Lesson Video: Inverse of a Matrix: The Adjoint Method | Nagwa Lesson Video: Inverse of a Matrix: The Adjoint Method | Nagwa

Lesson Video: Inverse of a Matrix: The Adjoint Method Mathematics

In this video, we will learn how to find the inverse of 3 Γ— 3 matrices using the adjoint method.

17:49

Video Transcript

Inverse of a Matrix: The Adjoint Method

In this video, we’re going to discuss how to find the inverse of any square matrix with nonzero determinant by using the adjoint method. We’ll go through the method in detail, explaining how to find the matrix minors, how to construct the cofactor matrix, and then how to construct the adjoints of our matrix. We’ll then go through some examples explaining how we can apply this method to find inverses.

Before we discuss this new method of finding the inverse of a square matrix, we’re going to need to go through a few new concepts first. We’re going to start with defining a matrix minor. If we have a matrix 𝐴 which is of order π‘š times 𝑛, then the matrix minor, which we represent with capital 𝐴 𝑖𝑗, is the matrix 𝐴 where we remove the 𝑖th row and the 𝑗th column from matrix 𝐴. So our matrix minor 𝐴 𝑖𝑗 is exactly the same as matrix 𝐴. However, we removed the 𝑖th row and 𝑗th column. And we know if we remove one row and one column from matrix 𝐴, then our new order will be π‘š minus one by 𝑛 minus one. And the easiest way to understand this concept is through an example.

Let’s start with the matrix 𝐴, which is a three-by-four matrix given as follows. Now, let’s see how we would construct the matrix minor 𝐴 two three. From our definition of a matrix minor, we can see we need to remove the 𝑖th row and 𝑗th column from our matrix 𝐴. First, our value of 𝑖 is two. So we need to remove the second row from our matrix 𝐴. Next, we can see the value of 𝑗 is equal to three. So we need to remove the third column from matrix 𝐴. Then our matrix minor 𝐴 two three will be all the remaining elements we didn’t remove. 𝐴 two three will be the following matrix.

However, this is of course not the only matrix minor we could’ve constructed. Now, let’s construct the matrix minor 𝐴 three one. This time, we can see the value of 𝑖 is equal to three. So we need to remove the third row from our matrix 𝐴. And we can see the value of 𝑗 is one. So we need to remove the first column from matrix 𝐴. And then we can construct our matrix 𝐴 three one by using all of the remaining elements, the ones we didn’t remove from 𝐴. This gives us 𝐴 three one is the following matrix.

The methods we go through in this video will help us determine if we can find the inverse of any 𝑛-by-𝑛 matrix. However, we will need to find the determinant of any matrix we want to find the inverse of. And we’ve seen previously it’s difficult to calculate the determinant for large square matrices. So we’ll focus mainly on three-by-three matrices.

And before we do this, we’re going to need to recall how we find the determinant of a two-by-two matrix. Let 𝐴 be equal to the two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑. Then the determinant of 𝐴 is equal to π‘Žπ‘‘ minus 𝑏𝑐. We’re now ready to discuss the key part which will help us find the inverse of any square matrix with nonzero determinant. We need to discuss what the cofactor matrix of a matrix 𝐴 is.

First, because we’re using this to find the inverse of a matrix, we need our matrix 𝐴 to be a square matrix. Let’s say its order is 𝑛 by 𝑛. Next, to construct our cofactor matrix, we’re going to need to find all of the matrix minors of 𝐴. We’ll call these 𝐴 𝑖𝑗. We’re now ready to construct the cofactor matrix of 𝐴. We’ll call this matrix 𝐢. And we’ll define it element by element. The entry in row 𝑖 and column 𝑗 of our cofactor matrix will be given by negative one raised to the power of 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. And remember, our matrix 𝐴 is a square matrix of order 𝑛 by 𝑛. So our values of 𝑖 will be running through the rows of our matrix 𝐴. And the values of 𝑗 will be running through the columns of our matrix 𝐴. So our values of 𝑖 range from one to 𝑛, and our values of 𝑗 range from one to 𝑛. This means our cofactor matrix will be a square matrix. It will be of order 𝑛 by 𝑛.

It’s also worth pointing out sometimes you’ll see this written out in its entire matrix form. And if we were to do this, we would get the following matrix representation of our cofactor matrix. And it’s worth pointing out all we’ve done here is create an 𝑛 by 𝑛 square matrix where in row 𝑖 and column 𝑗 we’ve used our formula to construct the entry. But usually, it’s a lot easier to work with our definition for each entry individually. Before we discuss how we’re going to use the cofactor matrix to find the inverse of a matrix, let’s go through some examples.

