Video Transcript
Consider matrices 𝐴 and 𝐵, where
𝐴 is the three-by-three matrix one, negative one, one, one, one, two, zero, one,
zero and 𝐵 is the three-by-three matrix two, negative one, three, zero, zero, one,
negative one, one, negative two. By first calculating 𝐴 times 𝐵,
find the inverse of 𝐴.
In this question, we are given two
three-by-three matrices, 𝐴 and 𝐵. And we are asked to determine the
inverse of matrix 𝐴 by first finding the product of 𝐴 and 𝐵. To do this, we can start by
recalling that the inverse of matrix 𝐴 is the unique matrix that when multiplied by
𝐴 on either side gives the identity matrix of the same size.
There are many ways we can
calculate the inverse of a matrix; however, we are told to first calculate matrix 𝐴
times matrix 𝐵. So, let’s start by calculating the
product of the two given matrices. To do this, we can recall we need
to multiply each element of each row of matrix 𝐴 by the corresponding elements of
each column of matrix 𝐵 and then find the sum.
Let’s start with the first row of
matrix 𝐴 and the first column of matrix 𝐵. We have one times two plus negative
one times zero plus one multiplied by negative one. This will be the element in the
first row and first column of matrix 𝐴𝐵. It is also worth noting that the
matrix 𝐴𝐵 will have the same number of rows as matrix 𝐴 and the same number of
columns as matrix 𝐵. So, we know that we have a
three-by-three matrix.
Evaluating the sum of the products
of the corresponding entries in row one of 𝐴 and column one of 𝐵 gives us one. We can then add this into the
matrix 𝐴𝐵 and then follow this same process for the remaining rows of 𝐴 and
columns of 𝐵. Let’s follow this process again for
the first row of 𝐴 and second column of 𝐵. This will give us the element in
row one and column two of matrix 𝐴𝐵. We have one times negative one plus
negative one times zero plus one times one, which we can evaluate to give zero.
We can then write zero in this
position of matrix 𝐴𝐵. We need to follow this process for
all of the remaining rows of 𝐴 and columns of 𝐵. For row one of 𝐴 and column three
of 𝐵, we obtain zero. For row two of 𝐴 and column one of
𝐵, we get zero.
We can follow this process for the
remaining rows of 𝐴 and columns of 𝐵 to obtain the three-by-three matrix one,
zero, zero, zero, one, zero, zero, zero, one. This is the identity matrix of size
three; it is a square diagonal matrix with every entry on the main diagonal equal to
one.
Hence, we have shown that 𝐴 times
𝐵 is the identity matrix. We recall that the inverse of a
matrix is unique. Since 𝐵 is an inverse of 𝐴, it is
the inverse of 𝐴.
Therefore, we were able to show
that the inverse of 𝐴 is the three-by-three matrix two, negative one, three, zero,
zero, one, negative one, one, negative two.
