, actually, I kind of want to make it a little bit of an analogy. I like to think of a Kramer's rule almost as like the quadratic formula. We learn lots of different ways to solve quadratics. But ultimately, you could just plug a bunch of numbers into a formula and it'll just give you the answer. This is kind of the same way. We've learned lots of different ways to solve a system of equations. But ultimately, you could just take all these numbers over here, pop them into these equations and it'll just give you your answers. Now. It's tedious and it requires a little bit of work, but it'll always work, right. That's basically uh kind of the analogy there. All right. So let's go ahead and just get started here. Basically, the whole idea is that if we're given a system of two equations with two unknowns, then Kramer's rule is gonna involve calculating determinants of two by two matrices. And we've seen how to do this before. Let's go ahead and get started with this problem. So we've got two X plus Y equals five, negative four X plus six Y equals negative two. We want to solve this by using Kramer's rule. We're gonna have to calculate once we determine it. So we're gonna have to turn this into a matrix, right? That's what I'm gonna do over here when I calculate or determine my augmented matrix. And I'm gonna color code these things, right? So remember these are gonna be my X coefficients, then I've got my Y coefficients. Uh I'm sorry, then I've got my Y coefficients and then these are gonna be my constants over here, right? So these are gonna be my constants. OK? So here, what I've got is I've got the two and the negative four and then here I've got the one and the six and then remember there's a little black bar and then here I've got the five and the negative two. All right. So what Kramer's rule says is that to calculate X, you're gonna have to calculate the quotient of two. The, you're gonna have to calculate this one and then divided by this one. All right. So we're gonna have to calculate two determinants over here and then divide them the same thing by the way applies for the Y. So it's two determinants and then divide them. In fact, actually, the determinants of X and Y on the bottom are actually the same thing. So really only have to calculate three determinants. Let's take a look at the first one here. The first one is actually, or sorry, the bottoms are actually the easiest ones to calculate because they're really just the A and B terms that you have, which are really kind of just like the X and Y coefficients that you had over here. So you just take these numbers and you just copy them over, right? So this is gonna be two negative 42, negative four and then this is gonna be um one and six. So 1616. All right. So you just straight up copy those and we'll have to calculate the determinants of those. Now, on the tops is where it gets a little bit sort of more uh interesting. Now, what happens is when you're solving for the X, what you're gonna do is you're gonna take the constants that are on the right side of the equation in your matrix and you're gonna have to replace the X coefficients, right. So you're gonna basically just gonna take that column and swap them with the X and then you're gonna write that new matrix. So here what happens is I'm not gonna write two and negative four, I'm gonna write, this is gonna be five and negative two that takes the place of the X column. Now the Y columns are actually just gonna be the same, this is gonna be one and six, right? So that's what it is for X for the Y. What you're gonna see is that we're gonna do the same thing except now we're gonna replace the Y coe. All right. So now here what happens is the two and the four, we copy that over. But now instead of the one and the six, now we replace it with the five and the negative two. So five negative two, right? So that's really all there is to it. So you kind of just are replacing uh that column of those coefficients with the column of the constants on the right side. All right. And now we just have to go ahead and calculate these three determinants over here. So let's go ahead and get started. So, uh over here for X, what I've got is I've got um remember how, how to, how to calculate these, these determinants. Just we need a little bit of a refresher. You always just go down to the right and then down to the left and you subtract those products. All right. So for the first one of what you're gonna do here and this is gonna be five times six minus one times negative two. So this is gonna be five times six minus one. This is gonna be one times negative two and on the bottom, which you're gonna see is you're gonna have two times six and then one times negative four. So it's gonna be two times six minus one times negative four. All right. And on, on the y what you're gonna see is that on the bottom, you're actually gonna have the exact same thing two times six minus uh one times negative four. It's the same exact determinants on the top though. What you're gonna see is this ends up being two times negative two and then five times negative four. So this is gonna be two times negative two minus five times negative four. All right. So just again, be careful here, there's lots of ways that you can get tripped up with negative science. So just sort of uh so just be slow and methodical careful with your science. All right. OK. So for the X, what we're gonna see here is that this ends up being five times six, this is gonna be 30 minus one times negative two. So this is gonna be 30 minus negative two. So this is gonna be uh plus two. So you could basically just have done plus two. All right. And then on the bottom here, what we're gonna get is 12 minus again, negative four. So this is gonna be 12 plus four, which is gonna be 16. All right. So in other words, this really just becomes 32/16 and this X value is equal to two. All right. So that's basically what the X value turns out to be now for the Y which you'll see is that we already calculated this to be 16. So we don't have to recalculate it. What is the top end of being? So this is two times negative two. So that is negative four minus five times negative four. So this is negative two minus and this is gonna be negative 20. All right. So this is a little bit uh sorry, this is gonna be negative four. I'm sorry, let's be negative four. And then basically this is gonna be plus 20. All right. So what this actually ends up being when you work all this out, this is actually ends up being 16/16, which is just one. So in other words, the answer to this solution or sorry, the solution to the system of equations is X equals two and Y equals one. Now you can pause the video and just check this yourself. You can plug those numbers back into this equation and find that you'll actually get true statements for both. And again, there's probably a faster way that we could have done this. But again, some questions may ask you to use Kramer's rule. Again, it's just another tool in your tool belt. But that's how to solve these kinds of problems using Kramer's rule. Let me know if you have any questions.

