Determinants and Cramer's Rule - Online Tutor, Practice Problems & Exam Prep

Understanding determinants is crucial for solving systems of equations. For a 2 by 2 matrix, the determinant is calculated using the formula: $d^2=ad-bc$. Cramer's rule extends this concept to solve systems of equations, utilizing determinants of larger matrices. For 3 by 3 matrices, determinants involve calculating smaller 2 by 2 determinants, allowing for systematic solutions to equations with three variables.

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 are lots of letters and numbers flying around everywhere, but I'm going to break it down for you and I'm going to show you that the determinant is really just a number that you calculate by doing some subtraction and multiplication. So I'm going to show you that the determinant of this matrix over here is actually just the number 2. I'm going to break it down for you and show you exactly how we get that. Now later on, we're going to 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.

We're going to look at a 2 by 2 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. So we've got this matrix here, a, which is 3, 2, 5, and 4. So we've got columns and rows. 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 going to see some different notation for this, like det A or sometimes even these little straight bars, is you're going to multiply these diagonals. You're going to multiply these two numbers. You always do the blue ones first, and then you go do these two numbers. Alright? So you always go down to the right and then down to the left, and you subtract those. So really what happens is this just becomes a subtraction of 2 products. So I've got that's the first two, our numbers that multiplied, are going to be the 3 and the 4. So this is going to be 3×4−2×5. It doesn't matter which way you order you multiply these because the answer is the same. And so what this ends up being is 12 minus 10. And that's why we got our answer of 2. So the determinant of this matrix, which is really just a number, is just the number 2.

In fact, the general formula for doing this, for the determinant of some 2 by 2 matrix, is if you have these numbers arranged in a grid, then you again always just do these two numbers. So in other words, a×d−b×c. That's the general formula for any determinant of a 2 by 2 matrix. So hopefully, you guys remember that. That's really all there is to it. So let's go ahead and just get some more practice here. We're going to take these examples, and we're going to evaluate the determinants of these matrices. So here we have matrix A, which is given to us as 8, 4, 5, and 0. So the way we would write this is, again, the determinant of A, or you could also just use the square bracket notation where there are straight lines. The determinant of this is we're going to take a look at again, we're going to subtract these two products over here. Remember, these numbers will go first. So in other words, we're going to do 8×0−4×5. Now one of the things you might notice here is that if you do 8 times 0, that just ends up being 0. So this is just 0 minus, and then this over here is just going to be 20. So in other words, the determinant of this matrix is just the number negative 20.

Let's take a look at the second matrix. We have B, and here we have some negative numbers. So instead of using the det A notation, you also may see this written like this. You may see this written as just these little straight lines with B. And, again, this is just going to be the subtraction of 2 products. So, again, you're going to go down to the right and then down to the left, and you're going to do the multiplication in that order. So this is going to be −3×−2−−7×1. So just be careful with the negative signs here. What we're going to see here is that this first product ends up being negative 3 times negative 2, which actually just works out to just positive 6 because the negatives cancel. So this is just positive 6. And then we have minus, and this is just going to be negative 7. So here we have 6 minus negative 7. And so in other words, the determinant of B is just going to be the number 13.

That's your answer. That's all there is 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 2 by 2 matrix, and I mentioned that we would eventually use these to solve a system of equations. Well, that's exactly what we're going to do in this video. In this video, we're going to see how to solve a system of 2 equations with 2 unknowns using something called Cramer'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 going to 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 going to pop a bunch of numbers in, calculate determinants, and it'll just spit out our numbers for x and y, our 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 want to make a little bit of an analogy. I like to think of Cramer'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 learn 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. Alright? That's basically, kind of the analogy there. Alright. So let's go ahead and just get started here.

The whole idea is that if we're given a system of 2 equations with 2 unknowns, then Cramer's rule is going to involve calculating determinants of 2 by 2 matrices, and we've seen how to do this before. Let's go ahead and get started with this problem. So we've got \(2x + y = 5\), \(-4x + 6y = -2\). We want to solve this using Cramer's rule. We're going to have to calculate one of the determinants, so we're going to have to turn this into a matrix. Alright? That's what I'm going to do over here. I'm going to calculate or determine my augmented matrix, and I'm going to color code these things. Right? So remember, these are going to be my x coefficients, then I've got my y coefficients, and then these are going to be my constants over here. Right? So then these are going to be my constants.

