Question Video: Solving a System of Three Equations Using Determinants Mathematics

Video Transcript

Use determinants to solve the system five 𝑥 equals negative two 𝑦 minus five plus three 𝑧, negative three 𝑥 minus 𝑦 plus one equals two 𝑧, and two 𝑦 minus 𝑧 equals negative five 𝑥 plus three.

So, in a problem like this, the first thing we want to do is actually rearrange our equations so that we have our variables on the left-hand side. And then we have our answers on the right-hand side, which are numerical values or constants. So our first equation rearranged is gonna be five 𝑥 plus two 𝑦 minus three 𝑧 equals negative five. Then, for the second equation, we’re gonna have negative three 𝑥 minus 𝑦 minus two 𝑧 equals negative one. And then finally, five 𝑥 plus two 𝑦 minus 𝑧 equals three.

Okay, great. We’ve got it like this, but why do we want it in this form? We want it in this form so we can set up a matrix equation. And when we do, what we get is the matrix five, two, negative three, negative three, negative one, negative two, five, two, negative one multiplied by the matrix for our variables, which is 𝑥, 𝑦, 𝑧. Then this is equal to our answer matrix negative five, negative one, three. Okay, great. But how does this help us meet our objective, which is to solve the system of equations using determinants? Well, what we’re gonna do is use Cramer’s rule. And what Cramer’s rule tells us is that we can find the variables or solutions to our system of equations using, for example, 𝑥 is equal to, then we’ve got the determinant of the matrix Δ sub 𝑥 over the determinant of the matrix Δ. And then this pattern continues for 𝑦 and 𝑧.

Okay, to use this then, what we need to do is work out our determinants. The first determinant we’re gonna work out is the determinant of Δ, which is gonna be our coefficient matrix. So what we’re gonna do is find out the determinant of the matrix five, two, negative three, negative three, negative one, negative two, five, two, negative one. So this is gonna be equal to five multiplied by the determinant of the submatrix negative one, negative two, two, negative one minus two multiplied by the submatrix negative three, negative two, five, negative one minus three multiplied by the submatrix negative three, negative one, five, two, remembering that when we find the determinants, the coefficients go positive, negative, positive. And to find our submatrices, we delete the column and row that our coefficient is in.

Okay, great. So now, we calculate this. And then remembering that when we work out determinants of two-by-two matrices, what we do is cross multiply and then subtract, we’re gonna get five multiplied by one plus four minus two multiplied by three plus 10 minus three multiplied by negative six plus five which is equal to two. So this is great cause it also tells us that the matrix is nonsingular. So therefore, we know that there is not gonna be an infinite number of solutions. And that’s because if there was, then the determinant would be equal to zero. So now, what we’re gonna do is clear a space and work out the other determinants we need to find.

So next, what we want to do is find the determinant of Δ sub 𝑥. And the way we do that is by substituting in the answer matrix values for the coefficient of 𝑥-value, so the first column in our matrix. So now what we’re gonna want to do is find the determinant of this matrix. And to do that, what we’re gonna do is to use the same methods as we used for the previous determinant, which is gonna give us a determinant value of negative 42. And you can see the working there. Okay, great. So once again, we’re gonna clear a bit of space and look at our next determinant.

So now what we’re gonna find is the determinant of the matrix Δ sub 𝑦. And this is gonna be where we substitute in our answer matrix for the 𝑦-coefficients in the matrix. So then once again, using the same method to find the determinant, we’re gonna have a determinant of 112. And again, the working is shown here. So once again, what we’re gonna do is clear a space for the final determinant. So for the final one, it’s gonna be the determinant of the matrix Δ sub 𝑧. So then once again, we go through the same method to find the determinant of our three-by-three matrix. And what it gives us is a value of eight.

So now what we have are all the determinants we need to use Cramer’s rule to find out our variables 𝑥, 𝑦, and 𝑧. So, first of all, we’re gonna start with 𝑥, which is gonna be equal to negative 42 over two. And we get that because it’s the determinant of Δ sub 𝑥 over the determinant of Δ. So this is gonna give us a value of 𝑥 equals negative 21. And then for 𝑦, we’re gonna have 112 over two which is gonna give us a 𝑦-value of 56. And then finally, we’re gonna get 𝑧 is equal to eight over two, and this is gonna give us 𝑧 is equal four. So therefore, we can say the solutions to our systems of equation are 𝑥 equals negative 21, 𝑦 equals 56, and 𝑧 equals four.