Video: Representing a System of Two Equations in Matrix Form

In this video, we will learn how to use the matrix multiplication to determine the square and cube of a square matrix.

14:17

Video Transcript

In this video, we will learn how to represent a system of two equations as a matrix equation. We will only be dealing with matrices up to two by two. However, the method can be extended for matrices of higher order.

A system of linear equations can be represented in matrix form using a coefficient matrix, a variable matrix, and a constant matrix. If we consider the two equations 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑝 and 𝑐𝑥 plus 𝑑𝑦 is equal to 𝑞, where 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, and 𝑞 are all constants, then the coefficient matrix is 𝑎, 𝑏, 𝑐, 𝑑. This is formed by aligning the coefficients of the variables of each equation in a row. It is important that each equation is written in standard form with the constant term on the right.

We have two variables 𝑥 and 𝑦. Therefore, the variable matrix can be written as shown. Whilst we will only deal with two variables in this video, this can be extended to multivariable equations. On the right-hand side of our equations, we have the constant terms 𝑝 and 𝑞. These correspond to the first and second equations, respectively. Therefore, the constant matrix is 𝑝, 𝑞. The order of our three matrices are two by two, two by one, and two by one, respectively.

The system of linear equations 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑝 and 𝑐𝑥 plus 𝑑𝑦 is equal to 𝑞 can be rewritten in matrix form as 𝑎, 𝑏, 𝑐, 𝑑 multiplied by 𝑥, 𝑦 is equal to 𝑝, 𝑞. We can use matrix multiplication to see that the matrix representation is equivalent to the system of equations. As previously mentioned, we could extend this to a system of three equations in three variables and actually generalize the result to 𝑛 variables. Let’s now consider some specific examples.

Express the simultaneous equations three 𝑎 plus two 𝑏 equals 13 and two 𝑎 plus three 𝑏 equal seven as a matrix equation.

We recall that we can represent a system of linear equations in matrix form using a coefficient matrix, a variable matrix, and a constant matrix. Before doing this, we need to ensure that our equations are written in standard form, 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑐, where 𝑎, 𝑏, and 𝑐 are constants. In this question, both of our equations are written in standard form. However, it is important to note that 𝑎 and 𝑏 are variables and not constants.

We know that the coefficient matrix can be formed by aligning the coefficients of the variables of each equation in a row. The coefficients of our first equation are three and two. In the second equation, we have two and three, giving us the two-by-two coefficient matrix three, two, two, three. The variables here are 𝑎 and 𝑏. So we can write the variable matrix as the two-by-one matrix 𝑎, 𝑏. On the right-hand side, we have the constant terms 13 and seven. As these correspond to the first and second equation, respectively, the constant matrix is 13, seven.

We now have a matrix equation made up of a coefficient matrix, a variable matrix, and a constant matrix. The simultaneous equations three 𝑎 plus two 𝑏 is equal to 13 and two 𝑎 plus three 𝑏 is equal to seven can be rewritten as a matrix equation three, two, two, three multiplied by 𝑎, 𝑏 is equal to 13, seven.

In our next example, we will begin by rearranging one of the equations.

Express the simultaneous equations one-third 𝑥 minus two-thirds 𝑦 is equal to five-thirds and three-quarters 𝑦 plus one-quarter 𝑥 is equal to seven-quarters as a matrix equation. In order to rewrite a system of linear equations as a matrix equation, we need to find a coefficient matrix, a variable matrix, and a constant matrix. Before starting, however, we need to ensure that all of our equations are written in standard form.

As the 𝑥-term comes first in our first equation, we can rewrite the second equation as a quarter 𝑥 plus three-quarters 𝑦 is equal to seven-quarters. This is because addition is commutative. The coefficients of our first equation are one-third and negative two-thirds, whilst the coefficients of our second equation are one-quarter and three-quarters. This means that the two-by-two coefficient matrix is one-third, negative two-thirds, one-quarter, three quarters. The two variables here are 𝑥 and 𝑦. Therefore, the variable matrix is simply 𝑥, 𝑦.

On the right-hand side of our equations, we have the constant terms five-thirds and seven-quarters. These make up the constant matrix. We now have a matrix equation made up of a coefficient matrix, a variable matrix, and a constant matrix as required. The two simultaneous equations expressed as a matrix equation is one-third, negative two-thirds, one-quarter, three-quarters multiplied by 𝑥, 𝑦 is equal to five-thirds, seven-quarters.

In our next example, we will need to write a matrix equation as a set of simultaneous linear equations.

