### Video Transcript

Express the simultaneous equations four π₯ minus two π¦ is equal to zero and three π¦ plus five π₯ is equal to negative 11 as a matrix equation.

In this question, weβre given a pair of simultaneous equations in two variables π₯ and π¦, and weβre asked to write this as a matrix equation. And we can do this by recalling how we write a pair of simultaneous equations in two variables as a matrix equation. We can start by recalling that to write our simultaneous equations as a matrix equation, we first want to check that the simultaneous equations are written in standard form. And thatβs the form where the constant values only appear on the right-hand side of our equation and each of the variable terms are on the left-hand side of our equation written in the same order.

In this case, we say ππ₯ plus ππ¦ is equal to π and ππ₯ plus ππ¦ is equal to π, where π, π, π, π, π, and π are constants. If this is the case, then we can rewrite this as the matrix equation. We have the two-by-two matrix π, π, π, π multiplied by the two-by-one matrix π₯, π¦ is equal to the two-by-one matrix π, π. And we can see why this matrix equation is the same as the simultaneous equations by evaluating the matrix multiplication on the left-hand side of the equation. For example, multiplying the first row of the first matrix with the only column in the second matrix would give us ππ₯ plus ππ¦, which is equal to the left-hand side of the first simultaneous equation. We can also notice from the matrix equation this needs to be equal to π, which is the right-hand side of the first simultaneous equation.

Before we use this to write our simultaneous equations as a matrix equation, we should always check that our simultaneous equations are given in standard form. First, letβs check that the terms are given in the same order in all of our simultaneous equations. In the first simultaneous equation, we have our π₯-term and then our π¦-term. This is alphabetical order. However, in the second simultaneous equation, these two terms are given in the opposite order. So, weβll reorder these two terms in our simultaneous equation. Weβll write this equation as five π₯ plus three π¦ is equal to negative 11. And weβll leave the first simultaneous equation unchanged. And we can see that the constants are alone on the right-hand side of the equations.

So now weβve rewritten the simultaneous equations in standard form, so we can use this to rewrite the simultaneous equations as a matrix equation. We can do this step by step by finding each matrix separately. Letβs start with the first matrix which is known as the coefficient matrix. This is because each entry in this matrix is the coefficients of the variables in our simultaneous equations. And itβs very important to remember that coefficients take the sign of the constant value in front of the variable. So by writing these coefficients in a two-by-two matrix, we get the two-by-two matrix four, negative two, five, three.

We can now move on to the second matrix in this equation, which is called the variable matrix because the entries are the variables of our simultaneous equations. In this case, itβs a column matrix, which means it only has one column and the number of rows is equal to the number of variables. And we need to write these in the order given in the simultaneous equations. In this case, our simultaneous equations have two variables π₯ and π¦. So, the variable matrix for our simultaneous equation will be the two-by-one matrix π₯, π¦.

Finally, we need to find the third matrix in this equation, which is often called the solution matrix or constant matrix. And thatβs because the entries of this matrix are given as the constants of the simultaneous equations. In this case, we can see that this is zero and negative 11. So, we set this equal to the two-by-one matrix zero, negative 11.

And this is our final answer. Itβs worth pointing out that we can check this answer by evaluating the matrix multiplication on the left-hand side of the equation. However, we were able to show that we can rewrite the simultaneous equations given to us in the question as the two-by-two matrix four, negative two, five, three times the two-by-one matrix π₯, π¦ is equal to the two-by-one matrix zero, negative 11.