Write each equation in standard form and use Cramer's Rule to solve the system.

$y=-3x+4$

$-2x=7y-9$

• A. $x=1,y=1$
• B. $x=-1,y=1$
• C. $x=1,y=-1$
• D. $x=-1,y=-1$

Write each equation in standard form and use Cramer's Rule to solve the system.

$y-9x=-3$

$-3x=4y-1$

• A. $y=3,x=0$
• B. $x=0,y=3$
• C. $x=-\frac13,y=1$
• D. $x=\frac13,y=0$

Determinants of 3×3 Matrices

Hey, everyone, welcome back. So we've already seen how to calculate the determinant of a two by two matrix. And the way we did this was by multiplying the diagonals and subtracting. So we go down to the right and then down to the left. Now we're gonna take a look at how to calculate the determinant of a three by three matrix. And I wish I could tell you that it's just as simple as just going down to the right and down to the left. But the bad news is, it's a little bit more complicated. The good news however, is that it actually does involve calculating a bunch of two by two determinants, which we do know how to solve. So I'm gonna break it down for you and show you the equation for how to solve a three by three determinants. And then we'll take a look at an example together. All right, let's get started. So if I have a three by three matrix that's organized over here, I'm gonna organize all the letters, you know, into subscripts like a one B A 123, B, 123, C 123. So on and so forth. Really all it is to calculate a three by three determinants is you're gonna have to calculate three numbers. So these three numbers, and I wanna mention something here. The signs of these numbers will actually alternate. Notice how in this one, we have a plus sign, a minus sign and a plus sign. So the signs will flip. All right. And now what I wanna do is sort of give you a, like an understanding of what these, what this equation actually comes from. So if you have a three by three determinants, here's what's basically going on, right? I'm gonna write this the same exact matrix out three times and we'll see a pattern that starts happening with how these numbers, uh where these numbers really come from or these terms. So for the first time, what you're gonna do is you're gonna take the A in the first row and then you're gonna strike out all of the other entries or all the other numbers that are in that column and row. So you could strike out the A column and the rest of the first row. And what you're left with is you're left with a smaller two by two matrix over here with the B and C numbers. This thing over here is basically what this becomes, you're gonna take this a number and you're gonna multiply it by the smaller matrix that you've just come up with. And that's the first term, let's take a look at the second one. The second one is, you're gonna take the B number in the first row and again, you're gonna strike out everything that's part of that column in that row, strike out the rest of the B numbers and then the rest of the first row. So now we're gonna take this number and multiply it by the smaller matrix that's left over, which is just made up of the A and C numbers, right. So just draw a little box over here. That's what the second number, the second number ends up being. It's B one times the smaller matrix of the smaller determinants of the four numbers that you're left with. Now, for the last one, you might have guessed it's actually just gonna be the C number in the first row. Now you strike out everything that's part of that row. And what you're left with is you're left with this matrix over here. So for the third term, you're gonna take that C and M supply it by the determinants of the smaller matrix that you're left with. That's what that third term ends of being right now. I know it's kind of complicated with all these letters and numbers, but that's just an easy sort of vote way to visualize what's going on here. And hopefully you'll sort of remember this pattern. You take the first numbers in each one of the rows A B and C and you strike out everything else that's part of that column in row and multiply by the matrix that smaller two by two matrix that you're left with. Let's go ahead and actually do this example real quick here. All right, we can see how this works. So here we got these numbers 31002, negative three, negative 14 and six. I wanna calculate