So here, what I've got is I've got the 2 and the -4, and then here I've got the 1 and the 6. And then remember there's a little black bar, and then here I've got the 5 and the -2. Alright? So what Cramer's rule says is that to calculate x, you're going to have to calculate the quotient of 2 determinants. You're going to have to calculate this one and then divide it by this one. Alright? So we're going to have to calculate 2 determinants over here and then divide them. The same thing, by the way, applies for the y. So, it's 2 determinants, and then divide them. In fact, actually, the determinants for x and y on the bottom are actually the same thing, so you really only have to calculate 3 determinants. Let's take a look at the first one here.

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 going to be 2, -4, and then this is going to be, 1, 6. Alright? 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 more interesting. Now what happens is when you're solving for the x, what you're going to do is you're going to take the constants that are on the right side of the equation in your matrix, and you're going to have to replace the x coefficients. Right? So you're basically just going to take that column and swap them with the x, and then you're going to write that new matrix. So here what happens is I'm not going to write 2 and -4. I'm going to write this is going to be 5 and -2. That takes the place of the x column. Now the y columns are actually just going to be the same.

This is going to be 1, 6. Alright? So that's what it is for x. For the y, what you're going to see is that we're going to do the same thing, except now we're going to replace the y coefficients. Alright? So now here, what happens is the 2 and the -4, we copy that over. But now instead of the 1 and the 6, now we replace it with the 5 and the -2. So, 2, -4; 5, -2. Alright? So that's really all there is to it. You just are replacing that column of those coefficients with the column of the constants on the right side. Alright? Now we just have to go ahead and calculate these three determinants over here.

| 5 1 −2 6 | | 2 1 −4 6 | | 2 5 −4 −2 | | 2 1 −4 6 |

For the x, what we're going to see here is that this ends up being 5 times 6. This is going to be 30 minus 1 times -2. So this is going to be 30 minus -2. So this is going to be, plus 2. So you could have just have done plus 2. Alright? And then on the bottom here, what we're going to get is 12 minus, again, -4. So this is going to be 12+4, which is going to be 16. Alright? So, in other words, this really just becomes 32 over 16, and this x value is equal to 2. Alright? So that's basically what the x value turns out to be.

Now, for the y, what you'll see is that we already calculated this to be 16, so we don't have to recalculate it. What does the top end up being? So this is 2 times -2, so that is -4, minus 5 times -4. So this is -2 minus, and this is going to be -20. Alright? So this is going to be -4, and then, basically, this is going to be plus 20. Alright? So what this actually ends up being when you work all this out, this actually ends up being 16 over 16, which is just 1. So, in other words, the solution to the system of equations is x equals 2 and y equals 1. 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 Cramer's rule. Again, it's just another tool in your tool belt. But that's how to solve these kinds of problems using Cramer'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 2 by 2 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 going to take a look at how to calculate the determinant of a 3 by 3 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 2 by 2 determinants, which we do know how to solve. So I'm going to break it down for you and show you the equation for how to solve a 3 by 3 determinant, and then we'll take a look at an example together. Alright? Let's get started. So if I have a 3 by 3 matrix that's organized over here, I'm going to organize all the letters, you know, into subscripts, like a₁ b a₁₂₃, b₁₂₃, c₁₂₃, so on and so forth. Really, all it is to calculate a 3 by 3 determinant is you're going to have to calculate three numbers, so these three numbers. And I want to 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. Alright? And now what I want to do is sort of give you, like, an understanding of what these what this equation actually comes from. So if you have a 3 by 3 determinant, here's what's basically going on. Right? I'm going to write this same exact matrix out three times, and we'll see a pattern that starts happening with how these numbers where these numbers really come from or these terms. So for the first time, what you're going to do is you're going to take the a in the first row, and then you're going to strike out all of the other entries or all the other numbers that are in that column and row. So you're going to 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 2 by 2 matrix over here with the b and c numbers. This thing over here is basically what this becomes. You're going to take this a₁ number and you're going to 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 going to take the b₁ number in the first row, and, again, you're going to 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 going to 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 up over here. That's what the second number or the second number ends up being. It's b₁ 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 going to 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 going to take that c and multiply it by ``` 2 1 4 6 ```

