Lesson Video: Solving a System of Two Equations Using a Matrix Inverse Mathematics • First Year of Secondary School

Video Transcript

In this video, we’ll learn how to solve a system of two linear equations using the inverse of the matrix of coefficients. Recall that the linear equation in two dimensions describes a straight line 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient of the line and 𝑏 is the 𝑦-intercept. That’s where the line meets the 𝑦-axis. A system of two linear equations is a set of two equations describing two straight lines, and a solution to the system is the single point in the 𝑥𝑦-plane where the two lines intersect. If two lines never meet, then there’s no solution and the lines are parallel.

Note that our equations may not necessarily be written in the form 𝑦 is equal to 𝑚𝑥 plus 𝑏. We may have something like the system shown. These equations both still describe straight lines, but now we have constants multiplying the 𝑦’s. The 𝑥’s and 𝑦’s are one side, and the constants are on the other. And the main thing is that we have two equations describing straight lines, and we want to find a solution, that is, where they intersect. And there are a few methods we could use. And in this video, we’re going to concentrate on using a matrix inverse to find a solution. So let’s see what steps we need to follow to do this.

Suppose we have a set of two simultaneous equations, for example, 𝑥 plus 𝑦 is equal to one and four 𝑥 plus five 𝑦 is equal to six. And let’s call these equations one and two. Our first step is to make sure that the variables 𝑥 and 𝑦 are vertically aligned on the left-hand side and our constants on the right-hand side. This is because we want to be able to read off our coefficients for our matrix. Our next step is to rewrite the system as a matrix equation. Since we have two equations in two unknowns, our coefficient matrix will be a two-by-two matrix. This multiplies a column matrix of our variables. And on the right-hand side, we have a column matrix of the constants.

The first row of our coefficient matrix consists of the coefficients of 𝑥 and 𝑦 in equation one, and these are both one, and the associated constant on the right-hand side is also one. The second row of our coefficient matrix is the coefficients of 𝑥 and 𝑦 in equation two, that’s four and five, and the associated constant on the right-hand side is six. We now have our system represented in matrix form, and if we were to multiply this out, we would regain our original equations.

Our third step in solving our equations is to find the inverse of our coefficient matrix. Recall that for an 𝑛-by-𝑛 matrix 𝐴 which is nonsingular, that means it has an inverse, 𝐴 inverse times 𝐴 is equal to 𝐴 times 𝐴 inverse is equal to the identity matrix, where the identity matrix has leading diagonal elements equal to one and every other element equal to zero. And recall also that for a two-by-two nonsingular matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, then the inverse of 𝐴 is one over 𝑎𝑑 minus 𝑏𝑐 times the matrix with the elements 𝑑, negative 𝑏, negative 𝑐, and 𝑎. That is the matrix where 𝑎 and 𝑑 have been swapped and we take the negative of 𝑏 and 𝑐.

And we note that the denominator of our fraction is actually the determinant of the matrix 𝐴. So how does this help us? Well, notice we have the equation 𝐴𝑥 equal to 𝑏 and we multiply our equation on the left by 𝐴 inverse. We know that 𝐴 inverse times 𝐴 is the identity matrix. So on the left-hand side we have 𝐼𝑥 and on the right-hand side 𝐴 inverse times 𝑏. The left-hand side simplifies to 𝑥. So we have 𝑥 equal to 𝐴 inverse 𝑏. And this gives us our solution since, remember, 𝑥 is the column matrix with elements 𝑥, 𝑦. And it’s this that we’re trying to find. So to solve our system of equations, we basically need to find the inverse of our coefficient matrix. So let’s apply this to our example.

We need to find the inverse of our matrix 𝐴. And if 𝑎 is equal to one, 𝑏 is equal to one, 𝑐 is equal to four, and 𝑑 is equal to five, we have the fraction one over one times five minus one times four multiplying the matrix with elements five, negative one, negative four, and one. Our denominator evaluates to one so that the inverse of our coefficient matrix has elements five, negative one, negative four, and one. And if we substitute this into our equation, we have 𝑥, 𝑦 on the left-hand side. And on the right-hand side, the inverse of our coefficient matrix, which has elements five, negative one, negative four, and one, multiplies the column matrix one, six of constants.

