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

Understanding determinants is crucial for solving systems of equations using Cramer's rule. For a 2 by 2 matrix, the determinant is calculated as $\left(a\right)\left(d\right)-\left(b\right)\left(c\right))$. For 3 by 3 matrices, determinants involve smaller 2 by 2 calculations. Cramer's rule allows for direct solutions of equations by substituting coefficients and constants into determinant formulas, facilitating efficient problem-solving in linear algebra.

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 about determinants, it may be kinda scary because there's lots of letters and numbers flying around everywhere, but I'm going to break it down for you and 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 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. Right? So we've got this matrix here, A, which is 3, 2, 5, and 4. So we've got columns and rows. Right? This is the rows, and this is the columns. 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 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. Alright? 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 3 and 4. So this is going to be 3 times 4 minus you're going to subtract, and then 2 times 5. You always do the blue ones first, the red ones second. Alright? So this is going to be 2 times 5. It doesn't matter which way you multiply these because the answer is the same. And so what this ends up being is this ends up being 12 minus, and this ends up being 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. Alright? So that's really all there is to it. 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 like a grid, then you again always 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. Alright? 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 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 times 0, and then we're going to do minus 4 times 5. Alright? 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. Alright? So in other words, the determinant of this matrix is just the number negative 20. Alright? That's really all there is to it. Just 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. Alright?

So, let's take a look at the second matrix. We have B, and here we have some negative numbers. 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. Alright? So, again, you're going to go down to the right and then down to the left, and you're going to do this multiplication in that order. Alright? So this is going to be negative 3 times negative 2, and then you're going to subtract negative 7 times 1. Alright? So 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. Alright? So this is 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. Alright? So 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. And 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 some 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 kind of 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. All right? That's basically, 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 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 by 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. Okay. So here, what I've got is I've got the 2 and the negative 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 negative 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 of x and y on the bottom are actually the same thing, so really only have to calculate 3 determinants. Let's take a look at the first one here. The first one is 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, negative 4, 2, negative 4, and then this is going to be, 16. So 16, 16. 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 sort of 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 going to basically just 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 negative 4. I'm going to write this is going to be 5 and negative 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 16. 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 negative 2. So 5 negative 2. Alright? So that's really all there is to it. You kind of 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. So let's go ahead and get started. So, over here for x, what I've got is I've got, remember how to calculate these determinants? Just if you 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.

