Video Transcript
If 𝐴 is the two-by-one matrix six, seven and 𝐵 is the one-by-two matrix five, zero, what is 𝐵 times 𝐴 transposed?
In this question, we’re given two matrices, matrix 𝐴 and matrix 𝐵. And we’re asked to evaluate an expression involving these two matrices. We need to determine the matrix that represents 𝐵 multiplied by 𝐴 all transposed. And there’s two different ways we can evaluate this expression. First, we can just evaluate the expression directly. And we can do this by just substituting the matrices 𝐵 and 𝐴 into this expression. We get the transpose of matrix 𝐵 times 𝐴 is equal to the transpose of the one-by-two matrix five, zero multiplied by the two-by-one matrix six, seven.
Now, we need to evaluate the matrix multiplication inside the parentheses. And one way to do this is to note the first matrix is of order one by two and the second matrix is of order two by one. And we know for matrix multiplication to be valid, the number of columns in the first matrix, which is two, must be equal to the number of rows in the second matrix, which is also two. And then, the matrix resulting in this multiplication will have the number of rows of the first matrix, which is one, and the number of columns of the second matrix, which is also one.
So we know that the products of these two matrices should give us a one-by-one matrix. And we can find this matrix by just multiplying the two matrices together. To do this, we need to multiply the corresponding entries of the rows of the first matrix with the columns of the second matrix. And since the first matrix in this product only has one row and the second matrix only has one column, we’ll only get one entry since we have a one-by-one matrix. The entry of this matrix is five times six plus zero multiplied by seven. And remember, we need to take the transpose of this matrix. And if we evaluate this expression, we get 30. So we need to take the transpose of the one-by-one matrix whose only entry is 30. And we can do this by recalling the transpose of a matrix replaces all of the rows of the matrix with the corresponding columns.
However, our matrix only has one row and one column. So when we switch these rows and columns, we’ll end up with the same matrix, the one-by-one matrix 30. In fact, the transpose of any one-by-one matrix will just leave the matrix unchanged. And this is enough to answer the question. So we could stop here. However, there is a second method for answering this question.
We can recall if matrices 𝐵 and 𝐴 have orders such that we can multiply matrix 𝐵 by matrix 𝐴, then the transpose of 𝐵 times 𝐴 is equal to the transpose of 𝐴 multiplied by the transpose of 𝐵. In other words, if we can multiply matrices 𝐵 and 𝐴, then we can distribute the transpose over the products by taking the transpose of each matrix and reordering the multiplication. This means we can answer the question by instead evaluating the right-hand side of this equation, taking the transpose of 𝐴 and the transpose of 𝐵 separately and then finding the product of these two matrices.
So let’s start by finding the transpose of matrix 𝐴. That’s the transpose of the two-by-one matrix six, seven. And remember to take the transpose of this matrix, we need to write the corresponding rows of this matrix as the corresponding columns of our new matrix. And we can do this row by row. Let’s start with the first row which we can see is only the element six. Since this is the first row of matrix 𝐴, it will be the first column of matrix 𝐴 transpose.
We can do the exact same with the second row of matrix 𝐴, which is just the entry seven. We want to write the second row of matrix 𝐴 as the second column of its transpose. So our second column will only have the entry seven, which means the transpose of matrix 𝐴 is the one-by-two matrix six, seven. And it’s worth noting when we take the transpose of a matrix, we switch its dimensions around. Matrix 𝐴 is a two-by-one matrix, so its transpose will be a one-by-two matrix.
We can follow the exact same process to find the transpose of matrix 𝐵. We need to take the rows of matrix 𝐵 and write them as the corresponding columns of the transpose of 𝐵. If we start with the first row of 𝐵, we can see that 𝐵 only has one row. This means the transpose of 𝐵 will only have one column. If we write this row as the column, we see the transpose of 𝐵 is the two-by-one matrix five, zero.
We can now substitute the transpose of matrix 𝐴 and the transpose of matrix 𝐵 into our formula. This then gives us the one-by-two matrix six, seven multiplied by the two-by-one matrix five, zero. And we can now just evaluate this multiplication. We get that one-by-one matrix whose only entry is six times five plus seven times zero, which we can evaluate is also equal to the one-by-one matrix whose only entry is 30.
Therefore, we were able to show two different methods of determining the transpose of matrix 𝐵 times matrix 𝐴, where matrix 𝐴 is the two-by-one matrix six, seven and matrix 𝐵 is the one-by-two matrix five, zero. We were able to show the transpose of 𝐵 times 𝐴 is equal to the one-by-one matrix whose only entry is 30.