Multiplying out our right-hand side, we have five times one plus negative one times six and negative four times one plus one times six. And this evaluates to the column matrix with elements negative one, two so that by equality of matrices, 𝑥 is negative one and 𝑦 is equal to two. To find the solution to a set of two simultaneous equations then, we write our equations in matrix equation form, we find the inverse of our coefficient matrix, we multiply both sides of the equation on the left by the inverse, multiply out the right-hand side, and this gives us our solution. So let’s look at another example where we begin with two simultaneous equations and solve using matrices.

Consider the simultaneous equations four 𝑥 minus two 𝑦 is equal to zero and three 𝑦 plus five 𝑥 is negative 11. Express the given simultaneous equations as a matrix equation. Write down the inverse of the coefficient matrix. And multiply through by the inverse on the left-hand side to solve the matrix equation.

There are three parts to this question involving the set of simultaneous equations four 𝑥 minus two 𝑦 is equal to zero and three 𝑦 plus five 𝑥 is negative 11. And these three parts lead us to the solution of the equations via matrix methods. The first part is to express the simultaneous equations as a matrix equation. We must then write down the inverse of the coefficient matrix and use this by multiplying through on the left to solve the matrix equation. So let’s begin with the first part, which is to write the equations as a matrix equation.

The first thing we need to do is to make sure that our 𝑥’s and 𝑦’s are vertically aligned on the left-hand side. In our second equation then, we’ll need to swap the three 𝑦 and the five 𝑥. And now our 𝑥’s and 𝑦’s are vertically aligned. Let’s call our equations equations one and two so that equation one is four 𝑥 minus two 𝑦 is equal to zero and equation two is five 𝑥 plus three 𝑦 is negative 11. And this helps to read off our coefficients so we can put these into a two-by-two matrix, which then multiplies a column matrix of our variables 𝑥 and 𝑦. And we put this equal to the constants on the right-hand side.

The first row of our coefficient matrix contains the coefficients of 𝑥 and 𝑦 in equation one. That is four and negative two. Our associated constant on the right-hand side is zero. The second row of our coefficient matrix contains the constant coefficients of 𝑥 and 𝑦 in equation two. That is five and three. And our constant element on the right-hand side is negative 11. So now we have our equations in matrix equation form as required.

The second part of the question asks us to write down the inverse of the coefficient matrix. And to do this, we recall that for a nonsingular two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, the inverse of 𝐴 is one over 𝑎𝑑 minus 𝑏𝑐 times the matrix with elements 𝑑, negative 𝑏, negative 𝑐, and 𝑎. Recall that 𝑎𝑑 minus 𝑏𝑐 is the determinant of 𝐴. And notice that we’ve swapped the elements 𝑎 and 𝑑 and taken the negative of 𝑏 and 𝑐. In our case, our coefficient matrix has elements four, negative two, five, and three. So its inverse, if 𝑎 is four, 𝑏 is negative two, 𝑐 is five, and 𝑑 is three, is one over four times three minus negative two times five, that is one over 𝑎𝑑 minus 𝑏𝑐, times the matrix with elements three, two, negative five, and four. The inverse of our coefficient matrix is therefore one over 22 times the matrix with elements three, two, negative five, and four.

Our final part is to multiply through by the inverse on the left-hand side to solve the matrix equation. If we call our coefficient matrix 𝐴, we have 𝐴 inverse times 𝐴 times the column matrix 𝑥 is equal to 𝐴 inverse times the column matrix 𝑏, where 𝑥 is the matrix of variables and 𝑏 is the matrix of constants on the right-hand side. Remember, though, that for any nonsingular matrix 𝐴, that is, a matrix with an inverse, 𝐴 inverse times 𝐴 is equal to the identity matrix, which for a two-by-two matrix is the matrix with elements one, zero, zero, one. So that in our left-hand side, we have the identity matrix times 𝑥, 𝑦 and on our right, we have 𝐴 inverse times the column matrix 𝑏.

