The portal has been deactivated. Please contact your portal admin.

Question Video: Constructing the Transpose of a Matrix Given a General Equation for Its Elements Mathematics

Given that 𝐴 is a 3 Γ— 2 matrix, such that π‘Ž_(𝑖𝑗) = 3𝑖 + 5𝑗 + 9, find the matrix 𝐴^T.

03:46

Video Transcript

Given that 𝐴 is a three-by-two matrix such that π‘Ž 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 π‘Ž sub 𝑖𝑗. The transpose of this matrix will then be defined with 𝑖th row–𝑗th column elements π‘Ž 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 π‘Ž sub 𝑖𝑗, then the element in the first row and first column is π‘Ž sub one one. The element in the first row and second column is π‘Ž 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 π‘Ž sub 𝑗𝑖, then the element in the first row and first column is still π‘Ž sub one one. The element in the first row and second column, however, will be defined by π‘Ž sub two one. And the element in the first row and third column is π‘Ž sub three one. Then the second row has elements π‘Ž sub one two, π‘Ž sub two two, and π‘Ž 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 π‘Ž sub one one and π‘Ž sub one two. These are mapped onto the first column in the transpose of 𝐴. Similarly, the second row has elements π‘Ž sub two one and π‘Ž 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 π‘Ž 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 π‘Ž 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 π‘Ž 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 π‘Ž 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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.