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 2 by 2 determinant, and then we learned how to use Cramer's rule to solve a system of 2 equations with 2 unknowns. Well, similarly here, we've learned how to calculate 3 by 3 determinants. And what I'm going to show you in this video is how to solve a system of 3 equations with 3 unknowns using Cramer's rule. Now remember that Cramer's rule is really just a formula that directly gives you the solution to a system of equations. We're basically just going to plug in a bunch of numbers into some formulas, and it's going to spit out the answers for x, y, and z directly. Alright? The only thing that's different here is that because we have 3 equations with 3 unknowns, our determinants will just be a little bit more complicated. Instead, we're going to use determinants of 3 by 3 matrices. Let's just go ahead and get started here because there's actually not a whole lot that's different about Cramer's rule, and we'll go ahead and work this out together. Let's get started.

So we have the system of equations over here, and we want to solve this by using Cramer's rule. Now the first thing I'm going to do is I'm going to turn this into an augmented matrix just so I can get all my coefficients and stuff in the right place. Alright? So this just becomes a matrix like this. Remember, we're just going to copy over the coefficients. My x coefficients are going to be negative 5, and I have 0 over here. This is like 0x, and then I have 1. Now for the y column that's going to be in green, I'm going to have negative 1, 3, and then 1. And then for this z column over here, my z coefficients, this is going to be 4, 6, and 1. And then remember that my final, my final column over here with these purple numbers, these are going to be my constants, are just going to be other constants on the right side, 4, 21, and 6.

Alright? So what Cramer's rule says is that x, y, and z, I'm going to have to take basically the division of two numbers, something called dx, dy, and dz. So what is this capital D? This capital D really is just the 3 by 3 matrix that you get just by looking at the coefficients of your x, y, and z variables. Alright? Now because I'm going to use that variable 3 times in my equations, that's always just going to be a good place to start. Also, what happens is that if that capital D ends up being 0, then you'll actually have no solution. So you just want to make sure that it's not 0. So let's go ahead and start there. What capital D is is just the determinant of just this 3 by 3 matrix that you get over here. Right? So it's really just going to be the determinant of this thing.

So remember how to calculate determinants? Just really quickly as a brief refresher, what you have to do is you have to follow this formula over here where, remember, you're just going to have to take these numbers, cancel out strike out the rows and columns, and then focus on the smaller 2 by 2 matrices that you get. Alright? You can take a look there just in case you need a refresher. So I'm just going to go ahead and set this up, and I'm going to skip some steps here because we're a little bit sort of tight on space for this problem. What we're going to see here is that this is going to be negative 5 times the smaller determinant over here. So in other words, this is just going to be the 3, 1, and 6, 1. Now we flip signs. This will be minus. This is going to be negative one, the negative one times this smaller determinant over here, and we cancel out some stuff, and we're going to get 0, 1 and 6, 1. Alright. Now I'm going to flip signs again. This is going to become now 4. This is going to become now we're going to use smaller determinants that we get over here. So in other words, this is just going to be 1, 1 and 3, 1. Alright? So that's what this ends up being.

So if we calculate this, what we're going to end up seeing is that we end up getting, this is going to be negative 5. This is going to be negative 5, and then we're going to have 3 times 1 minus 6 times 1. So 3 minus 6, will end up being -3 so negative 5 times negative 3. that's what this whole thing ends up becoming, negative 3 over here. Alright? So now let's see. We have negative 1, so we're going to have minus. We're going to have negative one times and then what is this mess end up becoming? 0 times 1 is 0 minus 6 times 1, which is 6. So this is going to be let's see. This is going to be negative one times 0 minus 6. That's actually negative 6. Alright? That's what that ends up being. That's what this whole mess becomes. And then over here in this last term, we're going to get is we're going to get 0 times 1. So it's going to be 4 times let's see. 0 times 1 is 0, and then minus 3 times 1, which is 3. So in other words, it's going to be 0 minus 3, which is negative 3. Alright?