a determinants, right? So the determinant of this matrix, remember I'm gonna take three numbers over here. The first one is gonna be this three. So I'm gonna take this three and then uh what I'm gonna do is I'm gonna strike out all the numbers over here and here. And what I'm left with is I'm left with this smaller matrix. So that's the determinant or that's the smaller two by two matrix. I'm gonna multiply this thing by this is gonna be two, negative three, negative four and six. All right. That's the first number. So now what I'm gonna do is remember now the sign alternates. So I'm gonna have to put a minus sign here. And now what's the next number? Well, the next number here is gonna be the one. All right. So this is like the B number over here. Strike out everything in the, in that column and row. And what I'm left with is the smaller determinant over here. So this is gonna be 30 Sorry, this is gonna be one and then I'm gonna have the smaller determinants, which is gonna be zero, negative three and this is gonna be negative one and six. All right. And now, finally, so that's the second number. Now, finally, we just sort of undo all of that stuff. And now we're gonna take a look at this number over here. So this is gonna be, we're to flip the sign, it's gonna be plus. So we're gonna have zero and then we're gonna have the determinants of this matrix. Uh After we cross out all of these other numbers, so this is gonna be +02, negative one, negative four. Now, what's nice about this third term here is that we're gonna have zero times a determinant of which is gonna be some number. So zero times anything is just gonna be zero. All right. So that third term actually just completely goes away and that's good for us. This means that there's less calculating, but that's really all there is to it. We just reduced this three by three determinants into a bunch of multiplications and two by two determinants. Now, the rest you already know how to do. We really just have to calculate these smaller two by two matrices over here and we'll just go ahead and do that real quick. All right. So what this is gonna be is this is gonna end up being we're gonna expand this, this is gonna be three. And now remember we're gonna have a bracket here because now we're gonna need to evaluate these smaller determinants. So, remember you always go down to the right and then down to the left. So in other words, this is gonna be two times six minus negative, three times negative four. All right. So just keep track of the minus signs and that's what that first term will end up being. And we're gonna have minus and this is gonna be one times and then we're gonna have bracket again. This is gonna be zero times six, right? So there's gonna be zero times six minus negative three times negative one down to the right down to the left. All right. So now let's simplify one step further. So here, what we're gonna do is this is gonna be equals. So I've got three and then over here, what I've got is I've got 12, that's two times six minus and this is gonna be negative three times negative four, which is positive 12 as well. So in other words, 12 times 12. So what happens is this first term actually just goes away because this ends up just being three, this ends up being three times zero. So that whole first term actually just goes away as well. That whole thing just gets canceled out and now what we're left with is we're left with one and this is gonna be zero times six, which is just zero minus and then we have negative three times negative one. So this is just gonna be minus three. All right. So in other words, what this ends up being here is you end up just getting a grand total of zero minus uh one times negative three. All right. So there's a lot of negatives here. So it's minus one times negative three. So in other words, this ends up just being equals zero plus three. In which case, the determinant of this whole three by three matrix. Remember it's just a number, it's just gonna be the number three. All right. And the way we got here, it was we, we was, we reduced it to three smaller two by two matrices and calculated those. All right. That's really all there is to it, folks. Hopefully, this made sense. Thanks for watching and I'll see you in the next one.

