Question Video: Using Cramer’s Rule to Solve a System of Equations Defined Using Determinants of Two by Two Matrices | Nagwa Question Video: Using Cramer’s Rule to Solve a System of Equations Defined Using Determinants of Two by Two Matrices | Nagwa

# Question Video: Using Cramer’s Rule to Solve a System of Equations Defined Using Determinants of Two by Two Matrices Mathematics • First Year of Secondary School

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Solve, using Cramer’s rule, the simultaneous equations |−1, 𝑧 and −4, 𝑦| = 23, |2, 𝑦 and −5, 𝑥| = 13, |3, 𝑥 and 5, 𝑧| = 51.

10:24

### Video Transcript

Solve, using Cramer’s rule, the simultaneous equations the determinant of the matrix negative one, 𝑧, negative four, 𝑦 equals 23; the determinant of the matrix two, 𝑦, negative five, 𝑥 equals 13; and the determinant of the matrix three, 𝑥, five, 𝑧 equals 51.

We’ve been asked to use Cramer’s rule to solve this. It’s a little bit different to how we often see questions involving the use of Cramer’s rule, because our system of linear equations has been given to us in terms of determinants. But let’s begin by recalling Cramer’s rule. Cramer’s rule tells us that if the determinant of the coefficient matrix in a system of linear equations is nonzero, then there is a unique solution to the system given by 𝑥 equals Δ sub 𝑥 over Δ, 𝑦 equals Δ sub 𝑦 over Δ, and 𝑧 equals Δ sub 𝑧 over Δ.

But in order to use Cramer’s rule, we firstly need our system of linear equations to be in the form of a matrix equation. So we’re going to have to use the determinants that we’ve been given to put these simultaneous equations into a matrix equation. Let’s begin this question by evaluating the determinants of these two-by-two matrices that we were given in the question.

Recall that we find the determinant of a two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑 by calculating 𝑎𝑑 minus 𝑏𝑐. So let’s take a look at this first matrix negative one, 𝑧, negative four, 𝑦. The determinant of this matrix is going to be negative one multiplied by 𝑦. That gives us negative 𝑦. And then we subtract 𝑧 multiplied by negative four. That gives us negative four 𝑧. And because we’re subtracting a negative, we can write that as plus. So the determinant of this matrix is negative 𝑦 plus four 𝑧. And as we’re given in the question, this is equal to 23.

We can do the same for the second matrix. Let’s find the determinant of the matrix two, 𝑦, negative five, 𝑥. We firstly do two multiplied by 𝑥, which gives us two 𝑥. And we subtract 𝑦 multiplied by negative five, which is negative five 𝑦. And again, because we’re subtracting a negative, we can write this as plus. And as we’re given in the question, this is equal to 13.

So now, let’s do the third and final matrix. We multiply three with 𝑧 to give us three 𝑧. The, we subtract 𝑥 multiplied with five, which is five 𝑥. And we know that this is equal to 51.

So now, we’ve managed to take the three determinants we were given and find the three linear equations. But in order to apply Cramer’s rule, we need to rewrite the system as a matrix equation. To do this though, let’s rewrite our equations so that the 𝑥-coefficients, the 𝑦-coefficients, and the 𝑧-coefficients and the constants all line up. We do however need to be careful to include zero coefficients in our coefficient matrix. So to help us do this when we rewrite our system of equations, let’s include the zero coefficients before writing it as a matrix equation.

Our first equation is negative 𝑦 add four 𝑧 equals 23. But our variables are 𝑥, 𝑦, and 𝑧 for this system. So we can rewrite this as zero 𝑥 minus 𝑦 plus four 𝑧 equals 23. So we can do the same for our second equation. The second equation doesn’t include any coefficients for 𝑧. So we rewrite this as two 𝑥 plus five 𝑦 plus zero 𝑧 equals 13. Then, we can do the last equation. We write this as negative five 𝑥 plus zero 𝑦 plus three 𝑧 equals 51. We include the zero 𝑦 because the equation originally didn’t have any 𝑦-coefficients. Also, we rearrange this so the 𝑥’s come first, then the 𝑦’s, then the 𝑧’s. Writing it in this way makes it much easier to put this system into a matrix equation.

I’m going to clear some space so that we only have our equations which are in orange. Okay, so here our three equations, and we’re going to put this into a matrix equation. The matrix equation has three parts: the coefficient matrix, which is the matrix consisting of the coefficients of our variables 𝑥, 𝑦, and 𝑧; the variable matrix, which consists of the variables for our system; and the constant matrix, the matrix consisting of the constants.

