Video Transcript
Given that 𝐴 is equal to two,
five, three, negative three and 𝐵 is equal to negative two, two, five, negative
three, what is two times the transpose of 𝐴 plus 𝐵 in matrix form?
Our first step in solving this
problem is to find the sum of the matrices 𝐴 and 𝐵. When adding matrices, we first need
to check they’re of the same order. We can quite clearly see that 𝐴
and 𝐵 are both of the order two by two. So we’re okay to add them. Now what we need to do is add each
of the corresponding elements in the two matrices together. We can see that the first entry in
the matrix 𝐴 is two and the first entry in the matrix 𝐵 is negative two. Hence, the first entry in 𝐴 plus
𝐵 will be two plus negative two. Similarly, the second entry will be
five plus two; the third entry, three plus five; and the fourth and final entry,
negative three plus negative three.
Simplifying each of these entries,
we can see that the matrix 𝐴 plus 𝐵 will be zero, seven, eight, negative six. Our next step will be to find the
transpose of the matrix 𝐴 plus 𝐵. Now, in order to find the transpose
of a matrix, we simply take all the rows of that matrix and turn them into the
columns of the new matrix. So for the transpose of 𝐴 plus 𝐵,
we take the first row of 𝐴 plus 𝐵, which is zero, seven. And this becomes the first column
in the transpose matrix. Similarly, we take the second row,
which is eight, negative six, and this becomes the second column in the transpose
matrix. So now we have found that the
transpose of 𝐴 plus 𝐵 is equal to zero, eight, seven, negative six.
For our final step, we simply need
to multiply this matrix by two. When multiplying a matrix by a
constant, we simply take the constant and multiply each of the elements by this
constant. Since our constant here is two, we
simply need to double each element of our matrix. In doing so, we reach our solution,
which is that two timesed by the transpose of 𝐴 plus 𝐵 is equal to zero, 16, 14,
negative 12.
