Video Transcript
Given that the transpose of 𝐵 plus
the transpose of 𝐶 equals the square matrix four, four, 10, negative three, 𝐴
equals the square matrix negative four, one, zero, negative one, determine the
matrix 𝑋 that satisfies the relation 𝑋 equals the transpose of 𝐴𝐵 plus 𝐴𝐶.
We will begin by recalling the
definition of the transpose of a matrix. In particular, we’re interested in
the transpose of a two-by-two matrix, since both of the matrices we are given have
order two by two. Suppose 𝑀 is a two-by-two matrix
with elements 𝑎, 𝑏, 𝑐, 𝑑. The transpose is represented by 𝑀
with a superscript capital T. The transpose of 𝑀 is the
two-by-two matrix 𝑎, 𝑐, 𝑏, 𝑑. It is found by swapping the rows
and the columns. The elements on the main diagonal,
𝑎 and 𝑑, remain unchanged. But if we consider elements 𝑏 and
𝑐, they appear to have reflected or flipped across the diagonal.
In this example, we are looking for
the matrix 𝑋 defined by the transpose of 𝐴𝐵 plus 𝐴𝐶. In order to make sense of this
expression, we need to recall a few other relevant matrix properties. We recall that matrix
multiplication is distributive with respect to matrix addition, as long as 𝐴 has
order 𝑚 by 𝑛 and both 𝐵 and 𝐶 have order 𝑛 by 𝑝. Since we are working with
two-by-two matrices, the order requirement is fulfilled. Therefore, 𝐴 times the sum of 𝐵
and 𝐶 equals 𝐴𝐵 plus 𝐴𝐶 for all two-by-two matrices.
We also recall the distributive
property for the transpose of a sum of matrices, which says for two matrices 𝐵 and
𝐶 with the same order, the transpose of 𝐵 plus 𝐶 equals the transpose of 𝐵 plus
the transpose of 𝐶. Finally, we recall that matrix
multiplication combined with the transpose satisfies the following property. Given matrices 𝐴 and 𝐵 with a
well-defined product 𝐴𝐵, the transpose of 𝐴𝐵 equals the product of the transpose
of each matrix in the opposite order.
To begin making sense of the
expression defining 𝑋, we will use the first matrix distributive property. This allows us to rewrite 𝐴𝐵 plus
𝐴𝐶 as 𝐴 times the sum of 𝐵 plus 𝐶. Now we have a use for the transpose
and multiplication property. This property allows us to write
the transpose of the product of 𝐴 and 𝐵 plus 𝐶 as the transpose of 𝐵 plus 𝐶
times the transpose of 𝐴. In our next step, we will use the
distributive property for the transpose of a sum. So, we rewrite the transpose of 𝐵
plus 𝐶 as the transpose of 𝐵 plus the transpose of 𝐶. The sum of the transpose of 𝐵 and
the transpose of 𝐶 was given to us in the directions. So, we can now substitute the four,
four, 10, negative three matrix into our equation.
Now, all that remains is to find
the transpose of 𝐴 and to perform a matrix multiplication. We find the transpose of matrix 𝐴
by swapping the rows and columns. This means the first row of matrix
𝐴 with elements negative four and one becomes the first column of the transpose of
𝐴. Then, the second row of 𝐴 becomes
the second column in the transpose of 𝐴. Thus, the second column will
contain the elements zero and negative one. Now we can substitute the transpose
of matrix 𝐴 into our equation. We will use matrix multiplication
to find the product of four, four, 10, negative three and negative four, zero, one,
negative one. Let’s clear some space to make room
for this matrix multiplication.
We recall the procedure for matrix
multiplication involves the pairwise summation of elements from each matrix. In particular, the matrix
multiplication between two square matrices 𝑎, 𝑏, 𝑐, 𝑑 and 𝑒, 𝑓, 𝑔, ℎ is
defined as 𝑎𝑒 plus 𝑏𝑔, 𝑎𝑓 plus 𝑏ℎ, 𝑐𝑒 plus 𝑑𝑔, and 𝑐𝑓 plus 𝑑ℎ.
The results of this multiplication
are as follows: four times negative four plus four times one. The second element is calculated by
four times zero plus four times negative one. The third element is determined by
the sum of 10 times negative four and negative three times one. And our final element is the sum of
10 times zero and negative three times negative one. We use the standard order of
operations to perform the multiplications first. Then, we evaluate the four
remaining sums. The result is the matrix 𝑋 with
elements negative 12, negative four, negative 43, and three.
In summary, we used properties of
matrices and properties of the transpose to find 𝑋, the transpose of 𝐴𝐵 plus
𝐴𝐶.