Evaluate the determinant of the matrix.



• A. 165
• B. 9
• C. 63
• D. 25

Cramer's Rule - 3 Equations w/ 3 Unknowns

Hey, everyone, welcome back. So in an earlier video, we learned how to calculate a two by two determinant. And then we learned how to use Kramer's rule to solve a system of two equations with two and notes. Well, similarly, here we've learned how to calculate three by three determinants. And what I'm gonna show you in this video is how to solve a system of three equations with three knows using Kramer's rule. Now remember that Kramer's rule is really just a formula. It's a formula that directly gives you the solution to a system of equations. We're basically just gonna plug in a bunch of numbers into some formulas and it's gonna spit out the answers for XY and Z directly. All right. The only thing that's different here is that because we have three equations with three unknowns, our determinants will just be a little bit more complicated. Instead, we're gonna use determinants of three by three matrices. Let's just go ahead and get started here because there's actually not a whole lot that's different about Kramer's rule. Um And we'll go ahead and work out this out together. Let's get started. So we have the system of equations over here. Um And we wanna solve this by using Kramer's rule. Now, the first thing I'm gonna do is I'm gonna turn this into an augmented matrix just so I can get all my uh my coefficients and stuff in the right place. All right. So this is just becomes a matrix like this. Remember we're just gonna copy over the coefficients my X coefficients are gonna be negative five and I have zero over here, there's like a zero X and then I have one. Now for the Y column that's gonna be in green, I'm gonna have negative 13 and then one. And then for this C column over here, my Z coefficients, this is gonna be 46 and one. And then remember that my final um my final column over here with these purple numbers, these are be my constants are just gonna be other constants on the right side. 421 and six. All right. So what Kramer's rule says is that Xy and Z, I'm gonna have to take basically the division of two numbers, something called DX over DDY over DDZ over D. So what is this capital D capital D really is just the three by three matrix that you get just by looking at the coefficients of your Xy and Z variables? All right now because I'm gonna use that variable three times in my equations. That's always just gonna be a good place to start. Um Also what happens is that if you, if that capital D ends up being zero, then you'll actually have no solution. So you just wanna make sure that it's not zero. So let's go ahead and start there. What capital D is, is just the determinants of just this three by three matrix that you get over here, right? So it's really just gonna be the determinants of this thing. So remember how to calculate determinants just really quickly as a brief refresher. Uh what you have to do is you have to follow this formula over here where remember you're just gonna have to take these numbers cancel out the strike out the rows and columns and then focus on the smaller two by two matrices that you get. All right. So you can take a look there just in case you need a refresher. So I'm just gonna go ahead and set this up and I'm gonna, I'm gonna skip some steps here because we're a little bit sort of tight on space for this problem. Uh What we're gonna see here. And so this is gonna be negative five times the smaller determinant over here. So in other words, this is just gonna be the 31 and then this is gonna be the 61. Now we flip signs, this will be minus, this is gonna be negative one. So the negative one times this smaller determinant over here and we cancel out some stuff and we're gonna get 01 and 61 right now, I'm gonna flip signs again. This is gonna become now four and this is gonna become, now we're gonna use smaller determinants that we get over here. So in other words, this is just going to be 01 and then we have 31. All right. So that's what this ends up being. So if we calculate this, what we're gonna end up seeing is that we end up getting uh this is gonna be negative five, this is gonna be negative five and then we're gonna have three times one minus six times one. So three minus six will end up being negative three. So negative five times negative three, that's what this whole thing ends up becoming negative three over here. All right. So now let's see we have negative one. So we're gonna have minus, we're gonna have negative one times and then what is this mess end up becoming zero times? One is zero minus six times one, which is six. So this is gonna be, let's see, this is gonna be negative one times zero minus six. That's actually negative six. All right. That's what that ends up being. That's what this whole mess becomes. And then over here in this last term we're gonna get is we're gonna get zero times one. So this is gonna be four times. Let's see, zero times one is zero and then minus three times one which is three. So in other words, it's gonna be zero minus three which is negative three. All right. So what I end up getting here is that this is going to be um let's see. So this is going to be, this is gonna be negative 15. Actually this is gonna be positive 15. Uh This is gonna be now minus and this is gonna be six because the negative is cancel out plus 12. And if you work this out, we here, what you're gonna get is the capital D is equal to negative three. So in other words, what's gonna happen here is when I calculate XY and Z I'm gonna have to calculate some number and then just divided by now, we know this is just gonna be negative three. All right. So let's go ahead and do that. So this capital D here is negative three. So now that we're done with this, let's go ahead and now focus on X. So what X says is to calculate X, this is just gonna be something called DX divided by D. So I'm gonna have to calculate just some number here, some determinate number called DX and divided by this capital D. Some of the words I'm gonna have to take some number divided by negative three. What is this DX that we've been talking about? Well, really what it is is in a similar way that in rule, we replaced one of the constant columns for one of the X and Y coefficients. It's the same exact thing for DX. What this really means is that we're gonna take our normal matrix over here of A B and C. And if you're solving for the X components, you're gonna take the constants on the right side of your equation and you're going to replace them in the X column. So in other words, you no longer are dealing with the X coefficients. Now you just have your constants. So what DX is is it's gonna be a three by three determinants except instead of the negative 50 and one, I'm really just gonna plug in these numbers. This