### Video Transcript

Using the inverse matrix, solve the system of linear equations two 𝑥 plus 𝑦 equals negative one, three 𝑥 plus two 𝑦 minus 𝑧 is equal to negative five, two 𝑧 plus 𝑦 equals one.

When we solve a system of linear equations using matrices, we write this system in the form 𝐴𝑥 equals 𝑏, where 𝐴 is the coefficient matrix. 𝑥 is the variable vector. Here that’s the vector 𝑥𝑦𝑧. And 𝑏 is the constant vector. By considering the dot product, we can represent the left-hand side of this system of linear equations as shown.

This three-by-three matrix is our coefficient matrix. And it can be generated by looking at the coefficients of 𝑥, 𝑦, and 𝑧 in each of the system of linear equations. For example, the first element in the second row is the coefficient of 𝑥 in our second equation, the second element is the coefficient of 𝑦, and the third element is the coefficient of 𝑧.

The constant matrix 𝑏 is given by the constants on the right-hand side of this system. They are negative one, negative five, one. We solve this equation by multiplying both sides by the inverse of the coefficient matrix. Remember, matrix multiplication is not commutative. It has to be done in a specific order. So the inverse matrix needs to be on the left of each part of our matrix equation. This works because the inverse of 𝐴 multiplied by 𝐴 is the identity matrix. And we can see that 𝑥 will be equal to the inverse of 𝐴 multiplied by 𝑏.

So let’s find the inverse of our coefficient matrix. We can use any method we feel comfortable with. Let’s look at the row reduction method. We create this augmented matrix by drawing a vertical line to the right of our matrix and adding the identity matrix. Our aim is to create the identity matrix on the left-hand side of this augmented matrix by performing row reduction operations.

As we perform these to the matrix on the left, we also do the same to the matrix on the right. And that will eventually be our inverse. As we go through this process, I will call row one R one, row two R two, and row three R three to make our life a little bit easier. And we’re going to begin by subtracting row two from row one. When we do that, we get negative one, negative one, one on that top row. Row two and row three stay the same. On the right-hand side, once again, row two and three remain unchanged. But row one becomes one, negative one, zero.

Next, we’re going to multiply row one by negative one. And when we do that, we get one, one, negative one on the left-hand side and negative one, one, zero on the right. Next, we’re going to subtract three times row one from row two. And when we do that, row two becomes zero, negative one, two on the left-hand side and three, negative two, zero on the right.

Now we’re going to multiply row two by negative one. And we get zero, one, negative two, negative three, two, zero. Next, we subtract the elements from row two from row three. And we can see now that we have some multiple on that left-hand side of the bottom row in the identity matrix. In fact, if we divide each of the elements in this row by four, we will end up with zero, zero, one, which is what we’re looking for. And repeating this process on the right-hand side, we get three-quarters, negative two-quarters, and one-quarter.

We’ve now completed any operations on that final row. And next, we’ll subtract row two from row one and then row three from row one. And we can now see that row one and row three on the left-hand side of our augmented matrix are the same as row one and row three from the identity matrix.

Our final step is to add two times row three to row two. And we finally have the identity matrix on the left-hand side. Notice how some of the fractions on the right-hand side are not yet simplified. You’ll see why in a moment. We know that the inverse of our coefficient matrix is as shown. And we can then take out that factor of one-quarter. And we’re left with a quarter multiplied by five, negative two, negative one, negative six, four, two, three, negative two, one.

Earlier, we said that we can solve this system of linear equations by multiplying this inverse by the constant vector. We’re going to need to find the dot product of each row of our matrix and the constant vector. For the first row, we get a quarter multiplied by five multiplied by negative one plus negative two multiplied by negative five plus negative one multiplied by one. That’s a quarter of four, which is one. For the second row, we get a quarter multiplied by negative six multiplied by negative one plus four multiplied by negative five plus two multiplied by one. This time, that’s a quarter of negative 12, which is negative three. And for this final row, we get a quarter multiplied by three multiplied by negative one plus negative two multiplied by negative five plus one multiplied by one. That’s a quarter of eight, which is two. So the solution to this system of linear equations is 𝑥 equals one, 𝑦 equals negative three, and 𝑧 equals two.

We could of course perform a simple check by substituting these values into each of our equations. Substituting 𝑥 equals one and 𝑦 equals negative three into our first equation, we get two multiplied by one plus negative three, which is indeed negative one, as required. We could repeat this process for all three equations, although one or two actually is enough to just give us the indication that we’ve probably done it right.