Our left-hand side simplifies to the column matrix 𝑥, 𝑦. And if we multiplied the right-hand side, we have one over 22 times the matrix with elements three times zero plus two times negative 11, negative five times zero plus four times negative 11. That is one over 22 times the column matrix with elements negative 22, negative 44. Making some space and evaluating this gives us the column matrix with elements negative one and negative two. By equality of matrices, this then gives us 𝑥 is equal to negative one and 𝑦 is equal to negative two.

So for the simultaneous equations four 𝑥 minus two 𝑦 is equal to zero and three 𝑦 plus five 𝑥 is equal to negative 11, we have the matrix equation where the coefficient matrix has elements four, negative two, five, three multiplying the column matrix of variables equal to the column matrix with elements zero, negative 11 of constants on the right-hand side. Our matrix of coefficients has an inverse, which is one over 22 times the matrix with elements three, two, negative five, and four. And we use this to find our solution: 𝑥 is negative one and 𝑦 is negative two.

Now let’s use our method to solve a set of simultaneous equations already expressed as a matrix equation.

Given that the two-by-two matrix with elements five, eight, one, negative eight multiplied by the column matrix with elements 𝑥 and 𝑦 is equal to the column matrix with elements negative 43 and one, determine the values of 𝑥 and 𝑦.

We’re asked to find the values of 𝑥 and 𝑦 given a matrix equation. Our matrix equation consists of a coefficient matrix with elements five, eight, one, and negative eight multiplying a column matrix of our variables 𝑥 and 𝑦. And this is equal to a column matrix with elements negative 43 and one. So, in fact, we have a matrix equation of the form 𝐴𝑥 is equal to 𝑏. And in order to solve for 𝑥 and 𝑦, we can use the fact that for any square nonsingular matrix 𝐴 with inverse 𝐴 inverse, 𝐴 inverse times 𝐴 is equal to 𝐴 times 𝐴 inverse, which is equal to the identity matrix. And that’s the matrix where the elements in the leading diagonal all equal one and the rest are equal to zero. And so, if there is a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, the identity is the matrix with elements one, zero, zero, one.

Now we can use this in our matrix equation. Multiplying both sides on the left with 𝐴 inverse, on our left-hand side we’d have 𝐴 inverse times 𝐴 times 𝑥, which is actually the identity times 𝑥. And this simplifies to 𝑥 so that we have 𝑥 equal to 𝐴 inverse times our column matrix 𝑏. And this gives us our solution for 𝑥 and 𝑦. We can apply this to our problem by first finding the inverse of our two-by-two matrix of coefficients. Recall that an inverse of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is equal to one over 𝑎𝑑 minus 𝑏𝑐 times the matrix with elements 𝑑, negative 𝑏, negative 𝑐, and 𝑎.

Let’s just make some room here and note that our matrix of coefficients has elements five, eight, one, negative eight. And if we label these 𝑎, 𝑏, 𝑐, 𝑑, its inverse is given by one over the determinant, which is five times negative eight minus eight times one times the matrix with elements negative eight, negative eight, negative one, and five. That is negative one over 48 times the matrix with elements negative eight, negative eight, negative one, and five. Now if we multiply the left-hand side of each side of our equation with this inverse, we have our inverse matrix times the coefficient matrix times the column matrix of variables is equal to the inverse of the coefficient matrix times the column of constants.

We know that the inverse of the coefficient matrix times the coefficient matrix itself is equal to the identity. So on our left-hand side, this simplifies to the column matrix with elements 𝑥, 𝑦. And all we need to do now is to multiply out our right-hand side. This gives us negative one over 48 times the matrix with negative eight times negative 43 plus negative eight times one and negative one times negative 43 plus five times one. That is negative one over 48 times the column matrix with elements 336 and 48. This evaluates to elements negative seven and negative one. And by equality of matrices, this gives us 𝑥 is negative seven, 𝑦 is negative one.

And so given the matrix equation where the coefficient matrix has elements five, eight, one, and negative eight and the constant column matrix has elements negative 43 and one, the value of 𝑥 is negative seven and the value of 𝑦 is negative one.

Now let’s look at another example where we start from a system of two equations.

Use matrices to solve the system negative 𝑥 plus five 𝑦 is equal to eight and negative three 𝑥 plus 𝑦 is equal to eight.