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 b1 a2 b2 c1 c2 c3, 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 this equation actually comes from. So if you have a 3 by 3 determinant, here is 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 a 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 b number in the first row, and again, you're going to strike out everything that's part of that column and 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 3rd term, you're going to take that c and multiply it by the determinant of the smaller matrix that you're left with. That's what that third term ends up being. Alright? Now I know it's kinda complicated with all these letters and numbers, but that's just an easy sort of 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 and row and multiply it by the matrix, that smaller 2 by 2 matrix that you're left with. Let's go ahead and, actually, do this example real quick here. Alright? We can see how this works. So here, we got these numbers 3,1,0, -2, -3, -1, -4, 6. I want to calculate the determinants. Alright. So, the determinants of this matrix, remember, I'm going to take three numbers over here. The first one is going to be this 3. So, I'm going to take this 3, and then what I'm going to do is I'm going to 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, so that's the smaller 2 by 2 matrix I'm going to multiply this thing by. So it's going to be (2, -3), (-4, 6). Alright? That's the first number. So now what I'm going to do is, remember, now the sign alternates, so I'm going to have to put a minus sign here. And now what's the next number? Well, the next number here is going to be the 1. Alright? So this is like the b number over here. Strike out everything in that column and row, and what I'm left with is this smaller determinant over here. So this is going to be 1, and then I'm going to have the smaller determinant, which is going to be (0, -3) and this is going to be (-1, 6). Alright? And now finally so that's the second number. Now finally, we just sort of undo all of that stuff. And now we're going to take a look at this number over here. So this is going to be remember, we're going to flip the sign. It's going to be plus. We're going to have 0, and then we're going to have the determinants of this matrix, after we cross out all of these other numbers. So this is going to be (2, -1), and (-1, 4). Now what's nice about this 3rd term here is that we're going to have 0 times the determinant of which is going to be some number. So 0 times anything is just going to be 0. Alright? So that third term actually just completely goes away, and that's good for us. This means there's less calculating. But that's really all there is to it. We just reduced this 3 by 3 determinant into a bunch of multiplications and 2 by 2 determinants. Now the rest, you already know how to do. We really just have to calculate these smaller 2 by 2 matrices over here, and we'll just go ahead and do that real quick. Alright? So what this is going to be is this is going to end up being we're going to expand this. This is going to be 3, and then remember we're going to have a bracket here because now we're going to need to evaluate these smaller determinants. So remember, you always go down to the right and then down to the left. So then in other words, this is going to be 2 times 6 minus (-3) times (-4). Alright? So just keep track of the minus signs. And that's what that first term will end up being. I'm going to have minus, and this is going to be 1 times, and then we're going to have bracket again. This is going to be 0 times 6. Right? So this is going to be 0 times 6 minus (-3) times (-1). Go down to the right, down to the left. Alright? So now let's simplify one step further. So here, what we're going to do is this is going to be equals. So I've got 3. And then over here, what I've got is I've got 12, so that's 2 times 6, minus and this is going to be -3 times -4, 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 3. This ends up being 3 times 0, So that whole first term actually just goes away as well. So that whole thing just gets canceled out. And now what we're left with is we're left with 1, and this is going to be 0 times 6, which is just 0, minus, and then we have -3 times -1. So this is just going to be minus 3. Alright? So in other words, what this ends up being here is you end up just getting a grand total of 0 minus 1 times -3. Alright? So there's a lot of negatives here. So it's minus 1 times -3. So in other words, this ends up just being equals 0 +3, in which case the determinant of this whole 3 by 3 matrix, remember, it's just a number. It's just going to be the number 3. Alright? And the way we got here is we we was we reduced it to 3 smaller 2 by 2 matrices and calculated those. Alright. So 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 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. It's 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. 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 turn this into an augmented matrix just so I can get all my, my coefficients and stuff in the right place. 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. 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, 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. 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 your 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 this 3 by 3 matrix that you get over here. 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 the rows and columns, and then focus on the smaller 2 by 2 matrices that you get. 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 then this is going to be the 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 6,1. Now I'm going to flip signs again. This is going to now 4. This is going to now we're going to use smaller determinants that we get over here. So in other words, this is just going to be 1, and then we have 3,1. 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. 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 1 times 0 minus 6. That's actually negative 6. 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. So what I end up getting here is that this is going to be, let's see. 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 negative 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 negative 3. So let's go ahead and do that. So this capital D here is negative 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 negative 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, 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. 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 this is going to be 4, and then 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.

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 2x2 matrix?

To calculate the determinant of a 2x2 matrix, use the formula:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=(a*d)-(b*c)$

For example, if the matrix is:

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

The determinant is calculated as:

$(3*4)-(2*5)=12-10=2$

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

Cramer's Rule is a method used to solve systems 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:

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

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

Calculate the determinants and substitute them into the formulas to find x and y.

How do you calculate the determinant of a 3x3 matrix?

To calculate the determinant of a 3x3 matrix, use the following method:

$\left|\begin{array}{ccc}a11& a12& a13\\ a21& a22& a23\\ a31& a32& a33\end{array}\right|=a11(a22a33-a23a32)-a12(a21a33-a23a31)+a13(a21a32-a22a31)$

For example, if the matrix is:

$\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|$

The determinant is calculated as:

$1(5*9-6*8)-2(4*9-6*7)+3(4*8-5*7)=0$

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 of equations in matrix form:

$\left|\begin{array}{ccc}a11& a12& a13\\ a21& a22& a23\\ a31& a32& a33\end{array}\right|\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}b1\\ b2\\ b3\end{array}\right|$

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:

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

For example, if D = -3, Dx = -15, Dy = 15, and Dz = -18, then:

$x=\frac{-15}{-3}=5,y=\frac{15}{-3}=-5,z=\frac{-18}{-3}=6$

What are the applications of determinants in linear algebra?

Determinants have several important applications in linear algebra:

1. **Solving Systems of Equations**: Using Cramer's Rule, determinants can solve systems of linear equations.

2. **Matrix Inversion**: Determinants help determine if a matrix is invertible. A non-zero determinant indicates an invertible matrix.

3. **Eigenvalues and Eigenvectors**: Determinants are used in characteristic equations to find eigenvalues and eigenvectors of a matrix.

4. **Area and Volume Calculation**: Determinants can calculate the area of parallelograms and the volume of parallelepipeds in geometry.

5. **Transformations**: In linear transformations, determinants indicate scaling factors and whether the transformation preserves orientation.

Understanding these applications is crucial for advanced studies in linear algebra and related fields.