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Question Video: Evaluating the Determinant of an Element of a Matrix Mathematics • 10th Grade

Given that 𝐴 = [βˆ’5, 8, βˆ’7 and 6, 0, 1 and 5, βˆ’4, βˆ’8], determine the value of (βˆ’1)^(1 + 2)|𝐴_12|

06:55

Video Transcript

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.

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