Question Video: Solving a System of Three Equation Using the Inverse of a Matrix Mathematics

Video Transcript

Use the inverse of a matrix to solve the system of linear equations negative four 𝑥 minus two 𝑦 minus nine 𝑧 equals negative eight, negative three 𝑥 minus two 𝑦 minus six 𝑧 equals negative three, and negative 𝑥 plus 𝑦 minus six 𝑧 equals seven.

We’re told in the question that we’re going to be solving this using the inverse of a matrix. We also know that we can rewrite a system of linear equations in an equivalent way as a matrix equation. Our matrix equation will have three parts: a coefficient matrix, a variable matrix, and a constant matrix.

The coefficient matrix consists of the coefficients of each variable in the correct order. So the first row of the coefficient matrix will be negative four, negative two, negative nine. The second row of the coefficient matrix will be negative three, negative two, negative six. And the third row of the coefficient matrix will be negative one, one, and negative six. For this particular equation, we have to be careful because although the coefficients of 𝑥 and 𝑦 are not visible, they are negative one and one, respectively.

Okay, so let’s move on and consider what goes in the variable matrix. That’s going to be our variables for this system of linear equations. And that’s 𝑥, 𝑦, and 𝑧. So now let’s fill in the entries for the constant matrix. This will consist of the constants for our system of linear equations, that is, negative eight, negative three, and seven. So here is our system of linear equations but written in an equivalent way as a matrix equation.

The inverse matrix method of solving this is clearly going to involve the matrix inverse, but why? Well, let’s call this coefficient matrix 𝐴, the variable matrix 𝑋, and the constant matrix 𝐵. So we can represent this matrix equation as 𝐴𝑋 equals 𝐵. Remember, our goal is to find the entries of the variable matrix, that is, 𝑋. So in order to solve 𝐴𝑋 equals 𝐵, remember 𝐴, 𝑋, and 𝐵 are matrices, so we have to do only matrix operations. We start by multiplying from the left the inverse of the coefficient matrix on both sides of the equation. We do this because then on the left-hand side of the equation, we have 𝐴 inverse multiplied by 𝐴.

We know that 𝐴 inverse multiplied by 𝐴 just gives us the identity matrix. We also know that multiplying the identity matrix by another matrix just gives us that matrix. So that is 𝑋 equals 𝐴 inverse 𝐵. Once we get to this stage, we’ll be able to multiply 𝐴 inverse by 𝐵 because we’re going to find what the matrix 𝐴 inverse is and we already know the entries of 𝐵 because that’s our constant matrix. So this is the method that we’re going to use. So we’re going to start by finding the inverse of the coefficient matrix.

We can use the adjoint method to obtain the inverse of this matrix if it exists. Recall that a square matrix is invertible if its determinant is nonzero. So let’s begin by finding the determinant of this matrix and making sure that it’s nonzero. Recall that this is how we find the determinant for a three-by-three matrix, where these are the matrix minors obtained by taking the 𝑖th row and the 𝑗th column from the matrix 𝐴.

So let’s go ahead and apply this to find the determinant for our coefficient matrix. We start by taking the entry 𝑎 one one, that’s negative four. And we multiply it by the determinant of the matrix minor 𝐴 one one. The entry in the first row and first column of the matrix 𝐴 is negative four. Therefore, the matrix minor 𝐴 one one is the matrix negative two, negative six, one, negative six.

So we multiply negative four by the determinant of the matrix negative two, negative six, one, negative six. Then we subtract the entry 𝑎 one two. That is negative two. And we multiply it by the determinant of the matrix minor 𝐴 one two. Since the entry in the first row and second column is here, the matrix minor associated with this entry is negative three, negative six, negative one, and negative six.

And finally, we add the entry 𝑎 one three. That’s the entry negative nine. And the matrix minor associated with it is negative three, negative two, negative one, and one.

So we can now calculate each of these determinants. Taking the first one as an example, we compute this by doing negative two multiplied by negative six. That gives us 12. And then we subtract negative six multiplied by one. That gives us negative six. So this determinant is 12 minus negative six, and that gives us 18. Then we can compute the other two determinants in the same way. Those are 12 and negative five, respectively.

We can then multiply these terms together. Then by adding and subtracting, we find the determinant of the coefficient matrix to be negative three. So because we know that the determinant of this matrix is nonzero, we know that the inverse matrix does exist. So let’s go ahead and find this inverse. I’m going to clear some space to find the matrix inverse. Let’s remind ourselves of the adjoint method to find the matrix inverse.