Given that 𝐴 is equal to the three-by-three matrix negative five, eight, negative seven, six, zero, one, five, negative four, negative eight, determine the value of negative one to the power of one plus two multiplied by the determinant of the matrix minor 𝐴 one two.

We’re given a three-by-three square matrix. And we’re asked to determine the value of negative one to the power of one plus two times the determinant of the matrix minor 𝐴 one two. And while not necessary to answering this question, it’s worth pointing out this will be the entry in row one and column two of our cofactor matrix.

The first step to answering this question is to remember what we mean by the matrix 𝐴 one two. We call this a matrix minor. The matrix minor 𝐴 𝑖𝑗 means we remove row 𝑖 and column 𝑗 from our matrix 𝐴. In our case, we can see the value of 𝑖 is equal to one and 𝑗 is equal to two. We can then write this into our definition for the matrix minor. We see that the matrix minor 𝐴 one two means we remove row one and column two from matrix 𝐴.

So to find our matrix minor 𝐴 one two, we need to start with our matrix 𝐴 and then remove row one. This means we remove the following three entries. Then, we also need to remove column two. This means we need to remove the entire second column from matrix 𝐴. And we can see this leaves us with only four elements. Then, we can construct our matrix minor 𝐴 one two by constructing a matrix with the four remaining elements. This gives us 𝐴 one two is the two-by-two matrix six, one, five, negative eight.

But the question is not just asking us to find the matrix minor 𝐴 one two. We also need to find its determinant. And since 𝐴 one two is a two-by-two matrix, we can do this by recalling the formula for the determinant of a two-by-two matrix. We recall the determinant of the square matrix π‘Ž, 𝑏, 𝑐, 𝑑 is equal to π‘Žπ‘‘ minus 𝑏𝑐. In our case, we can see that π‘Ž is equal to six and 𝑑 is equal to negative eight. And we can also see that our value of 𝑏 is equal to one and 𝑐 is equal to five. So by using this formula, we have the determinant of 𝐴 one two is equal to six times negative eight minus one times five. And if we calculate this expression, we get negative 53.

We’re now ready to find the value of the expression given to us in the question. We have negative one to the power of one plus two multiplied by the determinant of 𝐴 one two is equal to negative one cubed, since one plus two is equal to three, and negative 53, since we already found the value of this determinant. And we can simplify this to give us 53. Therefore, we were able to show for the square matrix 𝐴 given to us in the question, the value of negative one to the power of one plus two times the determinant of the matrix minor 𝐴 one two is equal to 53.

Let’s now go through an example of finding a cofactor matrix of a three-by-three square matrix. We’ll start with the three-by-three square matrix 𝐴 is equal to three, zero, negative three, negative two, negative three, negative six, seven, three, negative five. And now we recall the element in row 𝑖 and column 𝑗 of our cofactor matrix will be negative one to the power of 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗.

So to find our cofactor matrix, we’re first going to need to find all of our matrix minors. Let’s start with the matrix minor 𝐴 one one. Remember, this means we’re going to need to remove the first row and the first column from matrix 𝐴. This gives us the following two-by-two matrix. This gives us 𝐴 one one is negative three, negative six, three, negative five. To find our cofactor matrix, we’re going to need to find all of our matrix minors.

Let’s now find the matrix minor 𝐴 one two. This means we need to remove row one and column two from our matrix 𝐴. And doing this leaves us with the following four elements. So the matrix minor 𝐴 one two is negative two, negative six, seven, negative five. Because our matrix 𝐴 is a three-by-three matrix, our values of 𝑖 and 𝑗 will range from one to three. So we’ll have nine total matrix minors to calculate. And we can find all nine of these by using the same method. We remove row 𝑖 and column 𝑗 from our matrix 𝐴. This gives us the following nine matrix minors.

We can now see from our definition of the cofactor matrix we’re now going to need to find the determinant of all of these matrix minors. And since all of these are two-by-two matrices, we can do this by recalling the determinant of the two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑 is equal to π‘Žπ‘‘ minus 𝑏𝑐. So let’s start by finding the determinant of the matrix minor 𝐴 one one. This is equal to negative three times negative five minus negative six multiplied by three. And if we calculate this expression, it’s equal to 33.

We can then do exactly the same to find the determinant of our matrix minor 𝐴 one two. It’s equal to negative two times negative five minus negative six multiplied by seven. And if we calculate this expression, we see it’s equal to 52. Using the exact same method, we could find the determinants of all of our matrix minors. We would get the following values.

