Given that 𝐴 is a three-by-two
matrix such that 𝑎 sub 𝑖𝑗 equals three 𝑖 plus five 𝑗 plus nine, find the matrix
𝐴 superscript 𝑇.
We’re given information about
matrix 𝐴 as a three-by-two matrix and asked to find its transpose. That’s represented by 𝐴 with a
superscript capital 𝑇. And we’re given a formula that will
allow us to generate each element in the original matrix. And so there are two ways we could
answer the question. We could use this formula to
generate matrix 𝐴 and then find its transpose by swapping its rows and columns. Alternatively, we can use the
definition of the transpose, and we’re going to use the latter method here.
That is, suppose 𝐴 is a matrix
whose 𝑖th row–𝑗th column element is defined by 𝑎 sub 𝑖𝑗. The transpose of this matrix will
then be defined with 𝑖th row–𝑗th column elements 𝑎 sub 𝑗𝑖. Now we’re told the matrix 𝐴 is a
three-by-two matrix, so it has three rows and two columns. Since the 𝑖th row–𝑗th column
element is defined by 𝑎 sub 𝑖𝑗, then the element in the first row and first
column is 𝑎 sub one one. The element in the first row and
second column is 𝑎 sub one two, and so on.
Now that we’ve defined the matrix
𝐴, let’s define the matrix the transpose of 𝐴. Since the 𝑖th row–𝑗th column
element is defined by 𝑎 sub 𝑗𝑖, then the element in the first row and first
column is still 𝑎 sub one one. The element in the first row and
second column, however, will be defined by 𝑎 sub two one. And the element in the first row
and third column is 𝑎 sub three one. Then the second row has elements 𝑎
sub one two, 𝑎 sub two two, and 𝑎 sub three two.
And for clarity, we can
double-check by making sure that the rows and columns have switched places. The first row in matrix 𝐴 has
elements 𝑎 sub one one and 𝑎 sub one two. These are mapped onto the first
column in the transpose of 𝐴. Similarly, the second row has
elements 𝑎 sub two one and 𝑎 sub two two, which go into the second column of our
transpose. Finally, the third row maps onto
the third column in our transpose.
So let’s work out the element 𝑎
sub one one by using the formula. We substitute 𝑖 equals one and 𝑗
equals one into the formula given. And we get three times one plus
five times one plus nine, which is equal to 17. And so the element in the first row
and first column of our transpose matrix is 17.
Next, we’ll substitute 𝑖 equals
two and 𝑗 equals one. That gives us 𝑎 sub two one equals
three times two plus five times one plus nine. And that’s equal to 20. So the second element in our first
row is 20. Let’s repeat this one more time for
the element 𝑎 sub three one, where we’re going to let 𝑖 be equal to three and 𝑗
be equal to one. That’s three times three plus five
times one plus nine, which is equal to 23. Repeating this process for the
remaining elements, and we find that element 𝑎 sub one two, that is, the element in
the second row and first column, is 22. The second element in this row is
25, and the third is 28.
And so we’ve generated the matrix
transpose of 𝐴. It’s the two-by-three matrix with
elements 17, 20, 23, 22, 25, and 28.