So let’s firstly fill in the entries for the coefficient matrix, that is, the coefficients for our variables. Remember to be careful with the negative 𝑦 in the first equation because the coefficient is negative one. You can see now why we rewrote the equations in this way to include the zero coefficients. The variable matrix consists of the variables in our system. That’s 𝑥, 𝑦, and 𝑧. And finally, the constant matrix consists of the constants for our system of equations. That’s 23, 13, and 51.

Now, we can start thinking about applying Cramer’s rule. To do this, we’re going to need Δ sub 𝑥, Δ sub 𝑦, Δ sub 𝑧, and Δ. At this point, we remember that Δ sub 𝑥, Δ sub 𝑦, and Δ sub 𝑧 are the determinants of the matrices that are formed as a result of substituting the elements of the constants matrix with the elements of the columns of the 𝑥-, 𝑦-, and 𝑧-coefficients. So let’s go ahead and find the values for Δ sub 𝑥, Δ sub 𝑦, and Δ sub 𝑧.

Δ sub 𝑥 is the determinant of the matrix we get when we take the coefficient matrix and substitute the elements from the constant matrix with the elements from the column of the 𝑥-coefficients, that is, the determinant of the matrix 23, negative one, four, 13, five, zero, 51, zero, three. At this point, we can recall how we find the determinant of a three-by-three matrix. We use this formula, which involves finding determinants of two-by-two matrices, known as the cofactors of the three-by-three matrix. So let’s use this to find Δ 𝑥.

Notice how we have a minus negative here, so we can write this as a plus. We can now work this out by calculating the determinants of each of these cofactor matrices. The determinant of the first cofactor matrix is found by doing five multiplied by three minus zero multiplied by zero. That gives us 15. Then, we can find the determinant of the second cofactor matrix. We do this by doing 13 multiplied by three minus zero multiplied by 51. That gives us 39. Finally, we find the determinant of the last cofactor matrix. And we do this by doing 13 multiplied by zero minus five multiplied by 51. That gives us negative 255.

So I’m now going to substitute in these values. So we just need to calculate 23 multiplied by 15 add one multiplied by 39 add four multiplied by negative 255. And we can calculate this to be negative 636. So we found Δ sub 𝑥 to be negative 636.

We can now calculate Δ sub 𝑦 using the exact same method. I’ll clear some space first. Δ sub 𝑦 is the matrix we get when we substitute the elements of the constant matrix with the coefficients for 𝑦 from the coefficient matrix. And we’ll use the same method we just used to find the determinant for this matrix. From here, we once again calculate the determinant of these cofactors. That’s going to be 39, six, and 167. So then we calculate zero multiplied by 39 minus 23 multiplied by six add four multiplied by 167. And that gives us 530. So Δ sub 𝑦 is 530.

And by the same process, we’ll now calculate Δ sub 𝑧. We find Δ sub 𝑧 by finding the determinant of the matrix that we get when we substitute the elements of the constant matrix with the coefficients of 𝑧 from the coefficient matrix. Then, we can use the same method to find the determinant. We then calculate the determinants of the cofactors, that is, 255, 167, and 25. So now, we can calculate zero multiplied by 255 add one multiplied by 167 add 23 multiplied by 25. And we find that gives us 742. So Δ sub 𝑧 equals 742.

So the final thing we need to find is Δ. Δ is the determinant of the coefficient matrix. So one more time, we’re going to use the same method to find the determinant of this three-by-three matrix. Again, we need to find the determinant of each of these cofactor matrices. We find the first one to be 15. The second one is six. And the third one is 25. So we then calculate zero multiplied by 15 add one multiplied by six add four multiplied by 25. And that gives us 106. So Δ equals 106.

Now we have all of the components that we need to use Cramer’s rule. We have Δ sub 𝑥, Δ sub 𝑦, Δ sub 𝑧, and Δ. So we can now substitute these values into the unique solution from Cramer’s rule to find our values for 𝑥, 𝑦, and 𝑧. Starting with 𝑥 equals Δ sub 𝑥 over Δ and as we found Δ sub 𝑥 to be negative 636 and Δ to be 106, 𝑥 is therefore equal to negative 636 over 106. As 106 goes into 636 six times, this is just negative six. Cramer’s rule also tells us that 𝑦 is equal to Δ sub 𝑦 over Δ. That’s 530 over 106. And as 530 divided by 106 is five, we have that 𝑦 is equal to five. And finally, we find 𝑧 by doing Δ sub 𝑧 over Δ. That gives us 742 over 106. And 742 divided by 106 is just seven.

So our final answer is 𝑥 equals negative six, 𝑦 equals five, and 𝑧 equals seven.

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