is gonna be 421 and six. So 421 and six and then the rest of the numbers are the same negative 131 and this is gonna be 46 and one. All right. Now what we have to do is just calculate the determinants of this three by three matrix. So let's go ahead and do that. So what this says here is that we're gonna, this is gonna be four. And then this is gonna be um this is just gonna be the smaller matrix over here. So I'm just gonna copy that over like that's minus, this is gonna be negative one and then this is gonna be uh 21 and six. So 21 6 and then this is gonna be six and one and then we have plus plus and we have four and this is gonna be um this, these smaller numbers over here. So this is gonna be 21 +36 and one. So that is gonna be your three by three determinants. Now, we just have to go ahead and do a bunch of calculations. All right. So this DX here will be, well, this is just gonna be four times three minus six. So this is gonna be uh four times negative three, it's four times negative three minus and this is gonna be, let's see, negative one. So this is gonna be oops. So what ha what happens is that we do 21 times one, that's 21 minus 36 which is negative 15 and then that's we multiply this by negative one, that's gonna be positive 15. So this is gonna be 15 over here. Now this is gonna be 21 times one which is 21 times a minus 18. So that's three and four times three is 12, right? So we're just do, we're just gonna sort of uh mental math our way through this. This is gonna be negative 12 minus 15 plus 12. All right. So this ends up being negative 12 minus 15 plus 12. And you can just double check that I've done this correctly. Work out the math yourself and you'll see that this is correct. So what happens is these, this 12 and negative 12 will cancel and you'll see that DX just leaves you with negative 15, right? So now we can plug this DX back into this formula over here and we'll see that's, this is just a number negative 15 over negative three, just equals positive five. And that is the first number that you have. So X is equal to five. Now, we just basically are gonna rinse and repeat. We're gonna do this exact same thing. We're gonna do it twice, uh You know, two more times for Y and Z. All right. So this is gonna be Y over here, right? So here, what I've got is Dy over D some of the words, some number divided by negative three. And when I calculate that, I'm just gonna get whatever the answer is, how do I figure out what Dy is? Well, basically, Dy is just gonna be now if I replace the constants into the Y column instead of the X. All right. So now the constants are gonna be in this middle column. And so let's go ahead and write out what this matrix is gonna be. So what this is gonna be is this is gonna be uh let's see. So I've got my negative 50 and one and then I've got in purple, I've got 421 and one, sorry and six. And then over here what I've got is 46 and one. All right. So let's calculate this determinant over here, this is gonna be negative five times and then we're gonna have these two smaller numbers over here that's gonna go in here. Then we switch signs, this is gonna be minus, then we're gonna have four and this is gonna be a smaller determinants in which we're gonna have 01, zero and one. And then we're gonna have six and one over here, six and one. And then we're gonna flip signs again and now we're gonna have four. And then in this smaller determinants, we're just gonna have these four numbers over here. So which you'll also notice here is that occasionally you're gonna have sort of the same sort of numbers that pop up. So you're gonna have, for example, 2166 and one, we actually already know what that determinant of that is that end up being 15 or negative 15. So you're gonna see the same sort of numbers pop up. So you can usually look up on up and down on your page and look for determinants that you've already calculated before just to make things a little bit easier. All right. So let's calculate this. So this is gonna be negative five and then this whole thing just becomes 21 minus uh 36 which is gonna be minus 15. All right. So what this ends up being here is this end being a four now with zero minus one, sorry, zero times one minus six is, is gonna be negative six and this is gonna be four and then zero times six is zero minus 21 times one, which is negative 21. All right. So that's really what that becomes. And if you work this out here with this into being, is this into being 75 plus 24 minus 84 right? And so if you work this out, which you'll get is a grand total over here of 15. So this actually Dy ends up being positive 15. All right. Now, if you plug this back into this formula, what you'll see is that 15 over negative three ends up being negative five. And that is your second answer. Now, we just have to do this one last time over here to calculate what Z is Z is gonna be DZ over D and this is gonna be some number divided by negative three. When you calculate this, you're just gonna get some number over here. Let's go ahead and work this out. So this is gonna be DZ is equal to and now we've got, we're just gonna replace now the Z column with the constants. So in other words, this is gonna be negative 50 and one. This is gonna be uh negative 13 and one. And then here we're gonna have the 421 and six. Now, if you work this out, which this determinant is gonna be, this is gonna be um negative five and then we're gonna do these two numbers. By the way, we've already, we've already calculated this determinant before and this is gonna be minus uh negative one. And then we're gonna have this smaller determinant over here, which is gonna be 01, 21 and six. So 21 6. And then finally, we swap signs one last time and this is gonna now be four and then we have this smaller determinant over here. This is gonna be like that. All right now, I'm actually just uh I'm actually just gonna go ahead and skip ahead to the answer here. You can go ahead and pause this and make sure that you've done it yourself. Once you actually work out all of this, what you're gonna get is you're gonna end up getting 15. This is gonna be negative 21 and this is gonna be negative 12. And so what you're gonna get here for this is you're gonna get negative 18. So when you plug this in negative 18, you're gonna get an answer of six. And so finally the answer to your solution is five, negative five and positive six. So this is the solution to your system of equations negative oh sorry five, negative five, negative five and then positive six. So that is the solution. Let me know if that makes sense. I know this is a little bit tedious. Thanks for sticking with me. Thanks for watching.

Solve the system of equations using Cramer's Rule.

$4x+2y+3z=6$

$x+y+z=3$

$5x+y+2z=5$

• A. $x=-2,y=-8,z=4$
• B. $x=1,y=4,z=-2$
• C. $x=2,y=8,z=-4$
• D. $x=-1,y=-4,z=2$