We’re going to use the following three steps to find the inverse of the coefficient matrix 𝐴. We’ll begin by finding its cofactor matrix. And then we’ll find the adjoint matrix by transposing the cofactor matrix. Then we multiply the adjoint matrix by the reciprocal of the determinant of 𝐴 to obtain the inverse matrix. So we’re going to begin by finding the cofactor matrix. Entries of the cofactor matrix are the determinants of the corresponding matrix minors multiplied by the alternating sign negative one to the power of 𝑖 add 𝑗.

So we’re going to compute the determinant of these nine matrix minors, which each have a corresponding sign. We get this corresponding sign from negative one to the power of 𝑖 add 𝑗. For example, for this first matrix minor, this gives us negative one to the power of one add one. As this is just negative one squared, that gives us one. But if we look at the matrix minor 𝐴 two one, for example, we find the corresponding sign by doing negative one to the power of two add one. That gives us negative one to the third power. And that is negative one, therefore telling us that this one has a negative sign.

So let’s continue now by writing out each of the determinants that we need to find. Let’s begin by finding the matrix minor 𝐴 one one. We can do this by crossing off the first row and the first column of our matrix. That leaves us with the matrix negative two, negative six, one, negative six. And that is the matrix minor 𝐴 one one. We can then use the same method to find the matrix minor 𝐴 one two. We cross off the first row and the second column, and that leaves us with the matrix negative three, negative six, negative one, negative six. Then by the same method, we can find the rest of the matrix minors.

The next step is to then actually calculate each of these determinants. Remember we do this for two-by-two determinants by subtracting the product of the diagonals. For instance, for this first one, we do negative two multiplied by negative six. That gives us 12. Then we subtract negative six multiplied by one. That’s negative six. So the determinant of this matrix is 12 minus negative six. And that gives us 18.

Then these determinants that we’ve calculated give us the entries for the cofactor matrix. I’ve underlined these values in orange. Now I’m going to clear these calculations and write out our cofactor matrix, consisting of the entries underlined with orange. So we’ve now completed step one using the adjoint method to find the matrix inverse, and that’s finding the cofactor matrix.

So now we move on to step two. That is, we need to find the adjoint matrix by transposing 𝐶. Remember that transposing a matrix means that the rows become the columns and the columns become the rows. So that’s step two done. We’ve transposed the matrix 𝐶, which has given us the adjoint matrix of 𝐴. We can now move on to the third and final step to find our matrix inverse.

For step three, we multiply this adjoint matrix we’ve just found by the reciprocal of the determinant, which we computed earlier. Remember, we found this determinant to be negative three. This final step then gives us the inverse of the matrix 𝐴. We still have a couple more steps that we need to do to finish this question. Going back to the start of our problem, remember we said that we could rewrite this matrix equation as 𝐴𝑋 equals 𝐵. And in order to find 𝑋, we would need to multiply from the left the inverse matrix of 𝐴. And because multiplying the inverse of a matrix by the matrix itself just gives us the identity matrix, therefore we can rearrange this matrix equation to give us 𝑋 equals 𝐴 inverse multiplied by 𝐵.

So to find 𝑋 and therefore the entries 𝑥, 𝑦, 𝑧, we need to do 𝐴 inverse multiplied by 𝐵. So I’m going to clear some space so that we can do this calculation. So now we can find 𝑋 by computing this matrix multiplication. We do this in the usual method for matrix multiplication. Let’s now simplify. And from here, we just need to multiply each of the three entries by negative one over three. That gives us 41, negative 24, negative 12. Hence, we found the matrix 𝑥, 𝑦, 𝑧 to be 41, negative 24, negative 12. Therefore, 𝑥 equals 41, 𝑦 equals negative 24, and 𝑧 equals negative 12.

Notice how we could check this answer by substituting the values we found into the variable matrix and multiplying the coefficient matrix by the variable matrix, which should then give us the constant matrix negative eight, negative three, seven.

As this was a fairly lengthy question, let’s go through the steps we took to find the answer as a quick recap. We started off by writing our system of linear equations as a matrix equation. We then recognized that if we multiplied on the left by the matrix inverse of the coefficient matrix, it would give us the variable matrix. In order to do that, we had to find the matrix inverse of the coefficient matrix using the adjoint method. Then finally, we just had to multiply the inverse we found by the constant matrix to get the variable matrix and therefore the values for 𝑥, 𝑦, and 𝑧.