Now that we found the determinants of all of our matrix minors, we can find all of the elements of our cofactor matrix. We’ll start with the entry in row one and column one of our cofactor matrix. It will be equal to negative one to the power of one plus one multiplied by the determinant of our matrix minor 𝐴 one one. Well, we know negative one to the power of one plus one is equal to one. And we’ve already shown that the determinant of 𝐴 one one is equal to 33. So 𝐢 one one will be equal to 33. So we’ve shown that 𝐢 one one is equal to 33. And in fact, we can add this into our cofactor matrix in row one, column one.

We can do the same to find the entry in row one, column two. It’s equal to negative one to the power of one plus two times the determinant of 𝐴 one two. And since the determinant of 𝐴 one two is 52, this simplifies to give us negative 52. And we can then add this into our cofactor matrix in row one, column two. And we could then do exactly the same to find all the remaining entries of our cofactor matrix. Doing this would give us the following values. And just as we did before, we can then add these into our cofactor matrix 𝐢.

There is one thing worth pointing out here. When we calculate 𝐢 𝑖𝑗, we multiply the determinant of our matrix minor by negative one to the power of 𝑖 plus 𝑗. This means in row one, column one, we’ll always multiply by positive one. And then in row one and column two, we’ll always multiply by negative one. And this pattern will continue. And some people prefer to use this rather than multiplying by negative one to the power of 𝑖 plus 𝑗. We just need to remember to multiply our determinant by the value.

Now that we’ve constructed our cofactor matrix, we can discuss how we can use this to find the inverse of our matrix. First, we need one last definition. The adjoint matrix of a matrix 𝐴, denoted adjoint 𝐴, is equal to the transpose of our matrix 𝐢, where 𝐢 is the cofactor matrix of 𝐴. And it’s also worth remembering when we take the transpose of a matrix, we switch the rows and columns around.

And now we’re finally ready to determine how to find the inverse of a square matrix. We have if 𝐴 is a square matrix and the determinant of 𝐴 is not equal to zero, then the inverse of 𝐴 will be equal to one divided by the determinant of 𝐴 multiplied by the adjoint of 𝐴. So to find the inverse of any square matrix, there are two parts. First, we need to find the determinant of 𝐴, and then we need to find the adjoint of 𝐴. And remember, if the determinant of 𝐴 is equal to zero, then the matrix does not have an inverse. So normally, we check this first.

Let’s now go through some examples of using the adjoint method to find the inverse of some matrices.

Consider the matrix one, zero, three, one, zero, one, three, one, zero. Determine whether the matrix has an inverse by finding whether the determinant is nonzero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

We’re given a three-by-three matrix. And the first thing we’re asked to do is determine whether this matrix has an inverse by first finding the determinant of our matrix and checking whether this is equal to zero. Remember, if the determinant of a matrix is equal to zero, then that matrix cannot be invertible. And for a square matrix, if its determinant is not equal to zero, then it is invertible. So we need to start by finding the determinant of our matrix. We’ll call this matrix 𝐴.

There’s a lot of different ways of finding the determinant of a matrix. The easiest way is to find which row or column contains the most number of zeros. For our matrix 𝐴, we can see this is column two. It has two zeros. Then, we need to recall our formula for the determinant of a three-by-three matrix where we choose a column 𝑗. This gives us the following expression, where lowercase π‘Ž one 𝑗, lowercase π‘Ž two 𝑗, and lowercase π‘Ž three 𝑗 are the entries in row one, column 𝑗; row two, column 𝑗; and row three, column 𝑗 of our matrix 𝐴. And capital 𝐴 one 𝑗, capital 𝐴 two 𝑗, and capital 𝐴 three 𝑗 are our matrix minors.

Since we chose the second column, we’ll set our value of 𝑗 equal to two. So by using 𝑗 is equal to two, we get the following expression. And we can simplify this. First, negative one to the power of one plus two is equal to negative one. Negative one to the power of two plus two is equal to one. And negative one to the power of three plus two is equal to negative one. So we can simplify our expression to give us the following.

We can then use our definition of the matrix 𝐴 to find some of these values. First, lowercase π‘Ž of one two is the entry in row one, column two of our matrix. We can see this is zero. Next, lowercase π‘Ž two two is the entry in row two, column two. We can see this is also zero. Finally, lowercase π‘Ž three two is the entry in row three, column two. We can see this is equal to one. So our first two terms have a factor of zero and, therefore, are equal to zero. So this entire expression simplifies to give us negative one times the determinant of the matrix minor π‘Ž three two. And we can find our matrix minor 𝐴 three two from our definition of 𝐴.