We’re given a system of two simultaneous equations which we want to solve for 𝑥 and 𝑦. That means to find values of 𝑥 and 𝑦 that satisfy both equations. And we’re asked to use matrices to do this. Our equations are negative 𝑥 plus five 𝑦 is equal to eight and negative three 𝑥 plus 𝑦 is equal to eight. And we’re going to need to express our system of equations as a matrix equation. Since we have two equations in two unknowns, our coefficient matrix will be a two-by-two matrix. This will multiply a column matrix of variables. And on the right-hand side, we’ll have a column matrix of constants.

The first thing we need to do to read off the coefficients is to check that the 𝑥’s and 𝑦’s are vertically aligned. And this is actually the case, so we’re good to go. The first row of our coefficient matrix is populated by the coefficients of 𝑥 and 𝑦 in equation one. That is negative one and five. And the elements in the second row are the coefficients of 𝑥 and 𝑦 in equation two. That’s negative three and one. The associated constants on the right-hand side are both eight. So, using vector notation, we have an equation of the form 𝐴 times 𝑥 is equal to 𝑏. And we can use the fact that if 𝐴 is an 𝑛-by-𝑛 nonsingular matrix, then 𝐴 inverse times 𝐴 is 𝐴 times 𝐴 inverse is equal to the identity matrix. That’s the matrix with leading elements one and all the rest zero.

How this helps us is if you find the inverse of our matrix 𝐴 and multiply this on the left, we have 𝐴 inverse 𝐴 times 𝑥 is equal to 𝐴 inverse times 𝑏. And we know that 𝐴 inverse times 𝐴 is the identity matrix. So we’re left with 𝑥 is equal to 𝐴 inverse times 𝑏. And this will give us our solution. So let’s find the inverse of our coefficient matrix and recall that for a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, the inverse of 𝐴 is one over 𝑎𝑑 minus 𝑏𝑐 times the matrix with elements 𝑑, negative 𝑏, negative 𝑐, and 𝑎. And that’s where 𝑎𝑑 minus 𝑏𝑐 is nonzero.

In our case, our matrix has elements negative one, five, negative three, one. So its inverse will be one over negative one times one minus five times negative three times the matrix with elements one, negative five, three, negative one. Our denominator evaluates to 14. And so the inverse of our coefficient matrix is one over 14 times the matrix with elements one, negative five, three, negative one. Now, if we multiply both sides of our equation on the left with this inverse, we know that on the left-hand side, our inverse times our matrix of coefficients 𝐴 is equal to the identity. And if we multiply this out, it simplifies to the column matrix with elements 𝑥, 𝑦.

And so now we simply have to multiply out our right-hand side. And this gives us one over 14 times the matrix with elements one times eight plus negative five times eight and three times eight plus negative one times eight. And this evaluates to one over 14 times the column matrix with elements negative 32, 16, which is the column matrix with elements negative 16 over seven and eight over seven. And by equality of matrices, this gives us 𝑥 is negative 16 over seven and 𝑦 is eight over seven. The solution to the system negative 𝑥 plus five 𝑦 is equal to eight and negative three 𝑥 plus 𝑦 is equal to eight is therefore 𝑥 is equal to negative 16 over seven and 𝑦 is equal to eight over seven.

In this video, we’ve seen how to solve a system of two linear equations using the inverse of the matrix of coefficients. So let’s finish by noting some of the key points for this method.

Given the system of two linear equations 𝑎𝑥 plus 𝑏𝑦 is equal to 𝑒 and 𝑐𝑥 plus 𝑑𝑦 is equal to 𝑓, where 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, and 𝑓 are constants, we first ensure that the variable terms are vertically aligned. We then form a matrix equation with the matrix of coefficients 𝐴 multiplying a column matrix of variables 𝑥 equal to a column matrix of constants 𝑏. We then find the inverse of our coefficient matrix, multiply both sides of the matrix equation on the left by 𝐴 inverse so that we have 𝑥 is equal to 𝐴 inverse times 𝑏 on the right-hand side since 𝐴 inverse 𝐴 is the identity matrix. We then multiply out the right-hand side of our equation for our solution 𝑥, 𝑦. And remember, this solution 𝑥, 𝑦 is where the two lines one and two intersect.