So what I end up getting here is that this is going to be, let's see. So this is going to be this is going to be negative 15. Actually, this is going to be positive 15. This is going to be now minus, and this is going to be 6 because the negatives cancel out, plus 12. And if you work this out here, what you're going to get is that capital D is equal to -3. So in other words, what's going to happen here is when I calculate x, y and z, I'm going to have to calculate some number and then just divide it by now we know this is just going to be -3. Alright? So let's go ahead and do that. So this capital D here is -3. 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 going to be something called Dx. So I'm going to have to calculate just some number here, some determinant number called Dx and divided by this capital D. So in other words, some number divided by -3. What is this Dx that we've been talking about? Well, really, what it is is in a similar way that in Cramer's 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 going to take our normal matrix over here of a, b, and c. And if you're solving for the x components, you're going to 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 going to be a 3 by 3 determinants, except instead of the negative 5, 0, and 1, I'm really just going to plug in these numbers. This is going to be 4, 21, and 6. So 4, 21, and 6, and then the rest of the numbers are the same, negative 1, 3, and 1. And this is going to be 4, 6, and 1. Alright? Now what we have to do is just calculate the determinant of this 3 by 3 matrix. So let's go ahead and do that.

So what this says here is that we're going to this is going to be 4, and then this is going to be, this is just going to be the smaller matrix over here. So I'm just going to copy that over like that. Minus this is going to be negative 1, and then this is going to be, 2, 1, 6. So 2, 1, 6, and then this is going to be 6, 1. And then we have plus plus, and then we have 4, and this is going to be, these smaller numbers over here. so this is going to be 21, 3, and 6, 1. So that is going to be your 3 by 3 determinants. Now we just have to go ahead and do a bunch of calculations. Alright? So this Dx here will be well, this is just going to be 4 times 3 minus 6. So this is going to be, 4 times -3. That’s 4 times -3 minus, and this is going to be let's see. Negative one. So this is going to be oops. So what happens is we do 21 times 1, that's 21, minus 36, which is -15. And then that's we multiplyyclerView by negative one, that's going to be positive 15. so this is going to be 15 over here. Now this is going to be 21 times 1, which is 21, times a minus 18, so that's 3, and 4 times 3 is 12. Alright? So So we're just going to do we're just going to sort of, mental math our way through this.

This is going to be -12 minus 15 plus 12. Alright? So this ends up being -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 -12 will cancel, and you'll see that Dxjust leaves you with -15. Alright? So now we can plug this Dx back into this formula over here, and we'll see that this is just a number, -fifteen over -three just equals positive 5. And that is the first number that you have. So x is equal to 5. Now we just basically are going to rinse and repeat. We're going to do this exact same thing. We're going to do it twice, you know, 2 more times for y and z. Alright? So this is going to be y over here. Right? So here, what I've got is Dy over D. So in other words, some number divided by -3. And when I calculate that, I'm just going to get whatever the answer is. How do I figure out what Dy is? Well, basically, Dy is just going to be now if I replace the constants into the y column instead of the x. Right? So now the constants are going to be in this middle column. And so let's go ahead and write out what this matrix is going to be. So what this is going to be is this is going to be, let's see. So I've got my negative 5, 0, and 1, and then I've got, in purple, I've got 4, 21, and 6. And then over here, what I've got is 4, 6, and 1. Alright? So let's calculate this determinant over here. This is going to be negative 5 times, and then we're going to have these 2 smaller numbers over here. That's going to go in here. Then we switch signs. This is going to be minus, then we're going to have 4, and then this is going to be a smaller determinant; in which we're going to have 0, 1, and 6, 1, and then we're going to flip signs again, and now we're going to have 4, and then in this smaller determinant, we're just going to have these four numbers over here. So what you'll also notice here is that, occasionally, you're going to have the same numbers that pop up. So you're going to have, for example, 21, 6, 6, and 1. We actually already know what the determinant of that is; that ends up being 15 or -15. So you're going to 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. Alright. So let's calculate this. So this is going to be negative 5, and then this whole thing just becomes 21 minus, 36, which is going to be minus 15. Alright? So what this ends up being here is this ends up being a 4. Now with 0 minus 1, sorry. 0 times 1 minus 6 is going to be negative 6, and this is going to be 4. And then 0 times 6 is 0 minus 21 times 1, which is -21. Alright? So that's really what that becomes. And if you work this out here, what this ends up being is this ends up being 75, plus 24 minus 84. Right?