Remember, we need to remove row three and column two from our matrix 𝐴. And this leaves us with only four elements: one, three, one, one. So our adjoint matrix 𝐴 three two is the two-by-two matrix one, three, one, one. So the determinant of 𝐴 is negative one times the determinant of the two-by-two matrix one, three, one, one. And we know how to find the determinant of a two-by-two matrix. The determinant of a two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑 is equal to π‘Žπ‘‘ minus 𝑏𝑐. And by using this, we can show the determinant of the two-by-two matrix one, three, one, one is equal to one times one minus three times one. And of course, we still need to multiply this by negative one. And if we simplify this expression, we see we get two. And therefore, because the determinant of our square matrix is not equal to zero, we can conclude it must have an inverse.

The next part of our question asked us to find the inverse of our matrix by using the formula which involves the cofactor matrix. So let’s clear some space and then recall how we would do this for our matrix 𝐴. We recall we can find the inverse of a matrix by following five steps.

First, we need to calculate the determinant of our matrix 𝐴. We do this first because if this is equal to zero, then we can’t find an inverse. And we already did this in the first part of our question. We found that the determinant of our matrix 𝐴 was equal to two.

The second thing we need to do is find all of our matrix minors. Remember, the matrix minor 𝐴 𝑖𝑗 is our matrix 𝐴 where we remove row 𝑖 and column 𝑗. Let’s start by finding 𝐴 one one. This means we need to remove the first row and first column from our matrix 𝐴. So we remove the first row and the first column from our matrix 𝐴. And by doing this, we’re left with only four elements. So the matrix minor 𝐴 one one is equal to the two-by-two matrix zero, one, one, zero. We need to find all of our matrix minors. We now need to find 𝐴 one two. That means we remove the first row and the second column from our matrix 𝐴. By doing this, we can see we’re left with four entries: one, one, three, and zero. So the matrix minor 𝐴 one two is the two-by-two matrix one, one, three, zero. And we can use exactly the same method to find all of our matrix minors. We get the following nine matrices.

The third thing we need to do is construct our cofactor matrix. And remember, the entry in row 𝑖, column 𝑗 of our cofactor matrix is equal to negative one to the power of 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. This means we need to find the determinants of all nine of our matrix minors. And we know how to find the determinant of two-by-two matrices. For example, the determinant of 𝐴 one one will be equal to zero times zero minus one times one. And we can evaluate this expression. It’s equal to negative one.

We can do exactly the same to find the determinant of our second cofactor matrix. We get it’s equal to one times zero minus one times three, which we can calculate is equal to negative three. And we can do the exact same thing to find the determinant of all the rest of our matrix minors. We get the following values. But remember, we need to multiply these by negative one to the power of 𝑖 plus 𝑗. In other words, we multiply the determinant of 𝐴 one one by one. We then multiply the determinant of 𝐴 one two by negative one, this gives us positive three, and multiply the determinant of 𝐴 one three by one. We multiply the determinant of 𝐴 two one by negative one. This gives us positive three. And we can continue this on for all of our matrix minors. This gives us the following values.

And remember, each of these values is an entry in our cofactor matrix. So by filling row 𝑖, column 𝑗 with each of these values, we get our cofactor matrix 𝐢 is the following three-by-three matrix. So let’s clear our working and move on to the fourth step. We now need to find our adjoint matrix. That’s the transpose of our cofactor matrix.

Remember, the transpose of a matrix means we need to switch the rows with the columns. So when we transpose our cofactor matrix, our first row will be negative one, three, zero. So we write in the first row negative one, three, zero. And our second row will be three, negative nine, two. And our third row will be one, negative one, zero. And this is the adjoint of our matrix 𝐴. All that’s left to do now is use our formula to find 𝐴 inverse.

Using our formula for 𝐴 inverse, we get that 𝐴 inverse is equal to the following expression. And then we can simplify that matrix to find the three-by-three matrix which is the inverse of 𝐴.

Let’s now go over the key points of this video. First, the matrix minor 𝐴 𝑖𝑗 of a matrix 𝐴 is obtained by removing row 𝑖 and column 𝑗 from 𝐴. We also know for a square matrix, its cofactor matrix will have the same order. And the entries of our cofactor matrix are generated by using the following formula. We know the adjoint of 𝐴 is the transpose of the cofactor matrix. And finally, for a square matrix whose determinant is nonzero, we found the inverse of 𝐴 is equal to one over the determinant of 𝐴 times the adjoint of 𝐴.

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