Write down the set of simultaneous equations that could be solved using the given matrix equation. Three, three, two, four multiplied by 𝑎, 𝑏 is equal to 10, 12.

In order to answer this question, we need to perform matrix multiplication. We begin by multiplying the elements in the first row of the coefficient matrix by the elements in the column variable matrix. Three multiplied by 𝑎 is equal to three 𝑎, and three multiplied by 𝑏 is equal to three 𝑏. The sum of these terms will be equal to the element in the first row of the constant matrix. This gives us the equation three 𝑎 plus three 𝑏 is equal to 10.

We then repeat this process for the second row of the coefficient matrix. Two multiplied by 𝑎 is two 𝑎, and four multiplied by 𝑏 is four 𝑏. This gives us the equation two 𝑎 plus four 𝑏 is equal to 12. We now have a set of simultaneous equations that could be solved. Three 𝑎 plus three 𝑏 is equal to 10, and two 𝑎 plus four 𝑏 equals 12. Whilst we do not need to solve the equations in this video, we could do so using the elimination or substitution methods.

We will now try and solve a similar problem where one of our coefficients is negative.

Write down the set of simultaneous equations that could be solved using the given matrix equation. 11, negative three, nine, four multiplied by 𝑥, 𝑦 is equal to eight, 13.

As with any problem of this type, we can solve it using matrix multiplication. When multiplying matrices, we need to multiply each row of the first matrix by each column of the second matrix. Multiplying 11 by 𝑥 gives us 11𝑥. Negative three multiplied by 𝑦 is equal to negative three 𝑦. This will be equal to the element in the top row of our constant matrix, in this case, eight. Our first equation is 11𝑥 minus three 𝑦 is equal to eight.

We then repeat this process with the second row of our two-by-two coefficient matrix. Multiplying nine by 𝑥 gives us nine 𝑥. Four multiplied by 𝑦 is four 𝑦. Our second equation is therefore nine 𝑥 plus four 𝑦 is equal to 13. We now have a pair of linear simultaneous equations that could be solved using the elimination or substitution methods. These would give us the values of 𝑥 and 𝑦 that solve the matrix equation.

In our final question, we will begin by rearranging our equations so they’re in standard form.

Express the simultaneous equations three 𝑥 minus 24 is equal to negative eight 𝑦 and 𝑥 is equal to three minus 𝑦 as a matrix equation.

In order to answer this question, we need to ensure that both of our equations are written in standard form. We need to rewrite the first equation so it is in the form 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑝 and the second equation so it is in the form 𝑐𝑥 plus 𝑑𝑦 is equal to 𝑞, where 𝑎, 𝑏, 𝑐, 𝑑 as well as 𝑝 and 𝑞 are constants.

Let’s begin with the equation three 𝑥 minus 24 is equal to negative eight 𝑦. We can add 24 and eight 𝑦 to both sides of our equation. This gives us three 𝑥 plus eight 𝑦 is equal to 24. This equation has now been written in standard form. Our second equation stated that 𝑥 is equal to three minus 𝑦. Adding 𝑦 to both sides of this equation and ensuring that our 𝑥- and 𝑦-terms are in the same order gives us 𝑥 plus 𝑦 is equal to three.

We now have a pair of linear simultaneous equations written in standard form. Our matrix equation will consist of a two-by-two coefficient matrix. The coefficients of 𝑥 and 𝑦 in our first equation are three and eight. These will make up the top row. The coefficients of 𝑥 and 𝑦 in our second equation are one and one. We therefore have the two-by-two coefficient matrix three, eight, one, one.

Our variables are 𝑥 and 𝑦. This means that the coefficient matrix can be multiplied by this column variable matrix. On the right-hand side of our two equations, we have the constants 24 and three. Our simultaneous equations can be expressed as the matrix equation three, eight, one, one multiplied by 𝑥, 𝑦 is equal to 24, three.

We will know summarize the key points from this video. We saw in this video that a matrix equation consists of a coefficient matrix, a variable matrix, and a constant matrix. We saw that we can express a system of linear simultaneous equations as a matrix equation and vice versa. The linear simultaneous equations 𝑎𝑥 plus 𝑏𝑦 equals 𝑝 and 𝑐𝑥 plus 𝑑𝑦 equals 𝑞 can be rewritten as the matrix equation 𝑎, 𝑏, 𝑐, 𝑑 multiplied by 𝑥, 𝑦 is equal to 𝑝, 𝑞. As previously mentioned, this can be extended to equations with three or more variables, matrices larger than two by two, and also help us solve the simultaneous equations.

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