### Video Transcript

Use matrices to solve the system of equations three π₯ plus four π¦ equals 20 and two π₯ plus two π¦ equals 12.

Because of the way that we multiply matrices by using the dot product of rows and columns, we can use matrices to represent a system of linear equations. We take the coefficient of π₯, π¦ in our pair of equations. And we form a matrix equation as shown. It is three, four, two, two times π₯, π¦ equals 20, 12. And if you still remain unconvinced that these represent the same thing, letβs look at what happens if we do multiply the matrices out.

We take the dot product of the first row in the first matrix and the column π₯, π¦. Thatβs three times π₯ plus four times π¦. Thatβs the three π₯ plus four π¦ that we weβre looking for. We then take the dot product of the second row in the first matrix and the column π₯, π¦. Thatβs two times π₯ plus two times π¦, which is two π₯ plus two π¦. And so, we have these two matrices which are equal three π₯ plus four π¦, two π₯ plus two π¦ is equal to 20, 12. And we know that for the matrices to be equal, their individual components must be equal. So, we get back to that our original set of linear equations.

Back to the task at hand though. How does this make it easier? Well, letβs call the vector three, four, two, two π΄. We know that the product of the inverse of a matrix and the original matrix is the identity matrix. So, if we can multiply both sides of our matrix equation by the inverse of π΄, weβll end up with the solution for π₯, π¦. Multiplying the left-hand side by the inverse of π΄ leaves us with the identity matrix multiplied by π₯, π¦, which is just π₯, π¦. And on the right-hand side, weβll have the inverse of π΄ multiplied by 20, 12. So, letβs find the inverse of π΄.

If we have a two-by-two matrix given by π, π, π, π, its inverse is one over the determinant of π΄ multiplied by π, negative π, negative π, π. Where the determinant of π΄ is equal to ππ minus ππ. Now, it should be quite clear that if the determinant of π΄ is equal to zero, the inverse doesnβt exist. Because we would be multiplying some matrix by one over zero which we know to be undefined. So, letβs begin by finding the determinant of our matrix π΄.

We multiply the top-left and bottom-right elements and then subtract the product of the top-right and bottom-left. So, thatβs three times two minus four times two, which is negative two. The inverse of π΄ is one over the determinant multiplied by some matrix. Well, one over negative two is negative one-half. Then, the matrix weβre interested in is achieved by switching the elements in the top left and bottom right. So, we switch the two and the three. And we change the sign of the other two elements. We get negative four and negative two. The inverse of π΄ is, therefore, negative one-half multiplied by two, negative four, negative two, three.

It makes sense to multiply each element by negative one-half. And that will make the next step a little easier. Negative one-half multiplied by two is negative one. Negative one-half multiplied by negative four is two. Negative one-half multiplied by negative two is one. And negative one-half multiplied by three is negative three over two. Now that we know what the inverse of π΄ is, we can go back to our matrix equation.

So, π₯, π¦ is negative one, two, one, negative three over two multiplied by 20, 12. To find the first element in the solution to this then, weβre going to find the dot product of the first row in the first matrix and the column 20, 12. Thatβs negative one times 20 plus two times 12, which is equal to four. We repeat this process for the second row. Itβs one times 20 plus negative three over two times 12, which is equal to two. And so, we know that π₯, π¦ is equal to four, two. Now, for these matrices to be equal, their individual components must be equal. So, we would say that π₯ is equal to four and π¦ is equal to two are the solutions to our system of equations.

Now, itβs always sensible when solving systems of equations to check our solutions by substituting them back into the original equations. Letβs substitute π₯ equals four and π¦ equals two into our first expression. Thatβs three π₯ plus four π¦. We get three times four plus four times two, which is 20 as required. Similarly, two times four plus two times two is 12, which is also what we were looking for. And so, weβve used matrices to solve our system of linear equations. We have π₯, π¦ equals four, two, or π₯ equals four, π¦ equals two.