### Video Transcript

Express the simultaneous equations three π₯ plus two π¦ equals 12 and three π₯ plus π¦ equals seven as a matrix equation.

For this question, we have been given a system which includes two linear equations. We might be used to questions asking us to solve such a system, that is, asking us to find the value of each of our variables π₯ and π¦. The first thing to note for this question is that we havenβt been asked to solve the system at all. Rather, we must reexpress both of these equations in matrix form. In order to do this, we can use the following rule. Consider two equations: ππ₯ plus ππ¦ equals π and ππ₯ plus ππ¦ equals π, where π, π, π, π, π, and π are all constant terms.

Note that the equations given in our question exactly match the form that has been shown here. You might be familiar with calling this the standard form of a linear equation. Such a system can be expressed as the following matrix equation. On the left-hand side, we have the two-by-two matrix π, π, π, π, which is multiplied by the two-by-one matrix π₯, π¦. And we have this equal to the two-by-one matrix π, π on the right-hand side. Here itβs also worth noting that this rule works in both directions. So had we been given the matrix equation shown below, we would be able to convert it into the pair of simultaneous equations shown above. To solve this question, we can simply substitute the values of our system into the given rule.

To make things really simple for ourselves, let us write these out, starting with the coefficients on each of our variables. Weβll say that three π₯ plus two π¦ equals 12 is equation one. This means that π and π are the coefficients three and two, respectively. Okay, we now say that three π₯ plus π¦ equals seven is equation two. And this means that the coefficients π and π are three and one, respectively. Note that for the value of π, coefficients of one arenβt usually shown when applied to variables, but we know that this is the value it takes.

Here itβs worth noting that, in our matrix equation, the first two-by-two matrix contains all four of the coefficients within our system of equations. For this reason, we call this the coefficient matrix in our matrix equation. Likewise, this two-by-one matrix contains both of the variables in our system, and hence we call it the variable matrix. The final two-by-one matrix contains both of the constant terms on the right-hand side of our two equations. And for this reason, we call it the constant matrix. In our case, the constant π is the right-hand side of equation one, which is 12. And finally, the constant π is the right-hand side of equation two, and its value is seven.

With this information, weβre ready to form our matrix equation. Weβll first substitute in the values of π, π, π, and π into our coefficient matrix, giving us the matrix three, two, three, one. We next note that our variable matrix is fine and we do not need to perform any substitutions since π₯ and π¦ match the variables in our two equations. Itβs worth noting that this method would also work for any other two variables as long as the order is preserved. Finally, we substitute in the values of π and π into our constant matrix. This gives us the two-by-one matrix 12, seven. With this final step, we have solved our question. We have expressed the simultaneous equations given in our question as a matrix equation.

As a final point, we again reiterate that we have not solved this system or found the value of any particular variable. What weβve done is to represent the two equations given in the question using different mathematical notation. If you want to prove this to yourself, you might want to perform the matrix multiplication of the coefficient matrix and the variable matrix on the left-hand side of our matrix equation. We wonβt be showing it in this video, but perhaps itβs an exercise for you.