Video Transcript
Use matrices to solve the system of linear equations three 𝑥 minus 24 equals negative eight 𝑦 and 𝑥 equals three minus 𝑦.
Remember, given a system of linear equations of the form 𝑎𝑥 plus 𝑏𝑦 equals 𝑐 and 𝑑𝑥 plus 𝑒𝑦 equals 𝑓, because of the way that we multiply a square matrix with a column matrix, we can write this equivalently in the form 𝑎, 𝑏, 𝑑, 𝑒 times the column matrix 𝑥, 𝑦 equals the matrix 𝑐, 𝑓. And then, if we alternatively write this as the product of matrices 𝐴 and capital 𝑋 equals matrix 𝐵, then we know if the inverse of the two-by-two matrix 𝐴 exists, we can multiply both sides of our matrix equation by the inverse of 𝐴. And when we do, we find that 𝑋 is equal to the inverse of 𝐴 times 𝐵.
And so what we’re going to do to solve our system of linear equations is first write it in matrix form. Then we’ll find the inverse of the two-by-two matrix that appears and multiply it by the matrix on the right-hand side. Now, we might also have noticed that our system of linear equations is not in the correct form. We need to isolate the 𝑥- and 𝑦-terms on one side of each equation and the constant on the other. For our first equation, we achieve this by adding eight 𝑦 to both sides and then adding 24 to both sides. Similarly, for our second equation, we will just add 𝑦 to both sides. And so we have the system of linear equations in the form that we need.
To find our two-by-two matrix, we simply find the coefficients of 𝑥 and 𝑦 in each equation. They are three and eight in our first equation and one and one in our second. So the two-by-two matrix is three, eight, one, one. We then multiply this by the column matrix 𝑥, 𝑦. And this expression will be equal to the constant column matrix with elements 24, three. Remember, these are simply the constants in each of our linear equations when they’re in the correct form. And so 𝐴 in this case is the two-by-two matrix three, eight, one, one. Capital 𝑋 is the column matrix 𝑥, 𝑦. And 𝐵 is the column matrix 24, three. We said that to solve a matrix equation in this form, we begin by taking the inverse of matrix 𝐴 if it exists.
And if we go back to the general form of a matrix 𝐴 𝑎, 𝑏, 𝑑, 𝑒, the inverse is one over the determinant of 𝐴 times the two-by-two matrix whose elements are 𝑒, negative 𝑏, negative 𝑑, 𝑎. Essentially, we swap the elements in the top left and bottom right and change the sign of the remaining two. And the determinant of 𝐴 is simply the product of the elements in the top left and bottom right minus the product of the elements in the top right and bottom left. And of course, if the determinant is equal to zero, the inverse does not exist. So it always makes sense to begin by calculating the determinant.
The determinant of our matrix 𝐴 is three times one minus eight times one, which is negative five. This is not equal to zero, so we can deduce that the inverse of 𝐴 must indeed exist. So it’s one over the determinant of 𝐴, one over negative five, and then we multiply that by the matrix with elements one, negative eight, negative one, three. Remember, we switch the elements in the top left and bottom right and change the sign of the remaining two. Now, at this stage, we could distribute negative one-fifth across this matrix. But doing so will result in a fraction heavy matrix, which will make the next step a little more complicated. So we’ll do that at the end.
So to find the matrix 𝑥, 𝑦, we’re going to multiply the inverse of 𝐴 by the constant matrix 𝐵. That’s the matrix with elements 24, three. And to multiply this pair of matrices, we find the dot product of the elements in the first row of our first matrix and in the column of our matrix here. That’s one times 24 plus negative eight times three, which is zero. We repeat this process with the elements in the second row of our first matrix and the elements in our column. It’s negative one times 24 plus three times three, which is negative 15.
And so the matrix 𝑥, 𝑦 is negative one-fifth times the matrix with elements zero, negative 15. All that’s left is to distribute negative one-fifth across this matrix. Negative one-fifth times zero is zero, and negative one-fifth times negative 15 is three. We could separate this out and write that the solution is 𝑥 equals zero and 𝑦 equals three or equivalently write this in matrix form as 𝑥, 𝑦 equals zero, three.