And so if you work this out, what you'll get is a grand total over here of 15. So this actually, Dy ends up being positive 15. Alright? Now if you plug this back into this formula, what you'll see is that 15 over negative 3 ends up being -5, 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 going to be Dz over D, and this is going to be some number divided by negative 3. When you calculate this, you're just going to get some number over here. Let's go ahead and work this out. So this is going to 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 going to be negative 5, 0, and 1. This is going to be, negative 1, 3, and 1. And then here, we're going to have 4, 21, and 6. Now if you work this out, which this determinant is going to be, this is going to be, negative 5, and then we're going to do these two numbers. By the way, we've already calculated determinant before. And this is going to be minus, negative 1, and then we're going to have this smaller determinant over here, which is going to be 0, 1, 21, and 6. So 2, 1, 6. And then finally, we swap signs one last time, and this is going to now be 4. And then we have this smaller determinant over here. This is going to be like that. Alright? Now I'm actually just, I'm actually just going to 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 going to get is you're going to end up getting 15. This is going to be -21, and this is going to be -12. And so what you're going to get here for this is you're going to get -18. So when you plug this in negative 18, you're going to get an answer of 6. And so, finally, the answer to your solution is 5, negative 5, and positive 6. So this is the solution to your system of equations. Negative oh, sorry. 5, negative 5, negative 5, and then positive 6. 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. $x=-1,y=-4,z=2$

Here’s what students ask on this topic:

How do you calculate the determinant of a 2 by 2 matrix?

To calculate the determinant of a 2 by 2 matrix, use the formula:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

For example, if you have a matrix:

$\left|\begin{array}{cc}3& 2\\ 5& 4\end{array}\right|$

The determinant is calculated as:

$34-25=12-10=2$

What is Cramer's Rule and how is it used to solve a system of equations?

Cramer's Rule is a method used to solve a system of linear equations using determinants. For a system of two equations with two unknowns:

$ax+by=ecx+dy=f$

The solutions for x and y are given by:

$x=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

$y=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

By calculating these determinants, you can find the values of x and y that satisfy both equations.

How do you calculate the determinant of a 3 by 3 matrix?

To calculate the determinant of a 3 by 3 matrix, use the following method:

$\left|\begin{array}{ccc}a\_11& a\_12& a\_13\\ a\_21& a\_22& a\_23\\ a\_31& a\_32& a\_33\end{array}\right|$

The determinant is calculated as:

$a\_11\left|\begin{array}{cc}a\_22& a\_23\\ a\_32& a\_33\end{array}\right|-a\_12\left|\begin{array}{cc}a\_21& a\_23\\ a\_31& a\_33\end{array}\right|+a\_13\left|\begin{array}{cc}a\_21& a\_22\\ a\_31& a\_32\end{array}\right|$

Each term involves a 2 by 2 determinant, which you already know how to calculate.

How do you use Cramer's Rule to solve a system of 3 equations with 3 unknowns?

To use Cramer's Rule to solve a system of 3 equations with 3 unknowns, follow these steps:

1. Write the system in matrix form:

$\left|\begin{array}{ccc}a\_11& a\_12& a\_13\\ a\_21& a\_22& a\_23\\ a\_31& a\_32& a\_33\end{array}\right|\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}b\_1\\ b\_2\\ b\_3\end{array}\right|$

2. Calculate the determinant of the coefficient matrix (D).

3. Replace the x-column with the constants and calculate the determinant (Dx).

4. Replace the y-column with the constants and calculate the determinant (Dy).

5. Replace the z-column with the constants and calculate the determinant (Dz).

6. Solve for x, y, and z using:

$x=\frac{D}{x},y=\frac{D}{y},z=\frac{D}{z}$

This method systematically provides the values of x, y, and z that satisfy all three equations.

What are the applications of determinants in solving systems of equations?

Determinants are crucial in solving systems of equations, particularly through methods like Cramer's Rule. They help determine the solvability of a system. If the determinant of the coefficient matrix is zero, the system has no unique solution. Determinants also simplify the process of finding solutions by converting complex systems into manageable calculations. Additionally, they are used in various fields such as physics, engineering, and computer science for solving linear equations, analyzing stability, and transforming coordinates.