Question Video: Solving a Set of Simultaneous Equations Using Matrices Mathematics

Video Transcript

Use matrices to solve the following system of equations.

Remember, if we can write our system of equations in the form 𝐴𝐱 equals 𝐛, where 𝐴 is the coefficient matrix, 𝐱 is the column vector that contains all of the variables, and 𝐛 is the constant vector, then we can multiply both sides of this equation by the inverse of 𝐴. And when we do, we find that the variable vector 𝐱 is equal to the inverse of 𝐴 times 𝐛.

And so to solve our system of equations, we need to set it up in this form and then find the inverse of our coefficient matrix. We have three equations with three variables. And so vector 𝐴 is going to be three by three. We’re going to list the coefficients of our variables in order from our first equation. It’s negative one, eight, and negative three. These are simply the coefficients shown.

In our second equation, they’re four, negative three, and eight. And in our third equation, there’re six, negative 12, and 19. Then the vector 𝐱 contains all of our variables. It’s 𝑥, 𝑦, 𝑧. And of course our constant vector contains our constants negative 10, 12, and 18. So our first job is to find the inverse of the matrix 𝐴.

There are of course a number of ways to do this. One is to augment it with the identity matrix and perform elementary row operations. We might even choose to use a calculator. But let’s recall how we can use the minors, cofactors, and adjugate method. We begin by finding the matrix of minors. We ignore the values on the row and column we’re looking at and calculate the determinant of the remaining values. So for the first element on the first row, we’re going to do negative three times 19 minus eight times negative 12, which is 39. We then ignore all elements on the first row and second column. And we work out four times 19 minus eight times six, which is 28. Finally, we ignore elements on the first row and third column. And we work out four times negative 12 minus negative three times six. And we get negative 30.

Continuing the process, and the matrix of minors is as shown. We then find the matrix of cofactors using that sort of checkerboard pattern. This involves changing the sign of the second element in the first row, the first and third elements in our second row, and the second element in our third row. Next, we find the adjugate. The elements on a diagonal stay the same, and then we simply swap the positions of all of the rest. We go directly across that diagonal. So we swap negative 116 and negative 28, 55 and negative 30, and negative four and 36.

Then our final step is to multiply this by one over the determinant of 𝐴. We can use the values of the matrix of minors to do so. We’re going to multiply each of the elements in the top row of our original matrix by the cofactor for the same location. So we multiply negative one by 39. We then subtract the product of eight and 28. And then we add the product of negative three and negative 30. And so the determinant of 𝐴 is negative 173. And so we find the inverse of 𝐴 is negative one over 173 times that adjugate.

Let’s clear some space, and we’re now able to solve our system of equations. 𝐱 is therefore equal to the inverse of 𝐴 times 𝐛, which is as shown. Now, we could multiply each of the elements inside the inverse of 𝐴 by negative one over 173. Or we can do that right at the end. Let’s begin by finding the dot product of the elements in the first row of our inverse and the vector 𝐛. When we do, we get negative one over 173 times 39 times negative 10 minus 116 times 12 plus 55 times 18, which is negative 792. So 𝑥 is negative one over 173 times negative 792, which is the same as 792 over 173.

We then do the same for 𝑦, this time finding the dot product of the elements on this row with our vector, giving us negative one over 173 times 196, which is negative 196 over 173. Let’s do this for 𝑧 by finding the dot product of the elements on the third row of the inverse and the column vector 𝐛. And this time, we get negative one over 173 times 210 or negative 210 over 173.

And so we’ve found the values of 𝑥, 𝑦, and 𝑧. Let’s put them back into vector form. It is in fact easier to do so by taking out that constant factor of one over 173. And so we’ve solved our system of equations. In vector form, we can say that 𝑥, 𝑦, 𝑧 is equal to one over 173 times the vector 792, negative 196, negative 210.